2.0 Preview:

We shall begin by defining Electrostatic potential at any point in an electrostatic field as the work done in carrying a unit positive charge from infinity to that point against the electrostatic force of the field without accelerating it. Consequently, we will derive an expression for potential due to a point charge.

Next, we will define electric dipole as the separation of positive and negative charges. We shall then derive an expression for electric field due to a dipole: (a) at any point on its axis and (b) at any point on its equatorial plane. We shall also obtain the torque on an electric dipole in a uniform electric field.

We will then describe equipotential surface as a surface with a constant value of potential at all points on the surface. We will then establish a relation between electric field and electric potential.

Next, we will define Potential energy of a system of charges as the total amount of work done in bringing the individual charges from their state of rest at infinity to constitute the system.

We shall then use a Gaussian surface which is a closed surface in three-dimensional space through which, the flux of a vector field is calculated—usually the gravitational field, the electric field, or magnetic field.

We will then study about dielectrics and capacitors. Dielectrics are insulators hence, do not conduct electricity; the maximum electric field that a dielectric medium can withstand without break-down is called its dielectric strength. A capacitor is a system of two conductors separated by an insulator. Capacitance is the ability of the system to store charges. We will then derive an equation for the capacitance of a parallel plate capacitor. A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.

We shall then derive expression for capacitances when capacitors are connected in series and in parallel and for; energy stored in a capacitor. With the study of Van De Graff Generator, we shall come to the end of this chapter. We will describe the principle of its working as well.

2.1 Introduction

Newton’s gravitational law states that:

“A particle attracts every other particle in universe using a force that is directly proportional to the product of their masses and inversely proportional to the square of distance between them.”

The particles are replaced by points in Coulomb’s law. Coulomb force is a conservative force between two masses and has inverse-square dependence on distance. At every point in electrostatic field, a particle with charge ‘q’ possesses certain electrostatic potential energy.

“The potential energy of a charge ‘q’ at any point in the electric field is equal to the work done by an external force in bringing the charge from infinity to that point without any acceleration.”

Questions for section 2.1

1. State Newton’s gravity law.
2. Define the potential energy of a charge in an electric field.

2.2 Electrostatic Potential

“Electrostatic potential at any point in an electrostatic field is the work done in carrying a unit positive charge from infinity to that point against the electrostatic force of the field without accelerating it.”

$$ V= \frac {W}{q} $$

Where, V – electrostatic potential at that point

W -work done in carrying unit positive charge ‘q’ from infinity to a point

q -positive charge

“The electrostatic potential difference between two points in an electrostatic field is the work done in carrying unit positive charge from one point to another against electrostatic force of field without acceleration.”

$$ ΔV=V_B-V_A= \frac {W_{AB}}{q} $$

Where ΔV – potential difference between A and B in volts

$V_A$ & $V_B$- electrostatic potentials at A and B

$W_{AB}$ – work done in moving ‘q’ from A to B

q – Positive charge

Questions for section 2.2

1. Define and derive an equation for the electrostatic potential at any point.

2.3 Potential due to a point charge

A point charge +q is considered at the origin. P is situated at r distance from O. To determine the potential at P, a unit positive charge at distance x from O is taken. The force acting on unit positive charge is electric intensity at that point and is given by

$$ E = \frac {1}{(4πϵ0 )}. \frac {q}{x^2} $$

It is in the direction away from O. Let it be moved by a small distance dx against the field. Then the amount of work done will be dW = -E.dx (negative sign indicates that work is done against the field) dx is so small that E can be assumed to be constant over it.

The total amount of work done in moving a unit positive charge from infinