### Learning Objectives

**In this unit, student is exposed to** 

- Historical background of electricity and magnetism

- The role of electrostatic force in day – to-day life

- Coulomb’s law and superposition principle 

- The concept of electric field 

- Calculation of electric field for various charge configurations

- Electrostatic potential and electrostatic potential energy 

- Electric dipole and dipole moment 

- Electric field and electrostatic potential for a dipole 

- Electric flux 

- Gauss law and its various applications 

- Electrostatic properties of conductors and dielectrics

- Polarisation 

- Capacitors in series and parallel combinations 

- Effect of a dielectric in a capacitor 

- Distribution of charges in conductors, corona discharge

- Working of a Van de Graaff generator

INTRODUCTION

Electromagnetism is one of the most important branches of physics. The technological developments of the modern 21st century are primarily due to our understanding of electromagnetism. The forces we experience in everyday life are electromagnetic in nature except gravity.

In standard XI, we studied about the gravitational force, tension, friction, normal force etc. Newton treated them to be independent of each other with each force being a separate natural force. But what is the origin of all these forces? It is now understood that except gravity, all forces which we experience in every day life (tension in the string, normal force from the surface, friction etc.) arise from electromagnetic forces within the atoms. Some examples are

(i) When an object is pushed, the atoms in our hand interact with the atoms in the object and this interaction is basically electromagnetic in nature.

(ii) When we stand on Earth’s surface, the gravitational force on us acts downwards and the normal force acts upward to counter balance the gravitational force. What is the origin of this normal force?

It arises due to the electromagnetic interaction of atoms on the surface of the Earth with the atoms present in the feet of the person. Though, we are attracted by the gravitational force of the Earth, we stand on Earth only because of electromagnetic force of atoms.

(iii) When an object is moved on a surface, static friction resists the motion of the object. This static friction arises due to electromagnetic interaction between the atoms present in the object and atoms on the surface. Kinetic friction also has similar origin.

From these examples, it is clear that understanding electromagnetism is very essential to understand the universe in a holistic manner. The basic principles of electromagnetism are dealt with in volume 1 at XII standard physics. This unit deals with the behaviour and other related phenomena of charges at rest. This branch of electricity which deals with stationary charges is called Electrostatics.

Historical background of electric charges

Two millenniums ago, Greeks noticed that amber (a solid, translucent material formed from the resin of a fossilized tree)

after rubbing with animal fur attracted small pieces of leaves and dust. The amber possessing this property is said to be ‘charged’. It was initially thought that amber has this special property. Later people found that not only amber but even a glass rod rubbed with silk cloth, attracts pieces of papers. So glass rod also becomes ‘charged’ when rubbed with a suitable material.

Consider a charged rubber rod hanging from a thread as shown in Figure 1.1. Suppose another charged rubber rod is brought near the first rubber rod; the rods repel each other. Now if we bring a charged glass rod close to the charged rubber rod, they attract each other. At the same time, if a charged glass rod is brought near another charged glass rod, both the rods repel each other.

From these observations, the following inferences are made

(i) The charging of rubber rod and that of glass rod are different from one another.

(ii) The charged rubber rod repels another charged rubber rod, which implies that ‘like charges repel each other’. We can also arrive at the same inference by observing that a charged glass rod repels another charged glass rod.

(iii) The charged rubber rod attracts the charged glass rod, implying that the charge in the glass rod is not the same kind of charge present in the rubber. Thus unlike charges attract each other.

Therefore, two kinds of charges exist in the universe. In the 18th century, Benjamin Franklin called one type of charge as positive (+) and another type of charge as negative (–). Based on Franklin’s convention, rubber and amber rods are negatively charged while the glass rod is positively charged. If the net charge is zero in the object, it is said to be electrically neutral.

Following the pioneering work of J. J. Thomson and E. Rutherford, in the late 19th century and in the beginning of 20th century, we now understand that the atom is electrically neutral and is made up of the negatively charged electrons, positively charged protons, and neutrons which have zero charge. The material objects made up of atoms are neutral in general. When an object is rubbed with another object (for example rubber with silk cloth), some amount of charge is transferred from one object to another due to the friction between them and the object is then said to be electrically charged. Charging the objects through rubbing is called triboelectric charging.

Basic properties of charges

(i) Electric charge

Most objects in the universe are made up of atoms, which in turn are made up of protons, neutrons and electrons. These particles have mass, an inherent property of particles. Similarly, the electric charge is another intrinsic and fundamental property of particles. The nature of charges is understood through various experiments performed in the 19th and 20th century. The SI unit of charge is coulomb.

(ii) Conservation of charges

Benjamin Franklin argued that when one object is rubbed with another object, charges get transferred from one to the other. Before rubbing, both objects are electrically neutral and rubbing simply transfers the charges from one object to the other. (For example, when a glass rod is rubbed against silk cloth, some negative charge are transferred from glass to silk. As a result, the glass rod is positively charged and silk cloth becomes negatively charged).

From these observations, he concluded that charges are neither created or nor destroyed but can only be transferred from one object to other. This is called conservation of total charges and is one of the fundamental conservation laws in physics. It is stated more generally in the following way.

The total electric charge in the universe is constant and charge can neither be created nor be destroyed. In any physical process, the net change in charge will always be zero.

(iii) Quantisation of charges

What is the smallest amount of charge that can be found in nature? Experiments show that the charge on an electron is −e and the charge on the proton is +e. Here, e denotes the fundamental unit of charge. The charge q on any object is equal to an integral multiple of this fundamental unit of charge e.

q = ne (1.1)

Here n is any integer (0, ±1, ±2, ±3, ±4………..). This is called quantisation of electric charge.

Robert Millikan in his famous experiment found that the value of e = 1.6 × 10^–19 C. The charge of an electron is −1.6 × 10^–19 C and the charge of the proton is +1.6 × 10^–19 C. When a glass rod is rubbed with silk cloth, the number of charges transferred is usually very large, typically of the order of 1010. So the charge quantisation is not appreciable at the macroscopic level. Hence the charges are treated to be continuous (not discrete). But at the microscopic level, quantisation of charge plays a vital role.

EXAMPLE 1.1

Calculate the number of electrons in one coulomb of negative charge.

Solution

According to the quantisation of charge,

q = ne

Here q = 1C. So the number of electrons in 1 coulomb of charge is \[n = \frac{q}{e} = \frac{1C}{1.6 × 10^{–19}} = 6.25 × 10^{18}\] electrons

COULOMB’S LAW

In the year 1786, Coulomb deduced the expression for the force between two stationary point charges in vacuum or free space. Consider two point charges \(q _1\) and \(q _2\) at rest in vacuum, and separated by a distance of r as shown in Figure 1.2. According to Coulomb, the force on the point charge \(q _2 \)exerted by another point charge \(q _1\) is

\( \vec {F}{21} = k \frac{q_1q_2}{r^2} \hat{r}{12} \)(1.2)

where \(\hat{r} _{12}\) is the unit vector directed from charge \(q _1\) to charge \(q _2\) and k is the proportionality constant.

Important aspects of Coulomb’s law

(i) Coulomb’s law states that the electrostatic force is directly proportional to the product of the magnitude of the two point charges and is inversely proportional to the square of the distance between the two point charges.

(ii) The force on the charge \(q _2\) exerted by the charge \(q _1\) always lies along the line joining the two charges. \(\hat{r} _{12}\) is the unit vector pointing from charge \(q _1\) to \(q _2\) .It is shown in the Figure 1.2. Likewise, the force on the charge \(q _1\) exerted by \(q _2\) is along \(\hat{-r} _{12}\) (i.e., in the direction opposite to \(\hat{r} _{12}\) ).

(iii) In SI units, k = \(\dfrac{1}{4π\epsilon_{o}}\) and its value is \(9 × 10^9 N m^2 C^{–2}\). Here \(\epsilon_{o}\) is the permittivity of free space or vacuum and its value is

\(\epsilon_{o} = \dfrac{1}{4πk} = 8.85×10^{−12} C^2 N^{−1}m^{−2}.\)

(iv) The magnitude of the electrostatic force between two charges each of one coulomb and separated by a distance of 1 m is calculated as follows:

\(|F| = \frac{9×10 ×1×1}{1^2} = 9×10^{9} N.\)

This is a huge quantity, almost equivalent to the weight of one million ton. We never come across 1 coulomb of charge in practice. Most of the electrical phenomena in day-to-day life involve electrical charges of the order of µC (micro coulomb) or nC (nano coulomb).

(v) In SI units, Coulomb’s law in vacuum takes the form \(\vec {F} _{21} = \dfrac{1}{4π\epsilon _{o}} \dfrac{q_1q_2}{r^2} \hat{r} _{12} .\) In a medium of permittivity e, the force between two point charges is given by \(\vec {F} _{21} = \dfrac{1}{4π\epsilon} \dfrac{q_1q_2}{r^2} \hat{r} _{12} .\)

.

Since \(\epsilon>\epsilon _o\), the force between two point charges in a medium other than vacuum is always less than that in vacuum. We define the relative permittivity for a given medium \(\epsilon _r\) = \(\dfrac{\epsilon}{\epsilon _o}\). For vacuum or air, \(\epsilon _r = 1\) and for all other media \(\epsilon _r > 1\).

(vi) Coulomb’s law has same structure as Newton’s law of gravitation. Both are inversely proportional to the square of the distance between the particles. The electrostatic force is directly proportional to the product of the magnitude of two point charges and gravitational force is directly proportional to the product of two masses. But there are some important differences between these two laws.

• The gravitational force between two masses is always attractive but Coulomb force between two charges can be attractive or repulsive, depending on the nature of charges.

• The value of the gravitational constant \(G = 6.67 × 10^{–11} N m^2 kg{–2}\). The value of the constant k in Coulomb law is \(k = 9 × 10^9 N m^2 C^{–2}\). Since k is much more greater than G, the electrostatic force is always greater in magnitude than gravitational force for smaller size objects.

• The gravitational force between two masses is independent of the medium. For example, if 1 kg of two masses are kept in air or inside water, the gravitational force between two masses remains the same. But the electrostatic force between the two charges depends on nature of the medium in which the two charges are kept at rest.

(vii) The force on a charge \(q _1\) exerted by a point charge \(q _2\) is given by

\(\vec {F} _{12} = \dfrac{1}{4π\epsilon _o} \dfrac{q_1q_2}{r^2} \hat{r} _{21} .\)

Here \(\hat{r} _{21}\) is the unit vector from charge \(q _2\) to \(q _1\).

\(\hat{r} _{21} = -\hat{r} _{12}\)

\(\vec {F} _{12} = \dfrac{1}{4π\epsilon _o} \dfrac{q_1q_2}{r^2} (-\hat{r} _{12}) .\) = \(- \dfrac{1}{4π\epsilon _o} \dfrac{q_1q_2}{r^2} (\hat{r} _{12}) .\)

Therefore, the electrostatic force obeys Newton’s third law.

(viii) The expression for Coulomb force is true only for point charges. But the point charge is an ideal concept. However we can apply Coulomb’s law for two charged objects whose sizes are very much smaller than the distance between them. In fact, Coulomb discovered his law by considering the charged spheres in the torsion balance as point charges. The distance between the two charged spheres is much greater than the radii of the spheres.

EXAMPLE 1.2

Consider two point charges \(q _1\) and \(q _2\) at rest as shown in the figure.

They are separated by a distance of 1m. Calculate the force experienced by the two charges for the following cases:

(a) \(q _1 = +2 μC and q _2 = +3 μC\)

(b) \(q _1 = +2 μC and q _2 = –3 μC\)

(c) \(q _1= +2 μC and q _2 = –3 μC kept in water (\epsilon _r = 80\))

Solution

(a) \(q _1 = +2 μC, q _2 = +3 μC,\) and \(r = 1m\). Both are positive charges. so the force will be repulsive.

Force experienced by the charge \(q _2\) due to \(q _1\) is given by

\(\vec {F} _{12} = \dfrac{1}{4π\epsilon _o} \dfrac{q_1q_2}{r^2} \hat{r} _{12} .\)

Here \(\hat{r} _{12}\) is the unit vector from \(q _1\) to \(q _2\). Since \(q _2\) is located on the right of \(q _1\), we have

\(\hat{r} _{12} = \hat{i}\) and \(\dfrac{1}{4π\epsilon _o} = 9 ×10^9\) so that

\(\vec {F} _{21} = \dfrac{9 × 10^9×2×10^{-6}×3×10^{-6}}{1^2} \hat{i}\)

\(= 54 × 10^{-3} \hat{i}\) \(N\)

According to Newton’s third law, the force experienced by the charge \(q _1\) due to \(q _2\) is \(\vec {F} _{12} = - \vec {F} _{21} \). Therefore,

\(\vec {F} _{12} \) = \(- 54 × 10^{-3} \hat{i} N\)

The directions of \(\vec {F} _{21} and \vec {F} _{12}\) are shown in the above figure in case (a)

(b) \(q _1 = +2 μC, q _2 = -3 μC,\) and \(r = 1m\).They are unlike charges. So the force will be attractive.

Force experienced by the charge q₂ due to q₁ is given by

\(\vec {F} _{21} = \dfrac{9 × 10^9×2×10^{-6}×3×10^{-6}}{1^2} \hat{r} _{12}\)

\(= 54 × 10^{-3} N\hat{i}\) (using \(\hat{r}_{12} = \hat{i} \)

The charge q₂ will experience an attractive force towards q₁ which is in the negative \(x\) direction.

According to Newton’s third law, the force experienced by the charge q₁ due to q₂ is \(\vec {F} _{21}\) = \(-\vec {F} _{21}\), Therefore,

\(\vec {F} _{12} = 54 × 10^{-3} \hat{i} N\)

The directions of \(\vec {F} _{21}\) and \(\vec {F} _{12} \)are shown in the figure (case (b)).

(c) If these two charges are kept inside the water, then the force experienced by q₂ due to q₁

\(\vec {F} _{21}^{W}\) =\(\dfrac{1}{4\pi\epsilon}\\)(\dfrac{q_1q_2}{r^2}\\)(\hat{r} _{12}\)

since \(\epsilon = {\epsilon}_r {\epsilon}_o\)

we have \(\vec {F} _{12}^{W}\) = \(\dfrac{1}{4\pi{\epsilon}_r{\epsilon}_r}\) \(\dfrac{q_1q_2}{r^2}\) \(\hat{r} _{12}\) = \(\dfrac{\vec {F} _{12}}{{\epsilon}_r}\)

Therefore,

\(\vec {F} _{21}^{W}\) = \(\dfrac{54 × 10^{-3}}{80}\hat{i}\) = \(-0.675 \) × \(\ 10^{-3}N\hat{i}\)

Note that the strength of the force between the two charges in water is reduced by 80 times compared to the force between the same two charges in vacuum. When common salt (NaCl) is taken in water, the electrostatic force between Na and Cl ions is reduced due to the high relative permittivity of water \\(\epsilon _{r} = 80\\). This is the reason water acts as a good solvent.

EXAMPLE 1.3

Two small-sized identical equally charged spheres, each having mass 1 g are hanging in equilibrium as shown in the figure. The length of each string is 10 cm and the angle θ is 30° with the vertical. Calculate the magnitude of the charge in each sphere.

(Take g = 10 ms−2)

Solution

If the two spheres are neutral, the angle between them will be \(\ 0\degree\) when hanged vertically. Since they are positively charged spheres, there will be a repulsive force between them and they will be at equilibrium with each other at an angle of 30° with the vertical. At equilibrium, each charge experiences zero net force in each direction. We can draw a free body diagram for one of the charged spheres and apply Newton’s second law for both vertical and horizontal directions. The free body diagram is shown below.

In the x-direction, the acceleration of the charged sphere is zero.

Using Newton’s second law \((\vec {F} _{tot} = m \hat{a}\)) we have

\(T sin\theta \hat{i} - F_{e}\hat{i} = 0\)

\(T sin\theta = F_{e}\) —–> equation 1

Here T is the tension acting on the charge due to the string and \(F _e\) is the electrostatic force between the two charges.

In the y-direction also, the net acceleration experienced by the charge is zero.

\(T cos \theta \hat{j} -mg \hat{j} = 0\)

\(T cos \theta = mg\) —–> equation 2

By dividing equation (1) by equation (2),

\(\tan \theta = \dfrac{F_e}{mg}\)

Since they are equally charged, the magnitude of the electrostatic force is

\(F _e = k \dfrac{q^2}{r^2}\) where k = \(\dfrac{1}{4\pi\epsilon _o}\)

Here \(r = 2a = 2Lsin\theta\). By substituting these values in equation (3),

\(\ tan\theta = k \dfrac{q^2}{mg(2Lsin\theta)^{2}}\)

Rearranging the equation (4) to get q

\(q = 2Lsin\theta \sqrt{\dfrac{mg tan\theta}{k}}\)

\( = 2 × 0.1 × sin30\degree\) × \(\sqrt{\dfrac{10^{-3}×10×tan30\degree}{9×10^9}}\)

\(q = 8.01 × 10^{-8} C = 80.1 nC\)

EXAMPLE 1.4

Calculate the electrostatic force and gravitational force between the proton and the electron in a hydrogen atom. They are separated by a distance of \(5.3 × 10^{–11}\) m. The magnitude of charges on the electron and proton are \(1.6 × 10^{–19} C\). Mass of the electron is \(m _e = 9.1 × 10^{–31}\) kg and mass of proton is \(m _p\) = \(1.6 × 10^{-27}\) kg.

Solution The proton and the electron attract each other. The magnitude of the electrostatic force between these two particles is given by

\(F _{e} = \dfrac{Ke^2}{r^2}\) = \(\dfrac {9×10^{9}×(1.6×10^{-19})}{(5.3×10^{-11})^2}\)

\(=\dfrac{9×2.56}{28.09}×10^{-7} = 8.2 ×10^{-8}N\)

The gravitational force between the proton and the electron is attractive. The magnitude of the gravitational force between these particles is

\(F_G = \dfrac{Gm_em_p}{r_2}\) \( = \dfrac{6.67 × 10^{-11}×9.1×10^{-31}×1.6×10^{-27}}{(5.3×10^{-11})^2}\)

\(\ = \dfrac{97.11}{28.09}×10^{-47} = 3.4 ×10^{-47}N\)

The ratio of the two forces

\(\dfrac{F_e}{F_G} = \dfrac{8.2 × 10^-8}{3.4×10^{-47}} = 2.41 × 10 ^{39}\)

Note that \(F_e \approx 10^{39} F_G\)

The electrostatic force between a proton and an electron is enormously greater than the gravitational force between them. Thus the gravitational force is negligible when compared with the electrostatic force in many situations such as for small size objects and in the atomic domain. This is the reason why a charged comb attracts an uncharged piece of paper with greater force even though the piece of paper is attracted downward by the Earth. This is shown in Figure 1.3

Superposition principle

Coulomb’s law explains the interaction between two point charges. If there are more than two charges, the force on one charge due to all the other charges needs to be calculated. Coulomb’s law alone does not give the answer. The superposition principle explains the interaction between multiple charges.

According to this superposition principle, the total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.

Consider a system of n charges, namely \(q _1, q _2, q _3 ….q _n\). The force on \(q _1\) exerted by the charge \(q _2\)

\( \vec {F}=k\dfrac{q_1q_2}{r^{2}_{21}} \hat{r} _{21}\)

where \(r^{21}\) is the unit vector from \(q _2\) to \(q _1\) along the line joining the two charges and \(r _{21}\) is the distance between the charges \(q _1\) and \(q _2\). The electrostatic force between two charges is not affected by the presence of other charges in the neighbourhood.

The force on \(q _1\) exerted by the charge \(q _3\) is

\( \vec {F}_{13} = k\dfrac{q_1q_3}{r^{2} _{31}} \hat{r} _{31}\)

By continuing this, the total force acting on the charge |(q _1\) due to all other charges is given by

\( \vec {F} ^{tot} _1\) = \(\vec {F} _{12} + \vec {F} _{13} + \vec {F} _{14} + …..\vec {F} _{1n}\)

\( \vec {F} ^{tot} _1\) = k{\(\dfrac{q _1q _2}{r^2 _{21}}\hat {r} _{21}\) + \(\dfrac{q _1q _3}{r^2 _{31}}\hat {r} _{31}\) + \(\dfrac{q _1q _4}{r^2 _{41}}\hat {r} _{41}\) + ….. + \(\dfrac{q _1q _n}{r^2 _{n1}}\hat {r} _{n1}\)}

Without the superposition principle, Coulomb’s law will be incomplete when applied to more than two charges. Both the superposition principle and Coulomb’s law form fundamental principles of electrostatics and explain all the phenomena in electrostatics. But they are not derivable from each other.

EXAMPLE 1.5

Consider four equal charges \(q _1, q _2, q _3\) and \(q _4\) \(= q = +1 μC\) located at four different points on a circle of radius 1m, as shown in the figure. Calculate the total force acting on the charge \(q _1\) due to all the other charges.

Solution

According to the superposition principle, the total electrostatic force on charge \(q _1\) is the vector sum of the forces due to the other charges,

\(\vec {F}^{tot} = \vec {F} _{12} + \vec {F} _{13} + \vec {F} _{14}\)

The following diagram shows the direction of each force on the charge q₁

The charges \(q _2\) and \(q _4\) a re equi- distant from \(q _1\). As a result the strengths (magnitude) of the forces \(\vec {F} _{12}\) and \(\vec {F} _{14}\) are the same even though their directions are different. Therefore the vectors representing these two forces are drawn with equal lengths. But the charge \(q _3\) is located farther compared to \(q _2\) and \(q _4\). Since the strength of the electrostatic force decreases as distance increases, the strength of the force \(\vec {F}^{13}\) and \(\vec {F}^{14}\). Hence the vector representing the force \(\vec {F}^{13}\) is drawn with smaller length compared to that for forces \(\vec {F}^{12}\) and \(\vec {F}^{14}\).

From the figure, \(r _{21} - \sqrt{2}m = r _{41}\) and \(r _{31} = 2m\)

The magnitudes of the forces are given by

\(\vec {F} _{13} = \dfrac{kq^2}{r _{31}^2} = \dfrac{9 × 10^9 × 20^{-12}}{4}\)

\(\vec {F}_{13} = 2.25 × 10^{-3} \N\)

\(\vec {F} _{12} = \dfrac{kq^2}{r _{21}^2} = \vec {F} _{14} = \dfrac{9 × 10^9 × 20^{-12}}{2}\)

\(= 4.5 × 10^{-3} \N\)

From the figure, the angle \(θ = 45^o\). In terms of the components, we have

\(\vec {F} _{12} = F _{12} cos\theta \hat{i} - F _{12} sin\theta \hat{j}\)

\(= 4.5 × 10^{-3} × \dfrac{1}{\sqrt{2}} \hat{i} - 4.5 × 10^{-3} × \dfrac{1}{\sqrt{2}} \hat{j}\)

\(\vec {F} _{13} = F _{13} \) \(\hat{i} = 2.25 × 10^{-3} \N \hat{i}\)

\(\vec {F} _{14} = F _{14} cos\theta \hat{i} - F _{14} sin\theta \hat{j}\)

\(4.5 × 10^{-3} × \dfrac{1}{\sqrt{2}} \hat{i} + 4.5 × 10^{-3} × \dfrac{1}{\sqrt{2}} \hat{j}\)

Then the total force on \(q _1\) is

\(\vec {F} _1^{tot} = (F _{12} cos\theta \hat{i} - F _{12} sin\theta \hat{j}) + F _{13} \hat{i} + (F _{14} cos\theta \hat{i} + F _{14} sin\theta \hat{j}) \)

\(\vec {F} _1^{tot} = (F _{12} cos\theta + F _{13} + F _{14} cos\theta) \hat{i} + (-F _{12} sin\theta \hat{i} + F _{14} sin\theta) \hat{j}\)

Since \(F _{12} = F _{14}\), the \(j^{th}\) component is zero. Both the electric and gravitational forces are Hence we have

\(\vec {F} _1^{tot} = (F _{12} cos\theta + F _{13} + F _{14} cos\theta) \hat{i}\)

substituting the values in the above equation,

\(= (\dfrac{4.5}{\sqrt{2}} + 2 + \dfrac{4.5}{\sqrt{2}}) × 10^{-3} \hat{i}\)

\(=(4.5\sqrt{2} + 2.25)×10^{-3} \hat{i}\)

\(\vec {F} _1^{tot} = 8.61 × 10^{-3} \hat{i} N\)

The resultant force is along the positive x axis.

ELECTRIC FIELD AND ELECTRIC FIELD LINES

Electric Field

The interaction between two charges is determined by Coulomb’s law. How does the interaction itself occur? Consider a point charge kept at a point in space. If another point charge is placed at some distance from the first point charge, it experiences either an attractive force or repulsive force. This is called ‘action at a distance’. But how does the second charge know about existence of the first charge which is located at some distance away from it? To answer this question, Michael Faraday introduced the concept of field.

According to Faraday, every charge in the universe creates an electric field in the surrounding space, and if another charge is brought into its field, it will interact with the electric field at that point and will experience a force. It may be recalled that the interaction of two masses is similarly explained using the concept of gravitational field (Refer unit 6, volume 2, XI physics). Both the electric and gravitational forces are non-contact forces, hence the field concept is required to explain action at a distance.

Consider a source point charge q located at a point in space. Another point charge \(q _o\) (test charge) is placed at some point P which is at a distance r from the charge q . The electrostatic force experienced by the charge \(q _o\) due to q is given by Coulomb’s law.

\(\vec{F} = \dfrac{kqq _o}{r^2} \hat{r} = \dfrac{qq _o}{r^2}\hat{r}\) where \(k = \dfrac{1}{4\pi\epsilon _o}\)

The charge q creates an electric field in the surrounding space within which its effect can be felt by another charge. It is measured in terms of a quantity called electric field intensity or simply called electric field \(\vec{E}\). The electric field at the point P at a distance r from the point charge q _ is defined as the force that would be experienced by a unit positive charge placed at that point P and is given by

\(\vec{E} = \dfrac{\vec{F}}{q _o} = \dfrac{kq}{r^2} \hat{r} = \dfrac{1}{4\pi\epsilon _o}\dfrac{q}{r^2} \hat{r}\) —>(1.4)

Here \(\hat{r}\) is the unit vector pointing from q to the point of interest P. The electric field is a vector quantity and its SI unit is newton per coulomb \((NC^{–1}\)).

Important aspects of Electric field

(i) If the charge q is positive then the electric field points away from the source charge and if q is negative, the electric field points towards the source charge q. This is shown in the Figure 1.4.

(ii) If the electric field at a point P is \(\vec{E}\), then the force experienced by the test charge q _o placed at the point P is

\(\vec{F} = q_0 \vec{E}\) (1.5)

This is Coulomb’s law in terms of electric field. This is shown in Figure 1.5

(iii) The equation (1.4) implies that the electric field is independent of the test charge \(q _o\) and it depends only on the source charge q.

(iv) Since the electric field is a vector quantity, at every point in space, this field has unique direction and magnitude as shown in Figures 1.6(a) and (b). From equation (1.4), we can infer that as distance increases, the electric field decreases in magnitude.

Note that in Figures 1.6 (a) and (b) the length of the electric field vector is shown for three different points. The strength or magnitude of the electric field at point P is stronger than at the points Q and R because the point P is closer to the source charge.

(v) In the definition of electric field, it is assumed that the test charge \(q _0\) is taken sufficiently small, so that bringing this test charge will not move the source charge. In other words, the test charge is made sufficiently small such that it will not modify the electric field of the source charge.

(vi) The expression (1.4) is valid only for point charges. For continuous and finite size charge distributions, integration techniques must be used (Refer Appendix A1.1). However, this expression can be used as an approximation for a finite-sized charge if the test point is very far away from the finite sized source charge. Note that we similarly treat the Earth as a point mass when we calculate the gravitational field of the Sun on the Earth (Refer unit 6, volume 2, XI physics).

(vii) There are two kinds of the electric field: uniform (constant) electric field and non-uniform electric field. Uniform electric field will have the same direction and constant magnitude at all points in space. Non-uniform electric field will have different directions or different magnitudes or both at different points in space. The electric field created by a point charge is basically a non uniform electric field. This non-uniformity arises, both in direction and magnitude, with the direction being radially outward (or inward) and the magnitude changes as distance increases. These are shown in Figure 1.7.

EXAMPLE 1.6

Calculate the electric field at points P, Q for the following two cases, as shown in the figure.

(a) A positive point charge +1 µC is placed at the origin

(b) A negative point charge –2 µC is placed at the origin

Solution

Case (a) The magnitude of the electric field at point P is

\(E _p = \dfrac{1}{4\pi\epsilon _0}\dfrac{q}{r^2} = \dfrac{9×10^9×1×10^{-6}}{4}\)

\(= 2.25 × 10^3 NC^{-1}\)

Since the source charge is positive, the electric field points away from the charge. So the electric field at the point P is given by

\(\vec{E}_p = 2.25 × 10^3 \hat{i} NC^{-1}\)

For the point Q

\(|\vec{E} _Q| = \dfrac{9×10^9×1×10^{-6}}{16} = 0.56 × 10^3 NC^{-1}\)

Hence \(\vec{E} _Q = 0.56 × 10^3 \hat{j} NC^{-1}\)

Case (b)

The magnitude of the electric field at point P

\(|\vec{E}_p| = \dfrac{kq}{r^2} = \dfrac{1}{4\pi\epsilon _0}\dfrac{q}{r^2} = \dfrac{9×10^9×2×10^{-6}}{4}\)

\(= 4.5 × 10^3 NC^{–1}\)

Since the source charge is negative, the electric field points towards the charge. So the electric field at the point P is given by

\(\vec{E} _P= −4.5 × 10^3 \hat{i} NC^{-1} \)

For the point Q, \(|\vec{E} _Q| = \dfrac{9×10^9×2×10^{-6}}{36}\)

\(= 0.5 × 10^3 NC^{-1}\)

At the point Q the electric field is directed along the positive x-axis.

Electric field due to the system of point charges

Suppose a number of point charges are distributed in space. To find the electric field at some point P due to this collection of point charges, superposition principle is used. The electric field at an arbitrary point due to a collection of point charges is simply equal to the vector sum of the electric fields created by the individual point charges. This is called superposition of electric fields.

Consider a collection of point charges \(q _1 ,q _2 ,q _3,…..,q _n\) located at various points in space. The total electric field at some point P due to all these n charges is given by

\(\vec{E} _{tot} = \vec{E} _1 + \vec{E} _2 + \vec{E} _3 + …… +\vec{E} _n\) —>(1.6)

\(\vec{E} _{tot} = \dfrac{1}{4\pi\epsilon} (\dfrac{q _1}{r _{1p}^2}\hat{r} _{1p} + \dfrac{q _2}{r _{2p}^2}\hat{r} _{2p} + \dfrac{q _3}{r _{3p}^2}\hat{r} _{3p} + …. + \dfrac{q _n}{r _{np}^2}\hat{r} _{np} ) \) —> (1.7)

where \(r _{1P} , r _{2P} , r _{3P} ……….r _{nP}\) are the distance of the the charges \(q _1 , q _2 , q _3 ……….q _n\) from the point P respectively. Also \(\hat{r} _{1P} , \hat{r} _{2P} , \hat{r} _{3P} ………. \hat{r} _{nP}\) are the corresponding unit vectors directed from \(q _1 , q _2 , q _3 ……….q _n\) to P. Equation (1.7) can be re-written as,

\( \vec{E} _{tot} = \dfrac{1}{4\pi\epsilon _0} \displaystyle\sum _{i=1}^n (\dfrac{q _i}{r _{ip}^2}\hat{r} _{ip}\))

For example in Figure 1.8, the resultant electric field due to three point charges \(q _1 , q _2 , q _3\) at point P is shown.

Note that the relative lengths of the electric field vectors for the charges depend on relative distances of the charges to the point P.

todo

EXAMPLE 1.7

Consider the charge configuration as shown in the figure. Calculate the electric field at point A. If an electron is placed at points A, what is the acceleration experienced by this electron? (mass of the electron = 9.1 × 10–31 kg and charge of electron = −1.6 × 10–19 C)

= q1 +1µC

= q2 +1µC

x

y

A

2mm

2mm

+

+

Solution

By using superposition principle, the net electric field at point A is 

 

E q r

r q r

rA A

A A

A= + 1

4 1

4 1

1 2 1

2

2 2 2π πe e

 

where _r_1A and _r_2A are the distances of point A from the two charges respectively.



E j iA = × × ×

×( ) ( )+ × × ×

×( ) ( )

−

−

−

−

9 10 1 10

2 10

9 10 1 10

2 10

9 6

3 2

9 6

3 2 



q+	E3P
E1P
rP3P3	Etot
r2P2	r+ 1P	E2P

01 ×10  91××01 ×96 − 96	()10 i−2
 91××Ej =A
() () ()+21× 0 21× 0−3 2 −3

= × + × = × +( )2 25 10 2 25 10 2 25 109 9 9. . ._j i i j_   

The magnitude of electric field 

EA = ×( ) + ×( ) = × × −

2 25 10 2 25 10

2 25 2 10

9 2 9 2

9 1

. .

. NC

The direction of 

EA is given by 



E E

i j i jA

A

= × +( ) × ×

= +( )2 25 10

2 25 2 10 2

9

9

.

. ,

   

which is

the unit vector along OA as shown in the figure.

y

A**+**

+ = q1 +1µC

= q1 +1µC

xO 2mm

2mm

E**q1**



E**q2**



E**A**



aA 

The acceleration experienced by an electron placed at point A is



 

a F m

qE m

i j

A A= =

= − ×( )× ×( ) +( )

×

−

−

1 6 10 2 25 10

9 1 10

19 9

31

. .

.

 

=− × +( ) −3 95 1020 1. _i j_  N kg

The electron is accelerated in a direction exactly opposite to



EA .

Electric field due to continuous charge distribution

The electric charge is quantized microscopically. The expressions (1.2), (1.3), (1.4) are applicable to only point charges. While dealing with the electric

field due to a charged sphere or a charged wire etc., it is very difficult to look at individual charges in these charged bodies. Therefore, it is assumed that charge is distributed continuously on the charged bodies and the discrete nature of charges is not considered here. The electric field due to such continuous charge distributions is found by invoking the method of calculus. (For further reading, refer Appendix A1.1).

EXAMPLE 1.8

A block of mass m carrying a positive charge q _ is placed on an insulated frictionless inclined plane as shown in the figure. A uniform electric field E is applied parallel to the inclined surface such that the block is at rest. Calculate the magnitude of the electric field E.

m

h L

q _



E 

Solution

Note: A similar problem is solved in XIth Physics volume I, unit 3 section 3.3.2. There are three forces that acts on the mass m:

(i) The downward gravitational force exerted by the Earth (mg)

(ii) The normal force exerted by the inclined surface (N)

(iii) The Coulomb force given by uniform electric field (qE)

The free body diagram for the mass m is drawn below.

AO + am2mmAq = +1µC+1

m	
L	E

Thtfeexac	=×22.. 51	02 ji+×25 10 =×22. 51 0()tude of electr ()ic field9922.. 51 02 +× 25 10.N52 ×10 Cection o9 2 f  i s g9() E()51×+0 ij 91 = −ij + w. 52×× 10 2ector along O A a A s show99Ay E E+1µCq EO + a2mm2mmAq = +1µC+leration exp erienced blaced at point A is 1qE ()m ()16.. 10 ××22 51 0 A ()91. ×1039. 51×+0 −19 ij Nk g 9ctron is accelera−t31ed in a dirposite t20o E . −1A	()ij +9

| e magniE =×=×22The dirAE 22. =E 22he uni t vigurA e.Aq = 1The accelectron p ()Fa ==m−×=A=−The eletly op | | 2 | | iven b y hich i s n in t he Aqxy a n | | ij + | | ection |

m,q

mg

m

θ

θ

x

y

mg cosθ mg sin θ

Free body diagram

Forces acting on the mass m

NqE

mg

N qE

mg

A convenient inertial coordinate system is located in the inclined surface as shown in the figure. The mass m has zero net acceleration both in x and y-direction.

Along x-direction, applying Newton’s second law, we have

_mg i qEi_sinθ − = 0

_mg qE_sinθ− = 0

E mg

q _ =

sinθ

Note that the magnitude of the electric field is directly proportional to the mass m and inversely proportional to the charge q _. It implies that, if the mass is increased by keeping the charge constant, then a strong electric field is required to stop the object from sliding. If the charge is increased by

keeping the mass constant, then a weak electric field is sufficient to stop the mass from sliding down the plane.

The electric field also can be expressed in terms of height and the length of the inclined surface of the plane.

E mg h qL

=

Electric field lines

Electric field vectors are visualized by the concept of electric field lines. They form a set of continuous lines which are the visual representation of the electric field in some region of space. The following rules are followed while drawing electric field lines for charges. - The electric field lines start from a positive charge and end at negative charges or at infinity. For a positive point charge the electric field lines point radially outward and for a negative point charge, the electric field lines point radially inward. These are shown in Figure 1.9 (a) and (b).

(a)

q _

For a positive point charge, the eld lines are directed

radially outward.

+

m

| θ |

|——|——| | qExmNθmg cosθ mg sin θmgmgmgθ |

keeping t he m ass co nstant, t hen a w eak electric f ield i s s ufficient t o s top t he m ass from sliding down the plane.The e lectric f ield a lso c an b e exp ressed in terms of height and the length of the inclined surface of the plane.mg hE =qL

|——|——|

q+

|——|——|

(a)

(b)

- -q_

For a negative point charge, the eld lines are directed

radially inward.

Figure 1.9 Electric field lines for isolated positive and negative charges

Note that for an isolated positive point charge the electric field line starts from the charge and ends only at infinity. For an isolated negative point charge the electric field lines start at infinity and end at the negative charge.

• The electric field vector at a point in space is tangential to the electric field line at that point. This is shown in Figure 1.10

P

**E**

_+

Figure 1.10 Electric field at a point P

• The electric field lines are denser (more closer) in a region where the electric field has larger magnitude and less dense in a region where the electric field is of smaller magnitude. In other words, the number of lines passing through a given surface area perpendicular to the lines is proportional to the magnitude of

the electric field in that region. This is shown in Figure1.11

A B

+

Figure 1.11 Electric field has larger magnitude at surface A than B

Figure 1.11 shows electric field lines from a positive point charge. The magnitude of the electric field for a point charge decreases

as the distance increases 

E r

∝ 

 



 

1 2 . So the

electric field has greater magnitude at the surface A than at B. Therefore, the number of lines crossing the surface A is greater than the number of lines crossing the surface B. Note that at surface B the electric field lines are farther apart compared to the electric field lines at the surface A. - No two electric field lines intersect each other. If two lines cross at a point, then there will be two different electric field vectors at the same point, as shown in Figure 1.12.

= Electric eld = Electric eld lines

P

Figure 1.12 Two electric field lines never intersect each other

|——|——|——|

|——|——|——|——|——|——|——|

| –_ |-q |

As a consequence, if some charge is placed in the intersection point, then it has to move in two different directions at the same time, which is physically impossible. Hence, electric field lines do not intersect. - The number of electric field lines that emanate from the positive charge or end at a negative charge is directly proportional to the magnitude of the charges.

For example in the Figure 1.13, the electric field lines are drawn for charges +q _ and –2q _. Note that the number of field lines emanating from +q _ is 8 and the number of field lines ending at –2q _ is 16. Since the magnitude of the second charge is twice that of the first charge,

Figure 1.13 Electric field lines and magnitude of the charge

+q

-2q

Electric eld lines

Electric eld lines

(a)

(b)

+

_

the number of field lines drawn for –2q _ is twice in number than that for charge +q _.

EXAMPLE 1.9

The following pictures depict electric field lines for various charge configurations.

A C

B

(b)

+ +

q1

q2

(a)

1

2

(c)

_ q2q1 q3

(i) In figure (a) identify the signs of two

charges and find the ratio q q

1

2

(ii) In figure (b), calculate the ratio of two positive charges and identify the strength of the electric field at three points A, B, and C

(iii) Figure (c) represents the electric field lines for three charges. If q _2 = –20 nC, then calculate the values of q _1 and q _3

|——|——|——|——|——|——|——|——|——|——|

| _ |

| -2q |

Solution

(i) The electric field lines start at q _2 and end at q _1. In figure (a), q _2 is positive and q _1

is negative. The number of lines starting from q _2 is 18 and number of the lines ending at q _1 is 6. So q _2 has greater magnitude. The

ratio of q q

N N

1

2

1

2

6 18

1 3

= = = . It implies that

_q q_2 13=

(ii) In figure (b), the number of field lines emanating from both positive charges are equal (N=18). So the charges are equal. At point A, the electric field lines are denser compared to the lines at point B. So the electric field at point A is greater in magnitude compared to the field at point B. Further, no electric field line passes through C, which implies that the resultant electric field at C due to these two charges is zero. (iii) In the figure (c), the electric field lines start at q _1 and q _3 and end at q _2. This implies that q _1 and q _3 are positive charges. The ratio of the number of field lines is

q q

q q

1

2

3

2

8 16

1 2

= = = , implying that q _1and q _3

are half of the magnitude of q _2. So q _1 = q _3 = +10 nC.

ELECTRIC DIPOLE AND ITS PROPERTIES

Electric dipole

Two equal and opposite charges separated by a small distance constitute an electric dipole. In many molecules, the centres of positive and negative charge do

not coincide. Such molecules behave as permanent dipoles. Examples: CO, water, ammonia, HCl etc.

Consider two equal and opposite point charges (+q _, –q ) that are separated by a distance 2_a as shown in Figure 1.14(a). The electric dipole moment is defined as

  _p qr q r_= + −+ −( ) (1.9)

where _r_+ is the position vector of +q _ from the origin and _r_- is the position vector of –q _ from the origin. Then, from Figure 1.14 (a),

+_

(b)

(a)

p

_-q +q_a a

y

x _ +





p qai qa i qai= − −( )=  2 (1.10)

The electric dipole moment vector lies along the line joining two charges and is directed from –q _ to +q _. The SI unit of dipole moment is coulomb metre (Cm). The electric field lines for an electric dipole are shown in Figure 1.14 (b). - For simplicity, the two charges are placed on the x-axis. Even if the two charges are placed on y or z-axis, dipole moment will point from –q _ to +q _.

• The magnitude of the electric dipole moment is equal to the product of the magnitude of one of the charges and the distance between them,



p qa= 2 (1.11)

• Though the electric dipole moment for two equal and opposite charges is defined, it is possible to define and calculate the electric dipole moment for a collection of point charges. The electric dipole moment for a collection of n point charges is given by

 _p q ri i_

i

n

= = ∑

1

(1.12)

where _ri_ is the position vector of charge qi

from the origin.

EXAMPLE 1.10

Calculate the electric dipole moment for the following charge configurations.

+q +q a a

y y

y y (a)

+q +q a

(b)

+q

a

(c)

-2q +q

+q aa a

(d)

-2q

2a

x x

x x

+ + + +

+

+ +

_

_

Solution Case (a) The position vector for the +q _ on the positive x-axis is _ai_and position vector for the +q _ charge the negative x axis is -_ai_ . So the dipole moment is,



p q ai q ai= +( )( )+ +( ) −( )=  0

Case (b) In this case one charge is placed at the origin, so its position vector is zero. Hence only the second charge +q _ with position vector ai_contributes to the dipole moment, which is _p qai= .

From both cases (a) and (b), we can infer that in general the electric dipole moment depends on the choice of the origin and charge configuration. But for one special case, the electric dipole moment is independent of the origin. If the total charge is zero, then the electric dipole moment will be the same irrespective of the choice of the origin. It is because of this reason that the electric dipole moment of an electric dipole (total charge is zero) is always directed from –q _ to +q _, independent of the choice of the origin.

Case (c) _p q aj q a j qaj_= − + − =−( ) ( )( ) .2 2 4   Note that in this case _p_ is directed from –2q _ to +q _.

Case (d) 

p qa i qaj qa j qa i

=− − + + − =

2 2

( ) ( )  



The water molecule (H2O) has this charge configuration. The water molecule has three atoms (two H atom and one O atom). The centres of positive (H) and negative (O) charges of a water molecule lie at different points, hence it possess permanent dipole moment. The electric dipole moment _p_ is directed from centre of negative charge to the centre of positive charge, as shown in the figure.

H

H

O +

+

+

Centre of positive charge

Centre of negative charge

– _p_1 

_p_2 

**p**

104

H

H

+q

-2q

+q

O

+

+

–

| pq |= 2 |

O	p p1 
–	p

Electric field due to a dipole

Case (i) Electric field due to an electric dipole at points on the axial line

Consider an electric dipole placed on the x-axis as shown in Figure 1.15. A point C is located at a distance of r from the midpoint O of the dipole on the axial line.

A a a

B E– E+

-q +q C

r

O _ +

→ →p→ Axial line

Figure 1.15 Electric field of the dipole along the axial line

The electric field at a point C due to +q _ is 

E q r a

+ = −( )

1 4 0

2πe along BC

Since the electric dipole moment vector 

p is from –q _ to +q _ and is directed along BC, the above equation is rewritten as



E q r a

p+ = −( )

1 4 0

2πe  (1.13)

where _p_ is the electric dipole moment unit vector from –q _ to +q _.

The electric field at a point C due to –q _ is 

E q r a

_p_− =− +( )

1 4 0

2πe  (1.14)

Since +q _ is located closer to the point C than –q _,



E+ is stronger than 

E- . Therefore, the length of the



E+ vector is drawn larger than that of



E- vector. The total electric field at point C is

calculated using the superposition principle of the electric field.

  

E E Etot = ++ −

= −( )

− +( )

1 4

1 40

2 0

2π πe e

q r a

p q r a

_p_ 



E q r a r a

ptot = −( )

− +( )











4 1 1

0 2 2πe

 (1.15)



E q ra

r a ptot =

−( )













1 4

4

0 2 2 2πe

 (1.16)

Note that the total electric field is along 

E+ , since +q _ is closer to C than –q _. The direction of



Etot is shown in Figure 1.16.

A a a B Etot

-q +q C

r

O _ +



Figure 1.16 Total electric field of the dipole on the axial line

If the point C is very far away from the dipole (r » a). Then under this limit the term _r a r_2 2

2 4−( ) ≈ . Substituting this into

equation (1.16), we get 

E aq r

ptot = 

 



 

1 4

4

0 3πe  (r » a)

since 2_aq p p_ = 





E p rtot =

1 4

2

0 3πe

(r » a) (1.17)

If the point C is chosen on the left side of the dipole, the total electric field is still in the direction of _p_ . We infer this result by examining the electric field lines of the dipole shown in Figure 1.14(b).

Case (ii) Electric field due to an electric dipole at a point on the equatorial plane

Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane as shown in Figure 1.17. Since the point C is equi-distant from +q _ and –q _, the magnitude of the electric fields at C due to +q _ and –q _ are the same. The direction

E =Etot =tot

of 

E+ is along BC and the direction of 

E- is along CA.



E+ and 

E- can be resolved into two components; one component parallel to the dipole axis and the other perpendicular to it. Since perpendicular components 

E+ sinθ and 

E- sinθ are eqaul in magnitude and oppositely directed, they cancel each other. The magnitude of the total electric field at point C is the sum of the parallel components of



E+ and 

E- and its direction is along -_p_ as shown in the Figure 1.17.

  

E E p E ptot =− −+ −cos cosθ θ  (1.18)

The magnitudes 

E+ and 

E- are the same and are given by

 

E E q r a+ −= = +( )

1 4 0

2 2πe (1.19)

By substituting equation (1.19) into equation (1.18), we get



E q r a

ptot =− +( ) 1

4 2

0 2 2π

θ e

cos 

=− +( )

1 4

2

0 2 2 3 2πe

qa

_r a p_

since cosθ= +

a

_r a_2 2

Figure 1.17 Electric field due to a dipole at a

–p

C

A B

a aO θθ

-q +q

r

+–

Etot →

E+ →

→

E– →

Equatorial





E p

r a tot =−

+( ) 1

4 0 2 2 3 2πe

since _p qap_= 2  (1.20)

At very large distances (r » a), the equation (1.20) becomes

 



E p r

r atot =− » 1

4 3πe ( ) (1.21)

Important inferences (i) From equations (1.17) and (1.21), it is inferred that for very large distances, the magnitude of the electric field at point on the dipole axis is twice the magnitude of the electric field at the point at the same distance on the equatorial plane. The direction of the electric field at points on the dipole axis is directed along the direction of dipole moment vector _p_ but at points on the equatorial plane it is directed opposite to the dipole moment vector, that is along -_p_ . (ii) At very large distances, the electric

field due to a dipole varies as 1 3_r_

. Note

that for a point charge, the electric field

point on the equatorial plane

E+ sinθ

E+ cosθ

E– sinθ

E– cosθ θ θ

E+ →

E– →

→

→

→

→

plane

θ
θ

E→	+	Eq
C
Et →θ–	r	θ+
–p→
aa	O	+q

varies as 1 2_r_

. This implies that the electric

field due to a dipole at very large distances goes to zero faster than the electric field due to a point charge. The reason for this behavior is that at very large distance, the two charges appear to be close to each other and neutralize each other. (iii) The equations (1.17) and (1.21) are valid only at very large distances (r>>a). Suppose the distance 2_a_ approaches zero and q _ approaches infinity such that the product of 2_aq_ = p is finite, then the dipole is called a point dipole. For such point dipoles, equations (1.17) and (1.21) are exact and hold true for any r.

Torque experienced by an electric dipole in the uniform electric field

Consider an electric dipole of dipole moment _p_ placed in a uniform electric field



E whose field lines are equally spaced and point in the same direction. The charge +q _ will experience a force q _



E in the direction of the field and charge –q _ will experience a force –q _



E in a direction opposite to the field. Since the external field



E is uniform, the total force acting on the dipole is zero. These two forces acting at different points will constitute a couple and the dipole experience a torque as shown in Figure 1.18. This torque tends to rotate the dipole. (Note that electric field lines of a uniform field are equally spaced and point in the same direction). The total torque on the dipole about the point O

       τ= × −( )+ ×_OA qE OB qE_ (1.22)

Using right-hand corkscrew rule (Refer XI, volume 1, unit 2), it is found that total torque is perpendicular to the plane of the paper and is directed into it.

The magnitude of the total torque τ θ θ= −( ) +OA qE OB qE      

sin sin

τ θ= ⋅_qE a_2 sin (1.23)

where θ is the angle made by _p_ with 

E . Since p = 2_aq_, the torque is written in terms of the vector product as

 



τ= ×_p E_ (1.24)

The magnitude of this torque is τ θ= pE sin and is maximum when θ = 90 .

This torque tends to rotate the dipole and align it with the electric field



E . Once 

p is aligned with 

E , the total torque on the dipole becomes zero.

If the electric field is not uniform, then the force experienced by +q _ is different from that experienced by –q _. In addition to the torque, there will be net force acting on the dipole. This is shown in Figure 1.19.

E B_+q_

-q

2a sinθ

-qE

a

a

Torque is into the paper A

θ

θ

O

+

→

**qE**→

→

–

+
a
θ
aO

| θ |

Net torque, Net force

Non uniform E

O

+q

-q

**qE1**→

**qE2**→

→

+

––

Figure 1.19 The dipole in a non-uniform electric field

EXAMPLE 1.11

A sample of HCl gas is placed in a uniform electric field of magnitude 3 × 104 N C–1. The dipole moment of each HCl molecule is 3.4 × 10–30 Cm. Calculate the maximum torque experienced by each HCl molecule.

Solution The maximum torque experienced

by the dipole is when it is aligned perpendicular to the applied field.

τmax sin .= = × × ×−_pE_ 90 3 4 10 3 1030 4

τmax .= × −10 2 10 26 Nm

Microwave oven works on the p torque acting on an electric dipole we consume has water molecules permanent electric dipoles. Oven

microwaves that are oscillating electromagnetic produce torque on the water molecules. Due to on each water molecule, the molecules rotate ve produce thermal energy. Thus, heat generated heat the food.

ELECTROSTATIC POTENTIAL AND POTENTIAL ENERGY

Introduction In mechanics, potential energy is

defined for conservative forces. Since gravitational force is a conservative force, its gravitational potential energy is defined in XI standard physics (Unit 6). Since Coulomb force is an inverse- square-law force, its also a conservative force like gravitational force. Therefore, we can define potential energy for charge configurations.

Electrostatic Potential energy and Electrostatic potential

Consider a positive charge q _ kept fixed at the origin which produces an electric field



E around it. A positive test charge q _′ is brought from point R to point P against the repulsive force between q _ and q _′ as shown in Figure 1.20. Work must be done to overcome the repulsion between the charges and this work done is stored as potential energy of the system.

rinciple of . The food which are produces fields and

this torque ry fast and is used to

EXAMPLE 1 . 1 1A s ample o f H C l ga s i s p l ace d ina unif or m e le c t r ic f ie ld o f m ag ni tude3 × 10 N C . Th e di p ole m oment of e ac hHC l mole c u le is 3.4 × 10 Cm. C a lc u l atet he m axim um t orque exp er ien ce d b ye ac h H4 C l m –1ole c u le .–30SolutionThe m aximum t orque exp erienced by the dipole is when it is aligned perpendicular to the applied field.τ ==pE sin.90 34 ××10 31× 0τ =×10.21 0 Nm −30 4max−26max

q _

q´

P

R

+

+

The test charge q _′ is brought from R to P with constant velocity which means that external force used to bring the test charge q _′ from R to P must be equal and opposite to the coulomb force

 

F Fext coulomb=−( ). The work done is

W F drext R

P

= ⋅∫ 

 (1.25)

Since coulomb force is conservative, work done is independent of the path and it depends only on the initial and final positions of the test charge. If potential energy associated with q _′ at P is _U_P and that at R is _U_R, then difference in potential energy is defined as the work done to bring a test charge q _′ from point R to P and is given as _U_P – U_R = W = Δ_U

∆_U F drext R_

P

= ⋅∫ 

 (1.26)

Since   

F F q Eext coulomb=− =− ′ (1.27)

∆_U q E dr q E dr R_

P

R

P

= − ′( )⋅ = ′ −( )⋅∫ ∫ 





 (1.28)

The potential energy difference per unit charge is given by

∆_U q_

q E dr

q E drR

P

R

P

′ =

′ −( )⋅

′ =− ⋅

∫ ∫







 (1.29)

The above equation (1.29) is independent

of q ′. The quantity ∆_U q

E dr R

P

′ =− ⋅∫



 is called

electric potential difference between P and R and is denoted as _V_P – V_R = ∆_V.

In otherwords, the electric potential difference is defined as the work done by an external force to bring unit positive charge from point R to point P.

V V V E drP R R

P

− = = − ⋅∫∆ 

 (1.30)

The electric potential energy difference can be written as ∆_U_ = q ′ ∆_V. Physically potential difference between two points is a meaningful quantity. The value of the potential itself at one point is not meaningful. Therefore the point R is taken to infinity and the potential at infinity is considered as zero _V_∞ =( )0 .

Then the electric potential at a point P is equal to the work done by an external force to bring a unit positive charge with constant velocity from infinity to the point P in the region of the external electric field



E . Mathematically this is written as

V E drP

P

=− ⋅ ∞ ∫ 



(1.31)

Important points 1. Electric potential at point P depends

only on the electric field which is due to the source charge q _ and not on the test charge q _′. Unit positive charge is brought from infinity to the point P with constant velocity because external agency should not impart any kinetic energy to the test charge.

2. From equation (1.29), the unit of electric potential is Joule per coulomb. The practical unit is volt (V) named after Alessandro Volta (1745-1827) who invented the electrical battery. The potential difference between two points is expressed in terms of volt.

The description of motion of objects using the concept of potential or potential energy is

simpler than that using the concept of field.

Note

Electric potential due to a point charge

Consider a positive charge q _ kept fixed at the origin. Let P be a point at distance r from the charge q _. This is shown in Figure 1.21.

q _

P

+ r

The electric potential at the point P is

V E dr E dr r r

= −( )⋅ =− ⋅ ∞ ∞ ∫ ∫









(1.32)

Electric field due to positive point charge q _ is



E q r

r= 1

4 0 2πe 

V q r

r dr r

= −

∞ ∫

1 4 0

2πe . 

The infinitesimal displacement vector, _dr drr_

= and using _r r_ . ,=1 we have

V q r

r drr q r

dr r r

=− ⋅ =− ∞ ∞ ∫ ∫

1 4

1 42 2π πe e

 

 

After the integration,

V q r

q r

r

=− −    

    =

∞

1 4

1 1 4π πe e

 

Hence the electric potential due to a point charge q _ at a distance r is

V q r

= 1

4 0πe (1.33)

Important points (i) If the source charge q _ is positive, V > 0. If q _ is negative, then V is negative and equal

to V q r

=− 1

4 0πe

(ii) From expression (1.33), it is clear that the potential due to positive charge decreases as the distance increases, but for a negative charge the potential increases as the distance is increased. At infinity ( )r =∞ electrostatic potential is zero (V = 0).

In the case of gravitational force, mass moves from a point of higher gravitational potential to a point of lower

Earth

Higher gravitational potential

Lower gravitational potential

Mass moves from higher gravitational potential to lower

gravitational potential

O

E =4π−V =4

gravitational potential (Figure 1.22). Similarly a positive charge moves from a point of higher electrostatic potential to a point of lower electrostatic potential. However a negative charge moves from lower electrostatic potential to higher electrostatic potential. This comparison is shown in Figure 1.23.

(iii) The electric potential at a point P due to a collection of charges q _1,q _2,q _3…..q _n is equal to sum of the electric potentials due to individual charges.

V kq r

kq r

kq r

kq r

q r

tot

n

n

i

i i

n

= + + +

+ = =∑

1

1

2

2

3

3

0 1

1 4

…

…. πe

(1.34)

where r1,r2,r3…….rn are the distances of q1, q2, q3 ….. qn respectively from P (Figure 1.24).

+q P Q R (a) Positive charge +q _´ moves from higher electric potential to lower electric potential (P Q R)

+q P Q R

+q´

–q´

(b) negative charge –q _´moves from lower electric potential to higher electric potential

(R Q P)

+

–

VP VQ VR> >

VP VQ VR> >

+

+

r1

r2

r3 rn

q1

q2

q3

qn

P

EXAMPLE 1.12

(a) Calculate the electric potential at points P and Q as shown in the figure below.

(b) Suppose the charge +9 µC is replaced by –9 µC find the electrostatic potentials at points P and Q

–q –q´

+q´

P Q R

-q P Q R (c) Positive charge +q _´ moves from higher electric potential to lower electric potential (R Q P)

(d) negative charge –q _´ moves from lower electric potential to higher electric potential

(P Q R)

–

– –

VP VQ VR< <

VP VQ VR< <

+

ectric potential

VQ	V electric poV tent «ial (R Q P)VR P Q	VR

+9µC 10m P

6m Q**+**

(c) Calculate the work done to bring a test charge +2 µC from infinity to the point Q. Assume the charge +9 µC is held fixed at origin and +2 µC is brought from infinity to P.

Solution

(a) Electric potential at point P is given by

V q rP

P

= = × × ×

= × −1

4 9 10 9 10

10 8 1 10

9 6 3

πe 

. V

Electric potential at point Q is given by

V q rQ

q _

= = × × ×

= × −1

4 9 10 9 10

16 5 06 10

9 6 3

πe 

. V

Note that the electric potential at point Q is less than the electric potential at point P. If we put a positive charge at P, it moves from P to Q. However if we place a negative charge at P it will move towards the charge +9 µC.

The potential difference between the points P and Q is given by

∆_V V VP Q_= − =+ ×3 04 103. V

(b) Suppose we replace the charge +9 µC by –9 µC, then the corresponding potentials at the points P and Q are,

V VP Q=− × =− ×8 1 10 5 06 103 3. , .V V

Note that in this case electric potential at the point Q is higher than at point P.

The potential difference between the points P and Q is given by

∆_V V VP Q_= − =− ×3 04 103. V

(c) The electric potential V at a point Q due to some charge is defined as the work done by an external force to bring a unit positive charge from infinity to Q. So to bring the q _ amount of charge from infinity to the point Q, work done is given as follows.

W = qV

WQ = × × × = ×− −2 10 5 06 10 10 12 106 3 3. . .J

EXAMPLE 1.13

Consider a point charge +q _ placed at the origin and another point charge -2q _ placed at a distance of 9 m from the charge +q _. Determine the point between the two charges at which electric potential is zero.

Solution

According to the superposition principle, the total electric potential at a point is equal to the sum of the potentials due to each charge at that point.

Consider the point at which the total potential zero is located at a distance x from the charge +q _ as shown in the figure.

_+q -2q_P

9m

9-xx+ _

Since the total electric potential at P is zero,

V q x

q xtot = − −( )



 





 =

1 4

2 9

0 πe



(or)

q x

q x

= −( ) 2

9 (or)

1 2 9_x x_

= −( )

Hence, x = 3 m

(c) The e lectric p otential V a t a point Q d ue t o s ome c harge i s def ined as t he w ork do ne b y a n ext ernal f orce to b ring a uni t p ositive c harge f rom infinity t o Q . S o to b ring the q a mount of charge from infinity to the point Q , work done is given as follows. W = qVW =×21 05 ××..06 10 =×10 12 10 J.−−63 3Q

V =−4q () 2=xtot 9 −12= ()Hxxence 9 , −

EXAMPLE 1 . 1 3Consider a p oint c harge + q p laced a t the o rigin a nd a nother p oint c harge -2 qplaced at a distance of 9 m from the charge +q. Determine the point between the two charges at which electric potential is zero.SolutionAccording to the superposition principle, the t otal e lectric p otential a t a p oint i s equal t o t he s um o f t he p otentials d ue t o each charge at that point.Consider t he p oint a t w hich t he t otal potential zer o i s lo cated a t a di stance xfrom the charge +q as shown in the figure. _9m+ x 9-x+q -2qSince the total eP lectric potential at P is zero,V =−4π1e  qx ()92−qx  = 0 (or)q () 2q (or)=xtot 9 − x12= ()Hxxence 9 , −x = 3 m

Electrostatic potential at a point due to an electric dipole

Consider two equal and opposite charges separated by a small distance 2_a_ as shown in Figure 1.25. The point P is located at a distance r from the midpoint of the dipole. Let θ be the angle between the line OP and dipole axis AB.

A B

r1

r2

180-θ θ

r

P

+q-q a a

O +_

_p_→

Let _r_1 be the distance of point P from +q _ and _r_2 be the distance of point P from –q _.

Potential at P due to charge +q _ = 1 4 1πe



q r

Potential at P due to charge –q _ =− 1 4 2πe



q r

Total potential at the point P

V q r r

= − 

 



 

1 4

1 1

1 2πe 

(1.35)

Suppose if the point P is far away from the dipole, such that r>>a, then equation (1.35) can be expressed in terms of r.

By the cosine law for triangle BOP

_r r a ra_1 2 2 2 2= + − cosθ

r r a r

_a r_1

2 2 2

21 2 = + − 

 



 

cosθ

Since the point P is very far from the

dipole (r>>a). As a result the term a r

2

2 is

very small and can be neglected. Therefore

_r r a r_1

2 2 1 2= − 

 



 

cosθ

(or) _r r a r_1

1 2

1 2 = − 

 



 cosθ

1 1 1 2

1

1 2

r r a r

= − 

 



 

−

cosθ

Since a r «1 , we can use binomial theorem

and retain the terms up to first order

1 1 1 1_r r_

a r

= + 

 



 cosθ (1.36)

Similarly applying the cosine law for triangle AOP,

_r r a ra_2 2 2 2 2 180= + − −( )cos θ

since cos cos180−( )=−θ θ we get

_r r a ra_2 2 2 2 2= + + cosθ

Neglecting the term a r

2

2 (because r>>a)

_r r a r_2

2 2 1 2 = + 

 



 

cosθ

PA_
180-θ	rr2 p→θ	rB1
-q O +qaa+

_r r a r_2

1 2

1 2 = + 

 



 

cosθ

Using Binomial theorem, we get

1 1 1 2_r r_

a r

= − 

 



 

cosθ (1.37)

Substituting equation (1.37) and (1.36) in equation (1.35),

V q r

a r r

a r

= + 

 



 − − 

 



 



 



 1

4 1 1 1 1

π θ θ

e 

cos cos 

V q r

a r

a r

= + − + 

 



 



 



 4

1 1 1 π

θ θ e 

cos cos

V aq r

= 1

4 2

2π θ

e 

cos

But the electric dipole moment p = 2_qa_ and we get,

V p r

= 

 



 

1 4 0

2π θ

e

cos

Now we can write p cosθ = _p r_× , where _r_ is the unit vector from the point O to point P. Hence the electric potential at a point P due to an electric dipole is given by

V p r r

= ⋅1

4 2πe

  (r>>a) (1.38)

Equation (1.38) is valid for distances very large compared to the size of the dipole. But for a point dipole, the equation (1.38) is valid for any distance.

Special cases Case (i) If the point P lies on the axial line of the dipole on the side of +q _, then θ = 0. Then the electric potential becomes

V p r

= 1

4 2πe 

(1.39)

Case (ii) If the point P lies on the axial line of the dipole on the side of –q _, then θ = 180o. Then

V p r

=− 1

4 2πe 

(1.40)

Case (iii) If the point P lies on the equatorial line of the dipole, then θ = 90o. Hence

V = 0 (1.41)

The potential due to an electric dipole falls as and the potential due to a single point

charge falls as . Thus the potential due to the dipole falls faster than that due to a monopole (point charge). As the distance increases from electric dipole, the effects of positive and negative charges nullify each other

Note

Equi-potential Surface

Consider a point charge q _ located at some point in space and an imaginary sphere of radius r is chosen by keeping the charge q _ at its centre (Figure 1.26(a)). The electric potential at all points on the surface of the given sphere is the same. Such a surface is called an equipotential surface.

An equipotential surface is a surface on which all the points are at the same electric potential. For a point charge the equipotential surfaces are concentric spherical surfaces as shown in Figure 1.26(b). Each spherical surface is an equipotential surface but the

q _

E

(b)

 E

EE

A B

C

D

E F

G

H

I

J

q _

All points in the surface of sphere are at same potential

r

(a)

Equipotential surfaces

E E

value of the potential is different for different spherical surfaces.

For a uniform electric field, the equipotential surfaces form a set of planes normal to the electric field



E . This is shown in the Figure 1.27.

Properties of equipotential surfaces (i) The work done to move a charge q between any two points A and B, W = q _ (VB – VA). If the points A and B lie on the same equipotential surface, work done is zero because VA = VB. (ii) The electric field is normal to an equipotential surface. If it is not normal, then there is a component of the field parallel to the surface. Then work must be done to move a charge between two points on the same surface. This is a contradiction. Therefore the electric field must always be normal to equipotential surface.

Relation between electric field and potential

Consider a positive charge q _ kept fixed at the origin. To move a unit positive charge by a small distance dx towards q _ in the electric field E, the work done is given by dW = −_E_ dx. The minus sign implies that work is done against the electric field. This work done is equal to electric potential difference. Therefore,

dW = dV. (or) dV = −_E dx_ (1.42)

Hence E dV dx

=− (1.43)

The electric field is the negative gradient of the electric potential. In vector form,



E V x

i V y

j V z

k=− ∂ ∂

+ ∂ ∂

+ ∂ ∂



 



 

 

 (1.44)

EXAMPLE 1.14

The following figure represents the electric potential as a function of x – coordinate. Plot the corresponding electric field as a function of x.

0 1 2 4 5 x(cm)

V(volts)

5

10

15

20

25

30

3

Solution

In the given problem, since the potential

depends only on x, we can use 

E dV dx

i=− 

(the other two terms ¶ ¶

¶ ¶

V y

V z

and are zero)

From 0 to 1 cm, the slope is constant and

so dV dx = −25 1Vcm . So



E i=− −25 1Vcm 

From 1 to 4 cm, the potential is constant,

V = 25 V. It implies that dV dx

= 0. So 

E = 0

From 4 to 5 cm, the slope dV dx =− −25 1Vcm .

So 

E i=+ −25 1Vcm  .

The plot of electric field for the various points along the x axis is given below.

0 1 2 4 5 x(cm)

-25

-12.5

12.5

25

37.5

3

-37.5

E(Vcm-1)

Electrostatic potential energy for collection of point charges

The electric potential at a point at a distance r from point charge q _1 is given by

V q r

= 1

4 1

πe 

This potential V is the work done to bring a unit positive charge from infinity to the point. Now if the charge q _2 is brought from infinity to that point at a distance r from q _1, the work done is the product of q _2 and the electric potential at that point. Thus we have

W q V= 2

This work done is stored as the electrostatic potential energy U of a system of charges q _1 and q _2 separated by a distance r. Thus we have

U q V q q

r = =2

1 21 4πe



(1.45)

The electrostatic potential energy depends only on the distance between the two point charges. In fact, the expression (1.45) is derived by assuming that q _1 is fixed and q _2 is brought from infinity. The equation (1.45) holds true when q _2 is fixed and q _1 is brought from infinity or both q _1 and q _2 are simultaneously brought from infinity to a distance r between them.

Three charges are arranged in the following configuration as shown in Figure 1.28.

To calculate the total electrostatic potential energy, we use the following procedure. We bring all the charges one by one and arrange them according to the configuration as shown in Figure 1.28.

|——|——|

|——| | 12 3 45 |

r23 r13

r12

q3

q1 q2

A B

C

(i) Bringing a charge q _1 from infinity to the point A requires no work, because there are no other charges already present in the vicinity of charge q _1.

(ii) To bring the second charge q _2 to the point B, work must be done against the electric field created by the charge q _1. So the work done on the charge q _2 is W = q _2 _V_1B. Here _V_1B is the electrostatic potential due to the charge q _1 at point B.

U q q rI =

1 4

1 2

12πe 

(1.46)

Note that the expression is same when q _2 is brought first and then q _1 later.

(iii) Similarly to bring the charge q3 to the point C, work has to be done against the total electric field due to both charges q _1 and q _2. So the work done to bring the charge q _3 is = q _3 (_V_1C + _V_2C). Here _V_1C is the electrostatic potential due to charge q _1 at point C and _V_2C is the electrostatic potential due to charge q _2 at point C.

The electrostatic potential energy is

U q q r

q q rII = +



 



 

1 4

1 3

13

2 3

23πe 

(1.47)

(iv) Adding equations (1.46) and (1.47), the total electrostatic potential energy for the system of three charges q _1, q _2 and q _3 is U = UI + UII

U q q r

q q r

q q r

= + + 

 



 

1 4

1 2

12

1 3

13

2 3

23πe 

(1.48)

Note that this stored potential energy U is equal to the total external work done to assemble the three charges at the given locations. The expression (1.48) is same if the charges are brought to their positions in any other order. Since the Coulomb force is a conservative force, the electrostatic potential energy is independent of the manner in which the configuration of charges is arrived at.

EXAMPLE 1.15

Four charges are arranged at the corners of the square PQRS of side a as shown in the figure.(a) Find the work required to assemble these charges in the given configuration. (b) Suppose a charge q _′ is brought to the centre of the square, by keeping the four charges fixed at the corners, how much extra work is required for this?

+q

+q

-q

-q

RS

P Q

Oa

a

a

a q́

+

+

–

–

Solution

(a) The work done to arrange the charges in the corners of the square is independent

q3
rrC q1323

| q | | 2 | | AB1 | | r12 |

of the way they are arranged. We can follow any order.

(i) First, the charge +q _ is brought to the corner P. This requires no work since no charge is already present, WP = 0

(ii) Work required to bring the charge –q _ to the corner Q = (–q _) × potential at a point Q due to +q _ located at a point P.

W q q a

q aQ =− × =−

1 4

1 4

2

π πe e  

(iii) Work required to bring the charge +q _ to the corner R = q _ × potential at the point R due to charges at the point P and Q.

W q q a

q a

q a

R = × − + 

 



 

= − + 

 



 

1 4 2

1 4

1 1 2

2

π

π

e

e





(iv) Work required to bring the fourth charge –q _ at the position S = q _ × potential at the point S due the all the three charges at the point P, Q and R

W q q a

q a

q aS =− × + −



 



 

1 4 2πe



W q aS =− − 

 



 

1 4

2 1 2

2

πe 

(b) Work required to bring the charge q _′ to the centre of the square = q _′ × potential at the centre point O due to all the four charges in the four corners

The potential created by the two +q _ charges are canceled by the potential created by the –q _ charges which are located in the opposite corners. Therefore the net electric potential at the centre O due to all the charges in the corners is zero.

Hence no work is required to bring any charge to the point O. Physically this implies that if any charge q _′ when brought close to O, then it moves to the point O without any external force.

Electrostatic potential energy of a dipole in a uniform electric field

Consider a dipole placed in the uniform electric field



E as shown in the Figure 1.29. A dipole experiences a torque when kept in an uniform electric field



E . This torque rotates the dipole to align it with the direction of the electric field. To rotate the dipole (at constant angular velocity) from its initial angle θ′ to another angle θ against the torque exerted by the electric field, an equal and opposite external torque must be applied on the dipole.

E B_+q_

-q

2a sinθ

2a cosθ

a

a

A

θ

θ

O

+

→

–

Figure 1.29 The dipole in a uniform electric field

The work done by the external torque to rotate the dipole from angle θ′ to θ at constant angular velocity is

W dext= ′ ∫ τ θ θ

θ

(1.49)

Since τ_ext_ is equal and opposite to  



τ_E p E_= × , we have

+
a
θ
a O

| θ |

  



τ τ_ext E p E_= = × (1.50)

Substituting equation (1.50) in equation (1.49), we get

W pE d= ′ ∫ sinθ θ θ

θ

W pE= ′−( )cos cosθ θ

This work done is equal to the potential energy difference between the angular positions θ and θ′.

_U U U pE pE_θ θ θ θ( )− ′( )= =− + ′∆ cos cos

If the initial angle is ′ =θ 90 and is taken as reference point, then U pE′( )= =θ cos90 0 .

The potential energy stored in the system of dipole kept in the uniform electric field is given by

U pE p E=− =− ⋅cosθ 



(1.51)

In addition to p and E, the potential energy also depends on the orientation θ of the electric dipole with respect to the external electric field.

The potential energy is maximum when the dipole is aligned anti-parallel (θ = π) to the external electric field and minimum when the dipole is aligned parallel (θ = 0) to the external electric field.

EXAMPLE 1.16

A water molecule has an electric dipole moment of 6.3 × 10–30 Cm. A sample contains 1022 water molecules, with all the dipole moments aligned parallel to the external electric field of magnitude 3 × 105 N C–1. How much work is required to rotate all the water molecules from θ = 0o to 90o?

Solution

When the water molecules are aligned in the direction of the electric field, it has minimum potential energy. The work done to rotate the dipole from θ = 0o to 90o is equal to the potential energy difference between these two configurations.

W U U U= = ( )− ( )∆ 90 0 

From the equation (1.51), we write U = − pE cos_θ_, Next we calculate the work done to rotate one water molecule from θ = 0o to 90o.

For one water molecule

W pE pE pE=− + =cos cos90 0 

W = × × × = ×− −6 3 10 3 10 18 9 1030 5 25. . J

For 1022 water molecules, the total work done is

Wtot = × × = ×− −18 9 10 10 18 9 1025 22 3. . J

GAUSS LAW AND ITS APPLICATIONS

Electric Flux

The number of electric field lines crossing a given area kept normal to the electric field lines is called electric flux. It is usually denoted by the Greek letter Φ_E_ and its unit is N m2 C–1. Electric flux is a scalar quantity and it can be positive or negative. For a simpler understanding of electric flux, the following Figure 1.30 is useful.

SolutionWhen t he wa ter m olecules a re a ligned in t he dir ection o f t he e lectric f ield, i t h as minimum p otential en erg y. Th e w ork done to rotate the dipole from θ = 0 to 90is equal to the potential energ y difference between these two configurations.o o() ()WU==∆ UU90 − 0From t he e quation (1.51), w e w rite U = − pE cosθ, Next w  e calculate the work done t o r otate o ne wa ter m olecule f rom θ = 0 to 90 .For one water moleculeWpo=− Epocosc90 +=Epos 0 EW =×63.. 10 ××31 01 =×89 10 JFor 10 wa ter m  olecules, t he t otal w ork done is−−30 52 5W =×1822 ..91 01 ×=01 89 ×10 J−−25 22 3tot

The electric field of a point charge is drawn in this figure. Consider two small rectangular area elements placed normal to the field at regions A and B. Even though these elements have the same area, the number of electric field lines crossing the element in region A is more than that crossing the element in region B. Therfore the electric flux in region A is more than that in region B. Since electric field strength for a point charge decreases as the distance increases, electric flux also decreases as the distance increases. The above discussion gives a qualitative idea of electric flux. However a precise definition of electric flux is needed.

Electric flux for uniform Electric field Consider a uniform electric field in a

region of space. Let us choose an area A normal to the electric field lines as shown in Figure 1.31 (a). The electric flux for this case is

Φ_E EA_= (1.52)

A

B

+q +

Figure 1.30 Electric flux

Suppose the same area A is kept parallel to the uniform electric field, then no electric field lines pass through the area A , as shown in Figure 1.31(b). The electric flux for this case is zero.

Φ_E_ = 0 (1.53)

If the area is inclined at an angle θ with the field, then the component of the electric field perpendicular to the area alone contributes to the electric flux. The electric field component parallel to the surface area will not contribute to the electric flux. This is shown in Figure 1.31 (c). For this case, the electric flux

Φ_E_ = (E cos_θ_) A (1.54)

Further, θ is also the angle between the electric field and the direction normal to the area. Hence in general, for uniform electric field, the electric flux is defined as

Φ_E E A EA_= ⋅ =  

cosθ (1.55)

Here, note that 

A is the area vector 

A An= . Its magnitude is simply the area A and its direction is along the unit vector n_ perpendicular to the area as shown in Figure 1.31. Using this definition for flux, Φ_E E A= ⋅

 

, equations (1.53) and (1.54) can be obtained as special cases.

In Figure 1.31 (a), θ = 0o. Therefore,

Φ_E E A EA_= ⋅ =  

In Figure 1.31 (b), θ = 90o. Therefore,

Φ_E E A_= ⋅ =  

0

B

| A |

EXAMPLE 1.17

Calculate the electric flux through the rectangle of sides 5 cm and 10 cm kept in the region of a uniform electric field 100 NC–1. The angle θ is 60o. If θ becomes zero, what is the electric flux?

Area A

 E→

Solution

The electric flux through the rectangular area

Φ_E E A EA_= ⋅ =

= × × × ×−

 



cos cos

θ 100 5 10 10 604

Φ_E_ = −0 25. Nm C2 1

For θ = 0o,

Φ_E E A EA_= ⋅ =

= × × ×

=

−

−

 

100 5 10 10 0 5

4

. Nm C2 1

Figure 1.31 The electric flux for Uniform elec

(b)Electric u

Here A = A

(a)Electric ux = EA A

E→

E→ →

Electric flux through an arbitrary area kept in a non uniform electric field

Suppose the electric field is not uniform and the area A is not flat surface (Figure 1.32). Then the entire area can be divided into n small area segments ∆ ∆ ∆ ∆

   

_A A A An_1 2 3, , ……… such that each area element is almost flat and the electric field over such area element can be considered uniform.

The electric flux for the entire area A is approximately written as

tric field

(c)Electric ux = (_E cos_x = 0



A

A

E→

E→ 

_E sin_

_E cos_

 E E



i

i

Figure 1.32 Electric flux for non- uniform electric Field



| os | | E c | | n |

Area A

|  |

EXAMPLE 1 . 1 7Calculate t he e lectric f lux t hrough t he rectangle of sides 5 cm and 10 cm kept in t he r egion o f a unif orm e lectric f ield 100  NC . Th e a ngle θ i s 60 . I f θ b ecomes zero, what is the electric flux?–1 oArea A→ ESolutionThe electric flux through the rectangular area Φ =⋅EA = EA cos θ=×100 51××01 06 ×cos 0Φ E = 02.N5 mCFor θ = 0 −4 21 −E  Φ =⋅EA = EAo,=×100 51××01 0= 05.N mCE−421 −

Φ ∆ ∆ ∆ ∆

∆

E n n

i i i

n

E A E A E A E A

E A

= ⋅ + ⋅ + ⋅ ⋅

= ⋅ = ∑

       

 

1 1 2 2 3 3

1

…….

(1.56)

By taking the limit ∆ 

Ai → 0 (for all i) the summation in equation (1.56) becomes integration. The total electric flux for the entire area is given by

Φ_E E dA_= ⋅∫  

(1.57)

From Equation (1.57), it is clear that the electric flux for a given surface depends on both the electric field pattern on the surface area and orientation of the surface with respect to the electric field.

Electric flux for closed surfaces

In the previous section, the electric flux for any arbitrary curved surface is discussed. Suppose a closed surface is present in the region of the non-uniform electric field as shown in Figure 1.33 (a).

  

 

 

dA 

(a)

(b)

  

 

 

dA 



Figure 1.33 Electric flux over a closed surface

The total electric flux over this closed surface is written as

Φ_E E dA_= ⋅∫    (1.58)

Note the difference between equations (1.57) and (1.58). The integration in equation (1.58) is a closed surface integration and for each areal element, the outward normal is the direction of dA



as shown in the Figure 1.33(b).

The total electric flux over a closed surface can be negative, positive or zero. In the Figure 1.33(b), it is shown that in one area element, the angle between dA



and 

E is less than 90o, then the electric flux is positive and in another areal element, the angle between dA



and 

E is greater than 90o, then the electric flux is negative.

In general, the electric flux is negative if the electric field lines enter the closed surface and positive if the electric field lines leave the closed surface.

Gauss law

A positive point charge q _ is surrounded by an imaginary sphere of radius r as shown in Figure 1.34. We can calculate the total electric flux through the closed surface of the sphere using the equation (1.58).

Φ_E E dA EdA_= ⋅ =∫ ∫     cosθ

The electric field of the point charge is directed radially outward at all points on the surface of the sphere. Therefore, the direction of the area element dA



is along the electric field



E and θ= °0 .

Φ_E EdA_= ∫ since cos0 10 = (1.59)

|  |

E is uniform on the surface of the sphere,

Φ_E E dA_= ∫ (1.60)

Substituting for dA r=∫ 4 2π 

and

E Q r

= 1

4 0 2πe

in equation (1.60), we get

Φ_E Q r_

r Q= × = 1

4 4 4 1

42 2

π π π

πe e  

Φ_E Q_

= e 

(1.61)

The equation (1.61) is called as Gauss’s law.

The remarkable point about this result is that the equation (1.61) is equally true for any arbitrary shaped surface which encloses the charge q _ and as shown in the Figure 1.35. It is seen that the total electric flux is the same for closed surfaces A1, A2 and A3 as shown in the Figure 1.35.

Spherical gaussian surface

r

q _

 

 d

When the charge is at the center of the sphere, the electric eld is

everywhere normal to the surface and constant in magnitude.

+

Figure 1.34 Total electric flux of point charge

A1

A2

A3

e net electric ux is the same through all surfaces.

++

Gauss’s law states that if a charge q _ is enclosed by an arbitrary closed surface, then the total electric flux Φ_E_ through the closed surface is

Φ_E enclE dA_

q _ = ⋅ =∫

 

  e

(1.62)

where Qencl denotes the charges within the closed surface.

Discussion of Gauss law (i) The total electric flux through the closed surface depends only on the charges enclosed by the surface and the charges present outside the surface will not contribute to the flux and the shape of the closed surface which can be chosen arbitrarily.

(ii) The total electric flux is independent of the location of the charges inside the closed surface.

|——|——|

(iii) To arrive at equation (1.62), we have chosen a spherical surface. This imaginary surface is called a Gaussian surface. The shape of the Gaussian surface to be chosen depends on the type of charge configuration and the kind of symmetry existing in that charge configuration. The electric field is spherically symmetric for a point charge, therefore spherical Gaussian surface is chosen. Cylindrical and planar Gaussian surfaces can be chosen for other kinds of charge configurations.

(iv) In the LHS of equation (1.62), the electric field



E is due to charges present inside and outside the Gaussian surface but the charge Qencl denotes the charges which lie only inside the Gaussian surface.

EXAMPLE 1.18

-10q

(a)

(b)

+

+

–

–

A1

A2 -q

+q

+Q

-2q

+5q +

–

(i) In figure (a), calculate the electric flux through the closed areas A1 and A2.

(ii) In figure (b), calculate the electric flux through the cube

Solution (i) In figure (a), the area A1 encloses the

charge q _. So electric flux through this

closed surface A1 is q _ e 

. But the closed

surface A2 contains no charges inside, so electric flux through A2 is zero.

(ii) In figure (b), the net charge inside the cube is 3q _ and the total electric flux in the

cube is therefore Φ_E q_

= 3 e 

. Note that the

charge -10 q _ lies outside the cube and it will not contribute the total flux through the surface of the cube.

Applications of Gauss law

Electric field due to any arbitrary charge configuration can be calculated using Coulomb’s law or Gauss law. If the charge configuration possesses some kind of symmetry, then Gauss law is a very efficient way to calculate the electric field. It is illustrated in the following cases. (i) Electric field due to an infinitely long

charged wire Consider an infinitely long straight

wire having uniform linear charge density λ(charge per unit length). Let P be a point located at a perpendicular distance r from the wire (Figure 1.36(a)). The electric field at the point P can be found using Gauss law.

We choose two small charge elements A1 and A2 on the wire which are at equal distances from the point P. The resultant electric field due to these two charge elements points radially away from the charged wire and the magnitude of electric field is same at all points on the circle of radius r. This is shown in the Figure 1.36(b). Since the charged wire possesses

|——| | ++Q |

a cylindrical symmetry, let us choose a cylindrical Gaussian surface of radius r and length L as shown in the Figure 1.37.

The total electric flux through this closed surface is calculated as follows.

Φ_E_

top surface

Curved surface

b

E dA

E dA E dA E dA

= ⋅

= ⋅ + ⋅ + ⋅

∫ ∫∫

 

      

ottom surface

∫ (1.63)

It is seen from Figure (1.37) that for the curved surface,



E is parallel to 

A and  

_E dA E dA_⋅ = . For the top and bottom surfaces,



E is perpendicular to 

A and  

_E dA_⋅ = 0 Substituting these values in the equation

(1.63) and applying Gauss law to the cylindrical surface, we have

Φ_E encl_

Curved surface

EdA Q

= =∫ e 

(1.64)

Since the magnitude of the electric field for the entire curved surface is constant, E is

Figure 1.36 Electric field due to infinite long

+ + + + + + + + + + + + + + + + + +



r P

q _E

(a)



+ +

+

L E

Top Surface

Bottom Surface

Curved Surface



E 

E 

r

+ + + + + + + + + + + + + +

taken out of the integration and Qencl is given by Q Lencl = λ , where λ is the linear charge density (charge present per unit length).

E dA L

Curved surface

=∫ λ e 

(1.65)

charged wire

+ + + + + + + + + + + + + + + + + + 

r P

A2

A1

E R

S

(b)



++++R++	A2r
Q
P
++++++	SA

++
r

| ed |

|  |

Here dA Curved surface

=∫ total area of the curved

surface = 2π_rL_. Substituting this in equation (1.65), we get

E rL L ⋅ =2π λ

e 

E r

= 1

2π λ

e 

(1.66)

In vector form,





E r

r= 1

2π λ

e  (1.67)

The electric field due to the infinite charged wire depends on 1

r rather than 1

2_r_

which is for a point charge. Equation (1.67) indicates that

the electric field is always along the perpendicular direction ( _r_) to wire. In fact, if λ > 0 then



E points perpendicularly outward ( _r_) from the wire and if λ < 0, then



E points perpendicularly inward ( )- _r_ .

The equation (1.67) is true only for an infinitely long charged wire. For a charged wire of finite length, the electric field need not be radial at all points. However, equation (1.67) for such a wire is taken approximately true around the mid-point of the wire and far away from the both ends of the wire

(ii) Electric field due to charged infinite plane sheet Consider an infinite plane sheet of

charges with uniform surface charge density σ (charge present per unit area). Let P be a point at a distance of r from the sheet as shown in the Figure 1.38.

+ + + + + + + + + + +

+ + + +

+ + + +

+ + + +

+ + + +

E P

P´

dA dA

dA E E



r

Gaussian surface

Figure 1.38 Electric field due to charged infinite planar sheet

Since the plane is infinitely large, the electric field should be same at all points equidistant from the plane and radially directed outward at all points. A cylindrical Gaussian surface of length 2_r_ and two flats surfaces each of area A is chosen such that the infinite plane sheet passes perpendicularly through the middle part of the Gaussian surface.

Total electric flux linked with the cylindrical surface,

Φ_E_

PCurved surface

encl

P

E dA

E dA E dA E dA Q

= ⋅

= ⋅ + ⋅ + ⋅ =

∫

∫∫ ′

 

      

e0 ∫

(1.68)

The electric field is perpendicular to the area element at all points on the curved surface and is parallel to the surface areas at P and ′P (Figure 1.38). Then, applying Gauss’ law,

Φ_E encl_

PP

EdA EdA Q

= + = ′ ∫∫ e0

(1.69)

Since the magnitude of the electric field at these two equal flat surfaces is uniform, E is taken out of the integration and Qencl is given by Q Aencl = σ , we get

++++ ++++ ++++++++++ ++++ +++++	GaussiansurfacdA
EP

| r |

2 0

E dA A

P

=∫ σ e

The total area of surface either at P or P′

dA A P

=∫

Hence 2 20 0

EA A E= = σ σ e e

or (1.70)

In vector form, 



E n= σ

2e  (1.71)

Here _n_ is the outward unit vector normal to the plane. Note that the electric field due to an infinite plane sheet of charge depends on the surface charge density and is independent of the distance r.

The electric field will be the same at any point farther away from the charged plane. Equation (1.71) implies that if σ > 0 the electric field at any point P is along outward perpendicular _n_ drawn to the plane and if σ < 0, the electric field points inward perpendicularly to the plane (-_n_ ).

For a finite charged plane sheet, equation (1.71) is approximately true only in the middle region of the plane and at points far away from both ends.

(iii) Electric field due to two parallel charged infinite sheets

Consider two infinitely large charged plane sheets with equal and opposite charge

– – – – – – – – –

– – – – – – – – –

– – – – – – – – –

+ + + + + + + + + +

+ + + + + + + + + +

+ + + + + + + + + +

+σ –σ

E+ P3

P1 P2

E– E+ E+ E–

E–

Figure 1.39 Electric field due to two parallel charged sheets

densities +σ and -σ which are placed parallel to each other as shown in the Figure 1.39.

The electric field between the plates and outside the plates is found using Gauss law. The magnitude of the electric field due to

an infinite charged plane sheet is σ 2e



and it

points perpendicularly outward if σ > 0 and points inward if σ < 0.

At the points P2 and P3, the electric field due to both plates are equal in magnitude and opposite in direction (Figure 1.41). As a result, electric field at a point outside the plates is zero. But between the plates, electric fields are in the same direction i.e., towards the right and the total electric field at a point P1 is

Einside = + = σ σ σ

2 2e e e   

(1.72)

The direction of the electric field between the plates is directed from positively charged plate to negatively charged plate and is uniform everywhere between the plates.

(iv) Electric field due to a uniformly charged spherical shell

Consider a uniformly charged spherical shell of radius R carrying total charge q _ as shown in Figure 1.40. The electric field at points outside and inside the sphere can be found using Gauss law.

Case (a) At a point outside the shell (r > R) Let us choose a point P outside the shell

at a distance r from the centre as shown in Figure 1.40 (a). The charge is uniformly distributed on the surface of the sphere (spherical symmetry). Hence the electric field must point radially outward if q _ > 0 and point radially inward if q _ < 0. So a spherical Gaussian surface of radius r is chosen and

the total charge enclosed by this Gaussian surface is q _. Applying Gauss law

  



E dA Q

Gaussian surface

∫ ⋅ = e

(1.73)

The electric field 

E and d A 

point in the same direction (outward normal) at all the points on the Gaussian surface. The magnitude of



E is also the same at all points due to the spherical symmetry of the charge distribution.

Hence E dA Q

Gaussian surface

 

∫ = e

(1.74)

But dA Gaussian surface

∫ = total area of Gaussian surface

= 4πr2. Substituting this value in equation (1.74)

E r Q ⋅ =4 2π

e 

E r Q ⋅ =4 2π

e 

(or) E Q r

= 1

4 2πe 

P

For points outside the spherical shell, a large, spherical gaussian surface is drawn concentric with the spherical shell.

Gaussian sphere

r

R

q _

(a)

Figure 1.40 The electric field due to a charge

In vector form,





E Q r

r= 1

4 2πe  (1.75)

The electric field is radially outward if q _ > 0 and radially inward if q _ < 0. From equation (1.75), we infer that the electric field at a point outside the shell will be the same as if the entire charge q _ is concentrated at the centre of the spherical shell. (A similar result is observed in gravitation, for gravitational force due to a spherical shell with mass M)

Case (b): At a point on the surface of the spherical shell (r = R)

The electrical field at points on the spherical shell (r = R) is given by





E Q R

r= 4 2πe

 (1.76)

Case (c): At a point inside the spherical shell (r < R)

Consider a point P inside the shell at a distance r from the centre. A Gaussian

Gaussian sphere_r_

R

q _

P

For points inside the spherical shell, a spherical gaussian surface smaller than the spherical shell is drawn.

(b)

d spherical shell

sphere of radius r is constructed as shown in the Figure 1.40 (b). Applying Gauss law

  



E dA Q

Gaussian surface

⋅ =∫ e

E r Q ⋅ =4 2π

e 

(1.77)

Since Gaussian surface encloses no charge, q _ = 0. The equation (1.77) becomes

E = 0 (r < R) (1.78)

The electric field due to the uniformly charged spherical shell is zero at all points inside the shell.

A graph is plotted between the electric field and radial distance. This is shown in Figure 1.41.

R

RO Er = 0

Er = 1 4e

Q r2

Er

r +

Figure 1.41 Electric field versus distance for a spherical shell of radius R

Gauss law is a powerful technique whenever a given charge configuration possesses

spherical, cylindrical or planar symmetry, then the electric field due to such a charge configuration can be easily found. If there is no such symmetry, the direct method (Coulomb’s law and calculus) can be used. For example, it is difficult to use Gauss law to find the electric field for a dipole since it has no spherical, cylindrical or planar symmetry.

Note

ELECTROSTATICS OF CONDUCTORS AND DIELECTRICS

Conductors at electrostatic equilibrium

An electrical conductor has a large number of mobile charges which are free to move in the material. In a metallic conductor, these mobile charges are free electrons which are not bound to any atom and therefore are free to move on the surface of the conductor. When there is no external electric field, the free electrons are in continuous random motion in all directions. As a result, there is no net motion of electrons along any particular direction which implies that the conductor is in electrostatic equilibrium. Thus at electrostatic equilibrium, there is no net current in the conductor. A conductor at electrostatic equilibrium has the following properties.

(i) The electric field is zero everywhere inside the conductor. This is true regardless of whether the conductor is solid or hollow.

This is an experimental fact. Suppose the electric field is not zero inside the metal, then there will be a force on the mobile charge carriers due to this electric field. As a result, there will be a net motion of the mobile charges, which contradicts the conductors being in electrostatic equilibrium. Thus the electric field is zero everywhere inside the conductor. We can also understand this fact by applying an external uniform electric field on the conductor. This is shown in Figure 1.42.

Eint

E 



E 

E 

-——–

+ + + + + + + + +

Figure 1.42 Electric field of conductors

Before applying the external electric field, the free electrons in the conductor are uniformly distributed in the conductor. When an electric field is applied, the free electrons accelerate to the left causing the left plate to be negatively charged and the right plate to be positively charged as shown in Figure 1.44.

Due to this realignment of free electrons, there will be an internal electric field created inside the conductor which increases until it nullifies the external electric field. Once the external electric field is nullified the conductor is said to be in electrostatic equilibrium. The time taken by a conductor to reach electrostatic equilibrium is in the order of 10–16s, which can be taken as almost instantaneous. (ii) There is no net charge inside the conductors. The charges must reside only on the surface of the conductors.

We can prove this property using Gauss law. Consider an arbitrarily shaped conductor as shown in Figure 1.43.

A Gaussian surface is drawn inside the conductor such that it is very close to the surface of the conductor. Since the electric field is zero everywhere inside the conductor, the net electric flux is also zero over this Gaussian surface. From Gauss’s law, this implies that there is no net charge inside the conductor. Even if some charge is introduced

inside the conductor, it immediately reaches the surface of the conductor.

(iii) The electric field outside the conductor is perpendicular to the surface of the

conductor and has a magnitude of σ e 

where

σ is the surface charge density at that point. If the electric field has components

parallel to the surface of the conductor, then free electrons on the surface of the conductor would experience acceleration (Figure 1.44(a)). This means that the

Gaussian surface

Qnet=0

- -

- - -

E = 0

E 

E 



E 

Conductor

Conductor





Figure 1.44 (a) Electric field is along the surface (b)Electric field is perpendicular to the surface of the conductor

E	———++++ +E+ +E +int +	E

|——|

conductor is not in equilibrium. Therefore at electrostatic equilibrium, the electric field must be perpendicular to the surface of the conductor. This is shown in Figure 1.44 (b).

We now prove that the electric field has

magnitude σ e 

just outside the conductor’s

surface. Consider a small cylindrical Gaussian surface, as shown in the Figure 1.45. One half of this cylinder is embedded inside the conductor.

E 

    

 

 

 

    





e ux through the gaussian surface is EA.

A

Figure 1.45 The electric field on the surface of the conductor

Since electric field is normal to the surface of the conductor, the curved part of the cylinder has zero electric flux. Also inside the conductor, the electric field is zero. Hence the bottom flat part of the Gaussian surface has no electric flux.

Therefore the top flat surface alone contributes to the electric flux. The electric field is parallel to the area vector and the total charge inside the surface is σA. By applying Gaus’s law,

EA A =

σ e 

In vector form,





E n= σ e  (1.79)

where _n_ represents the unit vector outward normal to the surface of the conductor. Suppose σ < 0, then electric field points inward perpendicular to the surface. (iv) The electrostatic potential has the same value on the surface and inside of the conductor.

We know that the conductor has no parallel electric component on the surface which means that charges can be moved on the surface without doing any work. This is possible only if the electrostatic potential is constant at all points on the surface and there is no potential difference between any two points on the surface.

Since the electric field is zero inside the conductor, the potential is the same as the surface of the conductor. Thus at electrostatic equilibrium, the conductor is always at equipotential.

Electrostatic shielding

Using Gauss law, we can prove that the electric field inside the charged spherical shell is zero, Further, we can show that the electric field inside both hollow and solid conductors is zero. It is a very interesting property which has an important consequence.

Consider a cavity inside the conductor as shown in Figure 1.46 (a). Whatever be the charges at the surfaces and whatever be the electrical disturbances outside, the electric field inside the cavity is zero. A sensitive electrical instrument which is to be protected from external electrical disturbance can be kept inside this cavity. This is called electrostatic shielding.




Faraday cage is an instrument used to demonstrate this effect. It is made up of metal bars as shown in Figure1.46 (b). If an artificial lightning jolt is created outside, the person inside is not affected.

During lightning accompanied by a thunderstorm, it is always safer to sit inside a bus than in open ground or under a tree. The metal body of the bus provides electrostatic shielding, since the electric field inside is zero. During lightning, the charges flow through the body of the conductor to the ground with no effect on the person inside that bus.

Electrostatic induction

In section 1.1, we have learnt that an object can be charged by rubbing using an appropriate material. Whenever a charged

Electric eld line

E=0

Figure 1.46 (a) Electric field inside the cavity (b) Faraday cage

rod is touched by another conductor, charges start to flow from charged rod to the conductor. Is it possible to charge a conductor without any contact? The answer is yes. This type of charging without actual contact is called electrostatic induction.

(i) Consider an uncharged (neutral) conducting sphere at rest on an insulating stand. Suppose a negatively charged rod is brought near the conductor without touching it, as shown in Figure 1.47(a).

The negative charge of the rod repels the electrons in the conductor to the opposite side. As a result, positive charges are induced near the region of the charged rod while negative charges on the farther side.

Before introducing the charged rod, the free electrons were distributed uniformly on the surface of the conductor and the net charge is zero. Once the charged rod is brought near the conductor, the distribution is no longer uniform with more electrons located on the farther side of the rod and positive charges are located closer to the rod. But the total charge is zero.

- - - - -

- - - - -

(c)

(a)

(d)

+ + + + +

+ +

(b)

- - - - -

+ + + +

– – – –

- - - - -

+ + + +

– – – –

(ii) Now the conducting sphere is connected to the ground through a conducting wire. This is called grounding. Since the ground can always receive any amount of electrons, grounding removes the electron from the conducting sphere. Note that positive charges will not flow to the ground because they are attracted by the negative charges of the rod (Figure 1.47(b)).

(iii) When the grounding wire is removed from the conductor, the positive charges remain near the charged rod (Figure 1.47(c))

(iv) Now the charged rod is taken away from the conductor. As soon as the charged rod is removed, the positive charge gets distributed uniformly on the surface of the conductor (Figure 1.47 (d)). By this process, the neutral conducting sphere becomes positively charged.

For an arbitrary shaped conductor, the intermediate steps and conclusion are the same except the final step. The distribution of positive charges is not uniform for arbitrarily-shaped conductors. Why is it not uniform? The reason for it is discussed in the section 1.9

EXAMPLE 1.19

A small ball of conducting material having a charge +q _ and mass m is thrown upward at an angle θ to horizontal surface with an initial speed _v_o as shown in the figure. There exists an uniform electric field E downward along with the gravitational field g. Calculate the range, maximum height and time of flight in the motion of this charged ball. Neglect the effect of air and treat the ball as a point mass.

 hmax

R x

y E

+

+ +

+

+



v0 

Solution

If the conductor has no net charge, then its motion is the same as usual projectile motion of a mass m which we studied in Kinematics (unit 2, vol-1 XI physics). Here, in this problem, in addition to downward gravitational force, the charge also will experience a downward uniform electrostatic force.

The acceleration of the charged ball due to gravity = -_g j_

The acceleration of the charged ball due

to uniform electric field = -qE m

_j_

The total acceleration of charged ball in

downward direction _a g qE_

m j=− +



 



  

It is important here to note that the acceleration depends on the mass of the object. Galileo’s conclusion that all objects fall at the same rate towards the Earth is true only in a uniform gravitational field. When a uniform electric field is included, the acceleration of a charged object depends on both mass and charge.

But still the acceleration a g qE m

= + 

 



  is

constant throughout the motion. Hence we use kinematic equations to calculate the

| v + h + +0+ max + |

range, maximum height and time of flight.

In fact we can simply replace g by g qE m

+ in

the usual expressions of range, maximum

height and time of flight of a projectile.

Without charge

With the charge +q

Time of flight T

2_v_ 

sinθ g

2_v_ 

sinθ

g qE m

+   

  

Maximum height hmax

v 

2 2

2 sin θ

g v 

2 2

2

sin θ

g qE m

+   

  

Range R v 

2 2sin θ g

v 

2 2sin θ

g qE m

+   

  

Note that the time of flight, maximum height, range are all inversely proportional to the acceleration of the object. Since

g qE m

g+ 

 



 > for charge +q _, the quantities T,

hmax, and R will decrease when compared to the motion of an object of mass m and zero net charge. Suppose the charge is –q _, then

g qE m

_g_− 

 



 < , and the quantities T, hmax and

R will increase. Interestingly the trajectory is still parabolic as shown in the figure.

θ

v0

x

y E

→ +

+

+

+

_

_

_

_

-q

-q+q q=0

Dielectrics or insulators

A dielectric is a non-conducting material and has no free electrons. The electrons in a dielectric are bound within the atoms. Ebonite, glass and mica are some examples of dielectrics. When an external electric field is applied, the electrons are not free to move anywhere but they are realigned in a specific way. A dielectric is made up of either polar molecules or non- polar molecules.

Non-polar molecules A non-polar molecule is one in which

centres of positive and negative charges coincide. As a result, it has no permanent dipole moment. Examples of non-polar molecules are hydrogen (H2), oxygen (O2), and carbon dioxide (CO2) etc.

When an external electric field is applied, the centres of positive and negative charges are separated by a small distance which induces dipole moment in the direction of the external electric field. Then the dielectric is said to be polarized by an external electric field. This is shown in Figure 1.48.

+– +–

_p_→

_E_→

Center of negative charge coincides with center of

positive charge

(a) (b)

Without charge
2v sin θg
v sin θ2 g22
v sin 2θg2

| -qv→ θ x0 + + + |

range, maximum height and time of f light. In fact we can simply replace g by qE in g +mthe u sual exp ressions o f ra nge, m aximum height and time of flight of a projectile.Without With the charge charge +qTime of ight 2v sin θ 2v sin θT g  g + qEm Maximum  v sin θheight hv si2 gn θ 2 g + qEm 22 22R ange R v sin 2θmax  v sign 2θ  g + qEm 2 2Note t hat t he t ime o f f light, m aximumheight, range are all in versely pro  portionalto the acceleration of the object. Since g + qEm  > g for charge +q, the quantities T, h , and R will decrease when compared tothe motion of an object of mass m and zeronet c harge. S uppose t he c harge i s – q, t henmag −x qEm  < g , and the quantities T, h andR will increase. Interestingly the trajectory isstill parabolic as shown in the figure.maxEy-qv→ θ x0 + + +q q=0_ -q++ _

Polar molecules In polar molecules, the centres of the

positive and negative charges are separated even in the absence of an external electric field. They have a permanent dipole moment. Due to thermal motion, the direction of each dipole moment is oriented randomly (Figure 1.49(a)). Hence the net dipole moment is zero in the absence of an external electric field. Examples of polar molecules are H2O, N2O, HCl, NH3.

When an external electric field is applied, the dipoles inside the material tend to align in the direction of the electric field. Hence a net dipole moment is induced in it. Then the dielectric is said to be polarized by an external electric field (Figure 1.49(b)).

+ –

+–

+ –

+–

+ –

+ – + –

+ –

+–

+ –+

–

+ –

+–

+–

+–

+–

+–

+–

+–

+–

+–

+–

+–

+–

+ + + + + + + + + + + + +

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Eext



Eext



Eext

 Eint



(a) (b)

Polar molecules are randomly oriented in

the absence of an external electric eld.

When an external electric eld is applied, the molecules partially

align with the eld.

Polarisation In the presence of an external electric

field, the dipole moment is induced in the dielectric material. Polarisation



P is defined as the total dipole moment per unit volume of the dielectric. For most dielectrics (linear isotropic), the Polarisation is directly proportional to the strength of the external electric field. This is written as

 

P Ee ext= χ (1.80)

where χ_e_ is a constant called the electric susceptibility which is a characteristic of each dielectric.

Induced Electric field inside the dielectric

When an external electric field is applied on a conductor, the charges are aligned in such a way that an internal electric field is created which tends to cancel the external electric field. But in the case of a dielectric, which has no free electrons, the external electric field only realigns the charges so that an internal electric field is produced. The magnitude of the internal electric field is smaller than that of external electric field. Therefore the net electric field inside the dielectric is not zero but is parallel to an external electric field with magnitude less than that of the external electric field. For example, let us consider a rectangular dielectric slab placed between two oppositely charged plates (capacitor) as shown in the Figure 1.50.

The uniform electric field between the plates acts as an external electric field 

Eext which polarizes the dielectric placed between plates. The positive charges are induced on one side surface and negative charges are induced on the other side of surface.

But inside the dielectric, the net charge is zero even in a small volume. So the dielectric in the external field is equivalent to two oppositely charged sheets with the surface charge densities +σb and –σb. These charges are called bound charges. They are not free to move like free electrons in

++++++++++++	– + – + – +– + E – + E – +– + int – + ext – +– + – + – +	______________

+

conductors. This is shown in the Figure 1.50.

For example, the charged balloon after rubbing sticks onto a wall. The reason is that the negatively charged balloon is brought near the wall, it polarizes opposite charges on the surface of the wall, which attracts the balloon. This is shown in Figure 1.51.

+ _ + _ + _ + _ + _ + _ + _ + _ + _ + _ + _

_ _ _ _

_

(b)(a)

Wall

+–

+–

+–

+–

+–

+–

+–

+–

+–

+–

+–

+–

+ + + + + + + + + + + + +

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Eext



Eext



Eext

 Eint



When an external electric eld is applied, the molecules partially

align with the eld.

+σb –σb

Figure 1.50 Induced electric field lines inside the dielectric

Dielectric strength

When the external electric field applied to a dielectric is very large, it tears the atoms apart so that the bound charges become free charges. Then the dielectric starts to conduct electricity. This is called dielectric breakdown. The maximum electric field the dielectric can withstand before it breaksdown is called dielectric strength. For example, the dielectric strength of air is 3 × 106 V m–1. If the applied electric field increases beyond this, a spark is produced in the air. The dielectric strengths of some dielectrics are given in the Table 1.1.

Table 1.1 Dielectric strength

Substance	Dielectric strength (Vm–1)
Mica	100 × 10 6
Teflon	60 × 106
Paper	16 × 106
Air	3 × 106
Pyrex glass	14 × 106

CAPACITORS AND CAPACITANCE

Capacitors

Capacitor is a device used to store electric charge and electrical energy. It consists of two conducting objects (usually plates or sheets) separated by some distance. Capacitors are widely used in many electronic circuits and have applications in many areas of science and technology.

A simple capacitor consists of two parallel metal plates separated by a small distance as shown in Figure 1.52 (a).

Area = A

d

+ -+ -

–Q+Q

V

C -Q+Q

(a) (b) (c)

When a capacitor is connected to a battery of potential difference V, the electrons are transferred from one plate to the other plate by battery so that one plate becomes negatively charged with a charge of –Q and the other plate positively charged with +Q. The potential difference between the plates is equivalent to the battery’s terminal voltage. This is shown in Figure 1.52 (b). If the battery voltage is increased, the amount of charges stored in the plates also increase. In general, the charge stored in the capacitor is proportional to the potential difference between the plates.

_Q V_∝ so that q _ = CV

where the C is the proportionality constant called capacitance. The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between them.

C Q V

= (1.81)

The SI unit of capacitance is c_oulomb per volt_ or farad (F) in honor of Michael Faraday. Farad is a larger unit of capacitance. In practice, capacitors are available in the range of microfarad (1µF = 10–6 F) to picofarad (1pF = 10–12 F). A capacitor is represented by the symbol or . Note that the total charge stored in the capacitor is zero (Q – Q = 0). When we say the capacitor stores charges, it means the amount of charge that can be stored in any one of the plates.

Nowadays there are capacitors available in various shapes (cylindrical, disk) and types (tantalum, ceramic and electrolytic), as shown in Figure 1.53. These capacitors are extensively used in various kinds of electronic circuits.

Capacitance of a parallel plate capacitor Consider a capacitor with two parallel

plates each of cross-sectional area A and separated by a distance d as shown in Figure 1.54.

The electric field between two infinite parallel plates is uniform and is given by

E = σ e 

where σ is the surface charge density

on either plates σ= 

 



 

Q A

. If the separation

distance d is very much smaller than the size of the plate (d2 « A), then the above result can be used even for finite–sized parallel plate capacitor.

The electric field between the plates is

E Q A

= e 

(1.82)

Since the electric field is uniform, the electric potential difference between the plates having separation d is given by

V Ed Qd A

= = e0

(1.83)

Therefore the capacitance of the capacitor is given by

C Q V

Q Qd A

A d

= = 

 



 

=

e

e

0

0 (1.84)

From equation (1.84), it is evident that capacitance is directly proportional to the area of cross section and is inversely proportional to the distance between the plates. This can be understood from the following.

-——————-

+ + + + + + + + + + + + + + + + + + + +

+Q -Q d

E 

Area A

Area A

(i) If the area of cross-section of the capacitor plates is increased, more charges can be distributed for the same potential difference. As a result, the capacitance is increased.

(ii) If the distance d between the two plates is reduced, the potential difference between the plates (V = Ed) decreases with E constant. As a result, voltage difference between the terminals of the battery increases which in turn leads to an additional flow of charge to the plates from the battery, till the voltage on the capacitor equals to the battery’s terminal voltage. Suppose the distance is increased, the capacitor voltage increases and becomes greater than the battery voltage. Then, the charges flow from capacitor plates to battery till both voltages becomes equal.

EXAMPLE 1.20

A parallel plate capacitor has square plates of side 5 cm and separated by a distance of 1 mm. (a) Calculate the capacitance of this capacitor. (b) If a 10 V battery is connected to the capacitor, what is the charge stored in any one of the plates? (The value of eo = 8.85 × 10–12 N–1m–2 C2)

Solution

(a) The capacitance of the capacitor is

C

A d

= = × × × ×

= ×

− −

−

−

e0 12 4

3

13

8 85 10 25 10 1 10

221 2 10

.

. F C= × =−22 12 10 22 1212. .F pF

(b) The charge stored in any one of the plates is Q = CV, Then

q _= × × = × =− −22 12 10 10 221 2 10 221 212 12. . .C pC

+Q	d	-Q
++++++++++++++++++++	——————-

EXAMPLE 1 . 2 0A p ara l le l p l at e c ap aci t or h a s s qu arep l at es o f side 5 cm a n d s ep ara t e d b y adi s t an ce o f 1 mm. (a) C a lc u l at e t h ec ap aci t an ce of t hi s c ap aci t or. (b) If a 10 Vb att er y i s co nn e c t e d t o t h e c ap aci t or,w h at i s t h e c h ar g e s t ore d in a ny o n e o ft h e p l at es? (Th e va lue o f e = 8.85 × 10N m C )Solution–12o–1 –2 2(a) The capacitance of the capacitor ise A 88. 51××02 51×C ==d 11× 0=×221.F21 0C =×22..12 10 Fp= −−2212 12 F0(b) The c harge s tored in a −n3y o nplates is Q = CV−13 , Then−1210 ×=10 221 21×=0 Cp−−12 12
04
e o f t he
221.2 C
Q =×22..12

Sometimes we notice that the ceiling fan does not start rotating as soon as it is switched on. But when we rotate the blades, it starts to rotate as usual. Why it is so? We know that to rotate any object, there must be a torque applied on the object. For the ceiling fan, the initial torque is given by the capacitor widely known as a condenser. If the condenser is faulty, it will not give sufficient initial torque to rotate the blades when the fan is switched on.

Energy stored in the capacitor

Capacitor not only stores the charge but also it stores energy. When a battery is connected to the capacitor, electrons of total charge –q _ are transferred from one plate to the other plate. To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor.

To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by

dW = V dQ (1.85)

where V Q C

=

The total work done to charge a capacitor is

W Q C

dQ Q C

q _

= =∫ 0

2

2 (1.86)

This work done is stored as electrostatic potential energy (UE) in the capacitor.

U Q C

CVE = = 2

2

2 1 2

( )∴ =Q CV (1.87)

where Q = CV is used. This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor.

But where is this energy stored in the capacitor? To understand this question, the equation (1.87) is rewritten as follows using the results C

A d

= e0 and V = Ed

U A

d Ed Ad EE =



 



 ( ) = ( )1

2 1 2

0 2 0

2e e (1.88)

where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined as energy

density u U VolumeE = . From equation (1.88),

we get

u EE = 1 2

2e 

(1.89)

From equation (1.89), we infer that the energy is stored in the electric field existing between the plates of the capacitor. Once the capacitor is allowed to discharge, the energy is retrieved.

It is important to note that the energy density depends only on the electric field and not on the size of the plates of the capacitor. In fact, expression (1.89) is true for the electric field due to any type of charge configuration.

Applications of capacitors

Capacitors are used in various electronics circuits. A few of the applications. (a) Flash capacitors are used in digital cameras for taking photographs. The flash which comes from the camera when we take photographs is due to the energy released from the capacitor, called a flash capacitor (Figure 1.55 (a)) (b) During cardiac arrest, a device called heart defibrillator is used to give a sudden surge of a large amount of electrical energy to the patient’s chest to retrieve the normal heart function. This is shown in Figure 1.55 (b).

(c) Capacitors are used in the ignition system of automobile engines to eliminate sparking

(d) Capacitors are used to reduce power fluctuations in power supplies and to increase the efficiency of power transmission.

However, capacitors have disadvantage as well. Even after the battery or power supply is removed, the capacitor stores charges and energy for some time. For example if the TV is switched off, it is always advisable to not touch the back side of the TV panel.

(a)

Effect of dielectrics in capacitors

In earlier discussions, we assumed that the space between the parallel plates of a capacitor is either empty or filled with air. Suppose dielectrics like mica, glass or paper are introduced between the plates, then the capacitance of the capacitor is altered. The dielectric can be inserted into the plates in two different ways. (i) when the capacitor is disconnected from the battery. (ii) when the capacitor is connected to the battery.

(i) when the capacitor is disconnected from the battery

Consider a capacitor with two parallel plates each of cross-sectional area A and are separated by a distance d. The capacitor is charged by a battery of voltage _V_0 and the charge stored is q _0. The capacitance of the capacitor without the dielectric is

_C Q V_0

0

0

= (1.90)

The battery is then disconnected from the capacitor and the dielectric is inserted between the plates. This is shown in Figure 1.56.

(b)

Heart defibrillator

++++++++++

d

E →

E0 →

– – – – – – – – – –

+ -

+++++

– – – – –

εr_d_

(a)

(b)

The introduction of dielectric between the plates will decrease the electric field. Experimentally it is found that the modified electric field is given by

E E

r

= 0

e (1.91)

where E_o is the electric field inside the capacitors when there is no dielectric and e_r is the relative permittivity of the dielectric or simply known as the dielectric constant. Since er > 1, the electric field E < _E_o.

As a result, the electrostatic potential difference between the plates (V = Ed) is also reduced. But at the same time, the charge q _o will remain constant once the battery is disconnected.

Hence the new potential difference is

V Ed E

d V

r r

= = =0 0

e e (1.92)

We know that capacitance is inversely proportional to the potential difference. Therefore as V decreases, C increases.

Thus new capacitance in the presence of a dielectric is

C Q V

Q V

Cr r= = =0 0

0 0e e (1.93)

Since e_r_ > 1, we have _C > C_o. Thus insertion of the dielectric increases the capacitance.

Using equation (1.84),

C A

d A d

r= = e e e0 (1.94)

where e = e_r_ e_o_ is the permittivity of the dielectric medium.

The energy stored in the capacitor before the insertion of a dielectric is given by

_U Q C_0

0 2

0

1 2

= (1.95)

After the dielectric is inserted, the charge q _0 remains constant but the capacitance is increased. As a result, the stored energy is decreased.

U Q C

Q C

U

r r

= = = 1 2

1 2

0 2

0 2

0

0

e e (1.96)

Since e_r >_ 1 we get U < Uo. There is a decrease in energy because,when the dielectric is inserted, the capacitor spends some energy in pulling the dielectric inside.

(ii) When the battery remains connected to the capacitor Let us now consider what happens when

the battery of voltage _V_0 remains connected to the capacitor when the dielectric is inserted into the capacitor. This is shown in Figure1.57.

The potential difference V_0 across the plates remains constant. But it is found experimentally (first shown by Faraday) that when dielectric is inserted, the charge stored in the capacitor is increased by a factor e_r.

+ + + + + + + + + +
d	→E
––––––––––
(a)
d	+ + + + +
→E	εr

| ––––– |

Q = e_r Qo_ (1.97)

Due to this increased charge, the capacitance is also increased. The new capacitance is

C Q V

Q V

Cr r= = = 0

0

0 0e e (1.98)

However the reason for the increase in capacitance in this case when the battery remains connected is different from the case when the battery is disconnected before introducing the dielectric.

Now, C A

_d_0 0=

e

and C A d

= e (1.99)

The energy stored in the capacitor before the insertion of a dielectric is given by

_U C V_0 0 0 21

2 = (1.100)

Note that here we have not used the

expression _U Q C_0

0 2

0

1 2

= because here, both

charge and capacitance are changed, whereas in equation (1.100), Vo remains constant.

++++++++++

d E0 →

→ E0

– – – – – – – – – –

+ -

+ - εr_d_

(a)

(b)

+++++++++ + + + + + + + +++

– – – – – – – – – – – –– – – – – – –

After the dielectric is inserted, the capacitance is increased; hence the stored energy is also increased.

U CV C V Ur r= = = 1 2

1 20

2 0 0

2 0e e (1.101)

Since e_r >_ 1 we have U > Uo. It may be noted here that since voltage

between the capacitor _V_0 is constant, the electric field between the plates also remains constant.

The energy density is given by

u E= 1 2 0

2e (1.102)

where e is the permittivity of the given dielectric material.

The results of the above discussions are summarised in the following Table 1.2

Computer keyboard keys are constructed using capacitors with a dielectric as shown in

the figure.

Key

Movable plate Insulator (dielectric)

Fixed plate

When the key is pressed, the separation between the plates decreases leading to an increase in the capacitance. This in turn triggers the electronic circuits in the computer to identify which key is pressed.

Table 1.2 Effect of dielectrics in capacitor

S.No	Dielectric is inserted	Charge Q	Voltage V	Electric field E	Capacitance C	Energy U
1	When the battery is disconnected	Constant	decreases	Decreases	Increases	Decreases
2	When the battery is connected	Increases	Constant	Constant	Increases	Increases

EXAMPLE 1.21

A parallel plate capacitor filled with mica having εr = 5 is connected to a 10 V battery. The area of each parallel plate is 6 cm2 and separation distance is 6 mm. (a) Find the capacitance and stored charge.

(b) After the capacitor is fully charged, the battery is disconnected and the dielectric is removed carefully.

Calculate the new values of capacitance, stored energy and charge.

Solution

(a) The capacitance of the capacitor in the presence of dielectric is

C A

d r= =

× × × ×

×

− −

−

e e0 12 4

3

5 8 85 10 6 10 6 10

.

= × =−44 25 10 4 42513. .F pF

The stored charge is

Q CV= = × ×

= × =

−

−

44 25 10 10 442 5 10 44 25

13

13

. . .C pC

The stored energy is

U CV C= = × × ×

= ×

−

−

1 2

1 2

44 25 10 100

2 21 10

2 13

10

.

. J

(b) After the removal of the dielectric, since the battery is already disconnected the total charge will not change. But the potential

S. No Dielectric is inserted

Charge q _

Volta V

1 When the battery is disconnected Constant decrea

2 When the battery is connected Increases Const

difference between the plates increases. As a result, the capacitance is decreased.

New capacitance is

C C

r 0

12

12

4 425 10 5

0 885 10 0 885

= = ×

= × =

−

−

ε .

. .F pF

The stored charge remains same and 44.25 pC. Hence newly stored energy is

U Q C

Q C

_Ur r_0

2

0

2

10 10

2 2 5 2 21 10 11 05 10

= = =

= × × = ×− −

e e

. .J J

The increased energy is

∆_U_ = − × = ×− − ( . . ) .11 05 2 21 10 8 84 10

10 10 J J

When the dielectric is removed, it experiences an inward pulling force due to the plates. To remove the dielectric, an external agency has to do work on the dielectric which is stored as additional energy. This is the source for the extra energy 8.84 × 10–10 J.

Capacitor in series and parallel

(i) Capacitor in series Consider three capacitors of capacitance

C1, C2 and C3 connected in series with a battery of voltage V as shown in the Figure 1.58 (a).

As soon as the battery is connected to the capacitors in series, the electrons of charge

s ge Electric field

E Capacitance

C Energy

U

ses Decreases Increases Decreases

ant Constant Increases Increases

A 58××.85 10 ××61−−120	04
eeC ==d=×44..r
61× 025 10 Fp= 4 425 F−3−13

difference b etween t he plates increases. As a result, the capacitance is decreased.New capacitance is C 4.425×10C ==ε 5=×0..885 10 Fp= 0 885 F−12The s tored c harge r emains s ame a nd 044.25 pC. Hr ence newly stored energ y is−12Q Q eU == = e U22C C=×52 ..21 ×=10 JJ 11 05 ×102 2The increased enerr g y is0 r0−−10 10∆U =−(.11 05 22.) 11×=08 JJ .84×10When t he die lectric i s r emoved, i t experiences a n in ward p −−10 ulling f orce d 10ue to t he p lates. T o r emove t he die lectric, an ext ernal a gency has t o do w ork on t he dielectric w hich i s s tored a s addi tional energ y. Thi s i s t he s ource f or t he ext ra energ y 8.84 × 10 J.–10

–q _ are transferred from negative terminal to the right plate of _C_3 which pushes the electrons of same amount –q _ from left plate of C3 to the right plate of _C_2 due to electrostatic induction. Similarly, the left plate of _C_2 pushes the charges of –q _ to the right plate of _C_1 which induces the positive charge +q _ on the left plate of _C_1. At the same time, electrons of charge –q _ are transferred from left plate of _C_1 to positive terminal of the battery.

By these processes, each capacitor stores the same amount of charge q _. The capacitances of the capacitors are in general different, so that the voltage across each capacitor is also different and are denoted as V1, V2 and V3 respectively.

V1 V2 V3

CS

C2C1 C3 + + + +

+ + + +

+ + + +

– – – –

– – – –

–

–

– – – –

V

V

(a)

(b)

The sum of the voltages across the capacitor must be equal to the voltage of the battery.

V = _V_1 + _V_2 + _V_3 (1.103)

Since, q _ = CV, we have V Q C

Q C

Q C

= + + 1 2 3

= + + 

 



 

Q C C C 1 1 1

1 2 3

(1.104)

If three capacitors in series are considered to form an equivalent single capacitor Cs

shown in Figure 1.58(b), then we have

V Q CS

= . Substituting this expression into

equation (1.104), we get

Q C

Q C C CS

= + + 

 



 

1 1 1

1 2 3

1 1 1 1

1 2 3_C C C CS_

= + + (1.105)

Thus, the inverse of the equivalent capacitance CS of three capacitors connected in series is equal to the sum of the inverses of each capacitance. This equivalent capacitance CS is always less than the smallest individual capacitance in the series.

(ii) Capacitance in parallel

Consider three capacitors of capacitance C1, C2 and C3 connected in parallel with a battery of voltage V as shown in Figure 1.59 (a).

Since corresponding sides of the capacitors are connected to the same positive and negative terminals of the battery, the voltage across each capacitor is equal to the battery’s voltage. Since capacitances of the capacitors are different,

the charge stored in each capacitor is not the same. Let the charge stored in the three capacitors be Q1, Q2, and Q3 respectively. According to the law of conservation of total charge, the sum of these three charges is equal to the charge q _ transferred by the battery,

q _ = Q1 +Q2 +Q3 (1.106)

Since Q = CV, we have

Q = C1V + C2V + C3V (1.107)

If these three capacitors are considered to form a single equivalent capacitance CP which stores the total charge q _ as shown in the Figure 1.59(b), then we can write Q = CPV. Substituting this in equation (1.107), we get

CPV = C1V + C2V + C3V

CP = C1 + C2 + C3 (1.108)

V

Q1

C1 C2 C3

Q2 Q3+ -

V + -

Q CP

(a)

(b)

Thus, the equivalent capacitance of capacitors connected in parallel is equal to the sum of the individual capacitances. The equivalent capacitance CP in a parallel connection is always greater than the largest individual capacitance. In a parallel connection, it is equivalent as area of each capacitance adds to give more effective area such that total capacitance increases.

EXAMPLE 1.22

Find the equivalent capacitance between P and q _ for the configuration shown below in the figure (a).

QQ Q QPPP P6µF 6µF

4µF8µF 8µF 8µF

2µF

2µF

4 4µF

4µF

3µF

1µF

(a) (b) (c) (d)

µF

Solution

The capacitors 1 µF and 3 µF are connected in parallel and 6 µF and 2 µF are also separately connected in parallel. So these parallel combinations reduced to equivalent single capacitances in their respective positions, as shown in the figure (b).

Ceq = 1 + 3 = 4 µF

Ceq = 6 + 2 = 8 µF

From the figure (b), we infer that the two 4 µF capacitors are connected in series and the two 8 µF capacitors are connected in series. By using formula for the series, we can reduce to their equivalent capacitances as shown in figure (c).

1 1 4

1 4

1 2

2 C

C eq

eq= + = ⇒ = µF

and

1 1 8

1 8

1 4

4 C

C eq

eq= + = ⇒ = µF

From the figure (c), we infer that 2 µF and 4 µF are connected in parallel. So the equivalent capacitance is given in the figure (d).

Ceq = 2 + 4 = 6 µF

Thus the combination of capacitances in figure (a) can be replaced by a single capacitance 6 µF.

DISTRIBUTION OF CHARGES IN A CONDUCTOR AND ACTION AT POINTS

Distribution of charges in a conductor

Consider two conducting spheres A and B of radii _r_1 and _r_2 respectively connected to each other by a thin conducting wire as shown in the Figure 1.60. The distance between the spheres is much greater than the radii of either spheres.

B

q1

_q2_r1 r2

A

in conducting wire

If a charge q _ is introduced into any one of the spheres, this charge q _ is redistributed into both the spheres such that the electrostatic potential is same in both the spheres. They are now uniformly charged and attain electrostatic equilibrium. Let q1 be the charge residing on the surface of sphere A and q2 is the charge residing on the surface of sphere B such that Q = q1 + q2. The charges are distributed only on the surface and there is no net charge inside the conductor.

The electrostatic potential at the surface of the sphere A is given by

V q rA =

1 4

1

1πe 

(1.110)

The electrostatic potential at the surface of the sphere B is given by

V q rB =

1 4

2

2πe 

(1.111)

The surface of the conductor is an equipotential. Since the spheres are connected by the conducting wire, the surfaces of both the spheres together form an equipotential surface. This implies that

VA = VB

or q r

q r

1

1

2

2

= (1.112)

Let the charge density on the surface of sphere A be σ1 and that on the surface of sphere B be σ2. This implies that q1 = 4π_r1_

2σ1 and

q2 = 4π_r2 2σ2_. Substituting these values

into equation (1.112), we get

σ1r1 = σ2r2 (1.113)

from which we conclude that

σr = constant (1.114)

atfin fc	11=+C 4eq	1 1=⇒ C = 24 2eq1 1=⇒ C = 48 4he f igure (c), w e inf er tre co nnected in p araeqlent c apacitance i s g iven in the co mbination o f c apacian b e r eplaced b y a since 6 µF.	µF

| and11=+C 8From tnd 4 µF ahe eeqquivaigure (d).C = 2 + 4 = 6 µFThus tigure (a) capacieq tan | | µF | | hat 2 µF llel. S o he |

| tances gle |

Thus the surface charge density σ is inversely proportional to the radius of the sphere. For a smaller radius, the charge density will be larger and vice versa.

EXAMPLE 1.23

Two conducting spheres of radius r1 = 8 cm and r2 = 2 cm are separated by a distance much larger than 8 cm and are connected by a thin conducting wire as shown in the figure. A total charge of q _ = +100 nC is placed on one of the spheres. After a fraction of a second, the charge q _ is redistributed and both the spheres attain electrostatic equilibrium.

Br1 r2

A

in conducting wire

(a) Calculate the charge and surface charge density on each sphere.

(b) Calculate the potential at the surface of each sphere.

Solution

(a) The electrostatic potential on the surface of the sphere A is V q

rA = 1

4 1

1πe 

The electrostatic potential on the surface

of the sphere B is V q rB =

1 4

2

2πe 

Since VA = VB. We have

q r

q r

q r r

q _1

1

2

2 1

1

2 2= ⇒ =



 



 

But from the conservation of total charge, Q = q1 + q2, we get q1 = Q – q2. By substituting this in the above equation,

Q q r r

q _− = 

 



 2

1

2 2

so that q Q r

_r r_2 2

1 2

= +



 



 

Therefore,

q _2 9100 10 2

10 20= × ×



  

 =

− nC

and q1 = Q – q2 = 80nC

The electric charge density on sphere A

is σ π1

1

1 24

= q _

r

The electric charge density on sphere B

is σ π2

2

2 24

= q _

r Therefore,

σ π1

9

4 680 10

4 64 10 0 99 10=

× × ×

= × −

− − −. C m 2

and

σ π2

9

4 620 10

4 4 10 3 9 10=

× × ×

= × −

− − −. C m 2

Note that the surface charge density is greater on the smaller sphere compared to the larger sphere (σ2 ≈ 4σ1) which confirms

the result σ σ

1

2

2

1

= r r

.

The potential on both spheres is the same. So we can calculate the potential on any one of the spheres.

V q rA = =

× × × ×

= −

−

1 4

9 10 80 10 8 10

91

1

9 9

2πe 

kV

Action of points or Corona discharge

Consider a charged conductor of irregular shape as shown in Figure 1.61 (a).

σ =and 46π1σ =44π2

sThisisThgttsa	Qq−=2	rr q=12 rr2 +r ,2210 ×12 = 80nC 102  = 20 nCctric c−9 harge density o n s2rctric c harge den sity o n s12r1,22802 ×10=×09.C91 0××41 0−9−−620×10 −4=×39.C 10 m×× 10at t he s urface c harge den−9the smaller sphere −−6 com−4phere (σ ≈ 4σ ) which coσ r . =σ rtential o n b oth s pheres i2 1e c an c alculate t he p ot1 2f the spheres.2 11 q 91××08 01× 0e r 81× 099 −1−2 1

| o that qQereforeq =×100and q = Q – qTh2 e e leqσ = 14πThe e leqσ1 =4πerefore2σ =and 46π1σ =44πNote t hreater on 2he larger she result The p oame. S o wny one oV ==4πA |

| phere A phere B | | m22 | | sity i s pared to nfirms s t he ential o n | | = 9 kV |

We know that smaller the radius of curvature, the larger is the charge density. The end of the conductor which has larger curvature (smaller radius) has a large charge accumulation as shown in Figure 1.61 (b).

As a result, the electric field near this edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge. This reduces the total charge of the conductor near the sharp edge. This is called action of points or corona discharge.

Lightning arrester or lightning conductor

This is a device used to protect tall buildings from lightning strikes. It works on the principle of action at points or corona discharge.

This device consists of a long thick copper rod passing from top of the building to the ground. The upper end of the rod has a sharp spike or a sharp needle as shown in Figure 1.62 (a) and (b).

The lower end of the rod is connected to copper plate which is buried deep into the ground. When a negatively charged cloud is passing above the building, it induces a positive charge on the spike. Since the induced charge density on thin sharp spike is large, it results in a corona discharge. This positive charge ionizes the surrounding air which in turn neutralizes the negative charge in the cloud. The negative charge pushed to the spikes passes through the copper rod and

is safely diverted to the Earth. The lightning arrester does not stop the lightning; rather it diverts the lightning to the ground safely.

Van de Graaff Generator

In the year 1929, Robert Van de Graaff designed a machine which produces a large amount of electrostatic potential difference, up to several million volts (107 V). This Van de Graff generator works on the principle of electrostatic induction and action at points.

A large hollow spherical conductor is fixed on the insulating stand as shown in Figure 1.63. A pulley B is mounted at the centre of the hollow sphere and another pulley C is fixed at the bottom. A belt made up of insulating materials like silk or rubber runs over both pulleys. The pulley C is driven continuously by the electric motor. Two comb shaped metallic conductors E and D are fixed near the pulleys.

The comb D is maintained at a positive potential of 104 V by a power supply. The upper comb E is connected to the inner side of the hollow metal sphere.

Due to the high electric field near comb D, air between the belt and comb D gets ionized by the action of points. The positive charges are pushed towards the belt and negative charges are attracted towards the comb D. The positive charges stick to the belt and move up. When the positive charges on the belt reach the point near the comb E, the comb E acquires negative charge and the sphere acquires positive charge due to electrostatic induction. As a result, the positive charges are pushed away from the comb E and they reach the outer surface of the sphere. Since the sphere is a conductor,

the positive charges are distributed uniformly on the outer surface of the hollow sphere. At the same time, the negative charges nullify the positive charges in the belt due to corona discharge before it passes over the pulley.

When the belt descends, it has almost no net charge. At the bottom, it again gains a large positive charge. The belt goes up and delivers the positive charges to the outer surface of the sphere. This process continues until the outer surface produces the potential difference of the order of 107 which is the limiting value. We cannot store charges beyond this limit since the extra charge starts leaking to the surroundings due to ionization of air. The leakage of charges can

Corona discharge

Corona discharge

Aluminum roller

Rubber belt

Rubber belt

Plastic roller

+

+ +

+ +

+ ++ ++

++ + +

+ + + + + + + + + + + + + +

+

+ + + + + + +

+ + +

+ ++

+ + + + + + + +

+ + +

+

+

++ + +

+++ + + +

++++ + +

+ +

++

––––

– –+

+ + + + + + + + + + + + +

+ + + + + + + + + + + + + +

B

B

C

C

D

D

E

E

Insulating stand

–

+

–

Figure 1.63 Van de Graaff generator

EB	+

| + ++ | | +++++ | | + – – | | – |

EB+ +++++++ ++++ ++	+ + ++ + +++ +

| +++++++++ +++++++++ ++++++++ |+ +D– | | ++++++ ––––– | | +++ +++ +++ |

| C |

be reduced by enclosing the machine in a gas filled steel chamber at very high pressure.

The high voltage produced in this Van de Graaff generator is used to accelerate positive ions (protons and deuterons) for nuclear disintegrations and other applications.

EXAMPLE 1.24

Dielectric strength of air is 3 × 106 V m–1. Suppose the radius of a hollow sphere in the Van de Graff generator is R = 0.5 m, calculate the maximum potential difference created by this Van de Graaff generator.

Solution

The electric field on the surface of the sphere is given by (by Gauss law)

E Q R

= 1

4 2πe 

The potential on the surface of the hollow metallic sphere is given by

V Q R

ER= = 1

4πe 

Since Vmax = EmaxR

Here _E_max = × −3 106 Vm .1 So the maximum

potential difference created is given by

Vmax = 3 × 106 × 0.5 = 1.5 × 106 V (or) 1.5 million volt

mSHp	1E =4π	Qe Rential on the surface of thpher2 e is given by QERe R = E R So the m=×31 0 Vm .ference created is given bx max6 −1× 10 × 0.5 V (or) 1.5 millio66

| e hollow | | The potetallic s1V ==4πince Vere Eotential difmaV = 3 max = 1.5 × 10max |

| aximum y | | n volt |

„ Like charges repel and unlike charges attr „ The total charge in the universe is conser „ Charge is quantized. Total charge in an

e is charge of the electron.

„ Coulomb’s law in vector form: 



F q

= 1

4πe

„ Electrostatic force obeys the superpositio

„ Electric field at a distance r from a point

„ Electric field lines starts at a positive charg

„ Electric field due to electric dipole at poin

„ Electric field due to electric dipole at poin

„ Torque experienced by a dipole in a unifo

„ Electrostatic potential at a distance r from

„ Electrostatic potential due to an electric d

„ The electrostatic potential is the same at a

„ The relation between electric field and ele



E V x

i V y

j V z

k=− ∂ ∂

+ ∂ ∂

+ ∂ ∂



 



 

 



„ Electrostatic potential energy for system arrange the charges in the given configur

„ Electrostatic potential energy of a dipole s

„ The total electric flux through a closed su enclosed by the surface

„ Electric field due to a charged infinite wi

„ Electric field due to a charged infinite pla

„ Electric field inside a charged spherical sh

S U M M A R Y

act ved object q = ne where n = 0,1,2,3… and

q r

_r_1 2 2  ( _r_is unit vector along joining q1, q2)

n principle.

charge: 



E q r

r= 1

4 2πe 

e and end at a negative charge or at infinity

ts on the axial line :  



E p rtot = 

  

 

1 4

2 3πe

ts on the equatorial line:  



E p rtot =− 

  

 

1 4 3πe

rm electric field:  



τ= ×_p E_

the point charge: V q r

= 1

4πe 

ipole: V p r r

= ⋅1

4 2πe

 

ll points on an equipotential surface.

ctrostatic potential:

of charges is equal to the work done to ation.

ystem in a uniform electric field: U p E=− ⋅ 



rface : Φ_E Q_

= e 

where q _ is the net charge

re : 



E r

r= 1

2π λ

e 

ne : 



E n= σ

2e  (_n_ is normal to the plane)

ell is zero. For points outside: 



E Q r

r= 4 2πe



„ Electric field inside a conductor is zero

conductor is normal to the surface and h

„ The surface of the conductor has the sam

„ Conductor can be charged using the proc „ A dielectric or insulator has no free elect

dielectric is polarised.

„ Capacitance of a conductor is given by C

„ Capacitance of a parallel plate capacitor:

„ Electrostatic energy stored in a capacitor

„ The equivalent capacitance for parallel co capacitance of the capacitors.

„ For a series combination: The inverse of inverse of individual capacitance of capac

„ The distribution of charges in the condu For sharper edge, the surface charge den lightning arrestor

„ Van de Graaff generator is used to produc

. The electric field at the surface of the

as magnitude E = σ e 

.

e potential, at all points on the surface.

ess of induction. rons. When an electric field is applied, the

Q V

= .

C A d

= e 

U CV= 1 2

2

mbination is equal to the sum of individual

equivalent capacitance is equal to sum of itors. ctors depends on the shape of conductor. sity is greater. This principle is used in the

e large potential difference (~107 V).

ELECTROS

Point charge Coloumb

Gauss law

Electric eld

Electric ux

Van de Gra Generator

Series Capac

With dielectric

C O N C E P

Conductors

TATICS

’s law Dipole

Electric potential

Electric Potential energy

itors Parallel

Without dielectric

T M A P

I Multiple choice questions

1. Two identical point charges of magnitude –q are fixed as shown in the figure below. A third charge +q is placed midway between the two charges at the point P. Suppose this charge +q _ is displaced a small distance from the point P in the directions indicated by the arrows, in which direction(s) will +q _ be stable with respect to the displacement?

-q _ -q _

+q _ A1 A2

B2

B1

P – –+

(a) A1 and A2 (b) B1 and B2 (c) both directions (d) No stable

2. Which charge configuration produces a uniform electric field? (a) point charge (b) uniformly charged infinite line (c) uniformly charged infinite plane (d) uniformly charged spherical shell

3. What is the ratio of the charges q q

1

2

for

the following electric field line pattern?

q1 q2

EVALUATION

(a) 1 5

(b) 25 11

(c) 5 (d) 11 25

4. An electric dipole is placed at an alignment angle of 30o with an electric field of 2 × 105 N C–1. It experiences a torque equal to 8 N m. The charge on the dipole if the dipole length is 1 cm is (a) 4 mC (b) 8 mC (c) 5 mC (d) 7 mC

5. Four Gaussian surfaces are given below with charges inside each Gaussian surface. Rank the electric flux through each Gaussian surface in increasing order.

+ +2q A

B

C D

+

+q

–q –

(a) D < C < B < A (b) A < B = C < D (c) C < A = B < D (d) D > C > B > A

6. The total electric flux for the following closed surface which is kept inside water

+

+–

+q

+2q–q Water

(a) 80q _ e 

(b) q _ 40e



(c) q _ 80e



(d) q _ 160e



7. Two identical conducting balls having positive charges q1 and q2 are separated by a centre to centre distance r. If they are made to touch each other and then separated to the same distance, the force between them will be (NSEP 04-05) (a) less than before (b) same as before (c) more than before (d) zero

8. Rank the electrostatic potential energies for the given system of charges in increasing order.

Q Q-Q -Q-Q -Q -2Q -2Qrr

(a) (b) (d)(c)

2r– – – –r –+ –+

(a) 1 = 4 < 2 < 3 (b) 2 = 4 < 3 < 1 (c) 2 = 3 < 1 < 4 (d) 3 < 1 < 2 < 4

9. An electric field 

E xi=10  exists in a certain region of space. Then the

VV

O O R Rr r

(b)(a)

potential difference V = Vo – VA, where Vo is the potential at the origin and VA is the potential at x = 2 m is: (a) 10 V (b) –20 V (c) +20 V (d) –10 V

10. A thin conducting spherical shell of radius R has a charge q _ which is uniformly distributed on its surface. The correct plot for electrostatic potential due to this spherical shell is

VV V

O O O R R Rr r

(b) (c)(a)

V V

O O R Rr r

(c) (d)

11. Two points A and B are maintained at a potential of 7 V and -4 V respectively. The work done in moving 50 electrons from A to B is (a) 8.80 × 10–17 J (b) -8.80 × 10–17 J (c) 4.40 × 10–17 J (d) 5.80 × 10–17 J

12. If voltage applied on a capacitor is increased from V to 2_V_, choose the correct conclusion. (a) q _ remains the same, C is doubled (b) q _ is doubled, C doubled (c) C remains same, q _ doubled (d) Both q _ and C remain same

|——|——|——|——|——|

13. A parallel plate capacitor stores a charge q _ at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change? (a) Capacitance (b) Charge (c) Voltage (d) Energy density

14. Three capacitors are connected in triangle as shown in the figure. The equivalent capacitance between the points A and C is

2µF

2µF

1µF

A B

C

(a) 1µF (b) 2 µF

(c) 3 µF (d) 1 4

µF

15. Two metallic spheres of radii 1 cm and 3 cm are given charges of –1 × 10–2 C

and 5 × 10–2 C respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is (AIIPMT -2012) (a) 3 × 10–2 C (b) 4 × 10–2 C

(c) 1 × 10–2 C (d) 2 × 10–2 C

Answers

1) b 2) c 3) d 4) b 5) a 6) b 7) c 8) a 9) c 10) b 11) a 12) c 13) d 14) b 15) a

II Short Answer Questions

1. What is meant by quantisation of charges?

2. Write down Coulomb’s law in vector form and mention what each term represents.

3. What are the differences between Coulomb force and gravitational force?

4. Write a short note on superposition principle.

5. Define ‘electric field’. 6. What is mean by ‘electric field lines’? 7. The electric field lines never intersect.

Justify. 8. Define ‘electric dipole’. Give the

expression for the magnitiude of its electric dipole moment and the direction.

9. Write the general definition of electric dipole moment for a collection of point charge.

10. Define ‘electrostatic potential”. 11. What is an equipotential surface? 12. What are the properties of an

equipotential surface? 13. Give the relation between electric field

and electric potential. 14. Define ‘electrostatic potential energy’. 15. Define ‘electric flux’. 16. What is meant by electrostatic energy

density?

17. Write a short note on ‘electrostatic shielding’.

18. What is polarisation? 19. What is dielectric strength?

20. Define ‘capacitance’. Give its unit.

21. What is corona discharge?

III Long Answer questions

1. Discuss the basic properties of electric charges.

2. Explain in detail Coulomb’s law and its various aspects.

3. Define ‘electric field’ and discuss its various aspects.

4. Calculate the electric field due to a dipole on its axial line and equatorial plane.

5. Derive an expression for the torque experienced by a dipole due to a uniform electric field.

6. Derive an expression for electrostatic potential due to a point charge.

7. Derive an expression for electrostatic potential due to an electric dipole.

8. Obtain an expression for potential energy due to a collection of three point charges which are separated by finite distances.

9. Derive an expression for electrostatic potential energy of the dipole in a uniform electric field.

10. Obtain Gauss law from Coulomb’s law.

11. Obtain the expression for electric field due to an infinitely long charged wire.

12. Obtain the expression for electric field due to an charged infinite plane sheet.

13. Obtain the expression for electric field due to an uniformly charged spherical shell.

14. Discuss the various properties of conductors in electrostatic equilibrium.

15. Explain the process of electrostatic induction.

16. Explain dielectrics in detail and how an electric field is induced inside a dielectric.

17. Obtain the expression for capacitance for a parallel plate capacitor.

18. Obtain the expression for energy stored in the parallel plate capacitor.

19. Explain in detail the effect of a dielectric placed in a parallel plate capacitor.

20. Derive the expression for resultant capacitance, when capacitors are connected in series and in parallel.

21. Explain in detail how charges are distributed in a conductor, and the principle behind the lightning conductor.

22. Explain in detail the construction and working of a Van de Graaff generator.

Exercises

1. When two objects are rubbed with each other, approximately a charge of 50 nC can be produced in each object. Calculate the number of electrons that must be transferred to produce this charge.

Ans: 31.25 × 1010 electrons 2. The total number of electrons in the

human body is typically in the order of 1028. Suppose, due to some reason, you

and your friend lost 1% of this number of electrons. Calculate the electrostatic force between you and your friend separated at a distance of 1m. Compare this with your weight. Assume mass of each person is 60 kg and use point charge approximation.

Ans: Fe = 23 × 1023 N, W = 588 N,

F W e = ×3 9 10

21 .

3. Five identical charges q _ are placed equidistant on a semicircle as shown in the figure. Another point charge q is kept at the centre of the circle of radius R. Calculate the electrostatic force experienced by the charge q _.

q _

q _

Q Q

q _

q x

y

R

Ans: 



F qQ R

i= +( )1 4

1 22πe N

4. Suppose a charge +q _ on Earth’s surface and another +q _ charge is placed on the surface of the Moon. (a) Calculate the value of q _ required to balance the gravitational attraction between Earth and Moon (b) Suppose the distance between the Moon and Earth is halved, would the charge q _ change?

(Take mE = 5.9 × 1024 kg, mM = 7.9 × 1022 kg) Ans: (a) q _ ≈ +5.87 × 1013 C, (b) no change

-kx

N

QE

mg

(a)

T

5. Draw the free body diagram for the following charges as shown in the figure (a), (b) and (c).

-- -

-—— – ———- + + + + + + ++ + + + +

+++++++++++

v0-q+q

E →

→

x = 0

k m,Q

(b) (c)(a)

-+

Ans:

-kx

N

QE

mg mg mg

(a) (b) (c)

T

qE

qE

mg mg

(b) (c)

qE

qE

6. Consider an electron travelling with a speed _v_o and entering into a uniform electric field



E which is perpendicular to _v_ as shown in the Figure. Ignoring gravity, obtain the electron’s acceleration, velocity and position as functions of time.

_v_0 E

+ + + + + + + + + + +

→

P

x

y

→

____________

– e

Ans :   

 _a eE m_

j i eE m

tj r t i eE m

t j=− = − = −    , ,v v v 1 2

2

k m,Q →E
+

|——|——|

7. A closed triangular box is kept in an electric field of magnitude E = 2 × 103 N C–1 as shown in the figure.

5cm

15cm

E →

60°

Calculate the electric flux through the (a) vertical rectangular surface (b) slanted surface and (c) entire surface.

Ans: (a) 15 Nm2 C–1 (b) 15 Nm2 C–1 (c) zero 8. The electrostatic potential is given as

a function of x in figure (a) and (b). Calculate the corresponding electric fields in regions A, B, C and D. Plot the electric field as a function of x for the figure (b).

0

2

4

6

0.2 0.4 0.6 x(m)

x(cm)

V V 30 20

10

0

-10

-20

-30

1 2 3

4 5

A

B

C

D

(a) (b)

Ans: (a) EX = 15 Vm–1 (region A), EX = -10 Vm–1 (region C)

EX = 0 (region B), EX = 30 Vm–1 (region D)

(b)

x(cm)

30

20

10

0

-10

-20

-30

2 3 4 51

V

9. A spark plug in a bike or a car is used to ignite the air-fuel mixture in the engine. It consists of two electrodes separated by a gap of around 0.6 mm gap as shown in the figure.

To create the spark, an electric field of magnitude 3 × 106 Vm–1 is required. (a) What potential difference must be applied to produce the spark? (b) If the gap is increased, does the potential difference increase, decrease or remains the same? (c) find the potential difference if the gap is 1 mm.

Ans: (a) 1800 V, (b) increases (c) 3000 V

10. A point charge of +10 µC is placed at a distance of 20 cm from another identical point charge of +10 µC. A point charge of -2 µC is moved from point a to b as shown in the figure. Calculate the change in potential energy of the system? Interpret your result.

10µC

–2µC

10µC5 cm

5 cm

15 cm a

b

++ –

–

15cm	→
E
60°

|——|——|——|——|——|——| | A |C |

| B | | D |

|——|——|——|——|——|

| 3 |5 | | 12 |4 |

Ans: ∆_U_ = +1.12 J, positive sign implies that to move the charge –2µC external work is required. 11. Calculate the resultant capacitances for

each of the following combinations of capacitors.

_C_0

_C_0

_C_0

_C_0

_C_0

_C_0

_C_0 _C_0

_C_0

C

C0 R

(a) (b) (c)

_C_0

_C_0

_C_0

_C_0

_C_0

_C_0

_C_0

_C_0

_C_1

_C_3 _C_4

_C_2

C0 R S P Q

(c) (d) (e)

P

Q

Ans: (a) 2 3

C 

(b) C 

(c) 3_C_ 

(d) across PQ:

C C C C C C C C C C C C C C C C

1 2 3 2 3 4 1 2 4 1 3 4

1 3 2 4

+ + + +( ) +( )

across RS:

C C C C C C C C C C C C C C C C

1 2 3 2 3 4 1 2 4 1 3 4

1 2 3 4

+ + + +( ) +( )

(e) across PQ: 2 Co

12. An electron and a proton are allowed to fall through the separation between the plates of a parallel plate capacitor of voltage 5 V and separation distance h = 1 mm as shown in the figure.

+ +++++++++

+ +++++++++

Electron Neutron Protony

x

+-

– – – – – – – – – –

– – – – – – – – – –

_C_0

_C_0

_C_0

_C_0

1

_C_3 _C_4

_C_2

S P Q

(d) (e)

P

Q

(a) Calculate the time of flight for both electron and proton (b) Suppose if a neutron is allowed to fall, what is the time of flight? (c) Among the three, which one will reach the bottom first? (Take mp = 1.6 × 10–27 kg, me = 9.1 × 10–31 kg and g = 10 m s–2)

Ans:

(a) t hm eEe

e= ≈ 2

1 5. ns (ignoring the gravity),

t hm eEp

p= ≈ 2

63ns (ignoring the gravity)

(b) t h gn = ≈

2 14 1. ms

(c) electron will reach first 13. During a thunder storm, the movement

of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air (3 × 106 Vm–1 ), lightning will occur.

+ + + + + + + + + + + + + + + + + + + + + +

+

– –– –– –– –– –– –– –– –– –

–

Positive charge

Negative charge

(a) If the bottom part of the cloud is 1000 m above the ground, determine the electric potential difference that exists between the cloud and ground.

(b) In a typical lightning phenomenon, around 25 C of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?

Ans: (a) V = 3 × 109 V, (b) U = 75 × 109 J 14. For the given capacitor configuration

(a) Find the charges on each capacitor (b) potential difference across them (c) energy stored in each capacitor

8µF 8µF2µF

6µF

9V + -

b

d

a

c

BOOKS FOR REFERENCE

1. Douglas C.Giancoli, , “Physics for Scientis Prentice Hall, Fourth edition.

2. James Walker, “Physics”, Pearson- Addison

3. Purcell, Morin, “Electricity and Magnetism

4. Serway and Jewett, “Physics for Scientist and Publishers, Eighth Edition.

5. Tipler, Mosca, “Physics for scientist and E Company, Sixth Edition.

6. Tarasov and Tarasova, “Questions and prob

7. H.C.Verma, “Concepts of Physics: Vol 2, Bh

8. Eric Roger, Physics for the Inquiring Mind,

Ans: Qa = 24 µC, Qb = 18 µC, Qc = 6 µC, Qd = 24 µC Va = 3V, Vb = 3V, Vc = 3V, Vd = 3V, Ua = 36 µJ, Ub = 27 µJ, Uc = 9 µJ, Ud = 36 µJ 15. Capacitors P and Q have identical cross

sectional areas A and separation d. The space between the capacitors is filled with a dielectric of dielectric constant er as shown in the figure. Calculate the capacitance of capacitors P and Q.

P Q

Ans : C A d

C A dP r Q

r

r

= +( ) = +



 



 

e e

e e

e  

2 1 2

1 ,

t & Engineers with Modern Physics”, Pearson

Wesley Publishers, Fourth Edition.

”, “Cambridge University Press, Third Edition.

Engineers with Modern Physics”, Brook/Coole

ngineers with Modern Physics”, Freeman and

lems in School Physics”, Mir Publishers.

arthi Bhawan Publishers.

Princeton University Press.

Electro

In this activity you will be able to learn about capacitor and the factors affecting

capacitance.

STEPS: - Open the browser and type “phet.colorado.ed

address bar. Go to the tab ‘Dielectric’. - Change the plate area, distance between the p

maximize or minimize to make a capacitor w - Explore the relationships between charge, volt

a capacitor system to store the greatest energy - Charge the capacitor with 1.0 v using the b

dielectric between the plates. Discus how elec dielectric is introduced.

• What is the effect of introducing a dielectric b

Note: Install Java application if it is not in your system. simulation and works in off line from https://phet.c

URL: https://phet.colorado.edu/en/simulation/legacy/capaci

* Pictures are indicative only. * If browser requires, allow Flash Player or Java Sc

ICT CORNER

Step1

Step3

Connect capacitors parallel and series combin

statics

u/en/simulation/legacy/capacitor-lab” in the

late and dielectric. Identify what you would ith the greatest capacitance. age, and stored energy for a capacitor. Design .

attery. Disconnect the battery. Now insert a tric field changes in between the plates when

etween plates? (Change dielectric materials)

You can download all the phet olorado.edu/en/offline-access .

tor-lab

ript to load the page.

Step4

Step2

Topic: Capacitor lab

ation and find the effective capacitance.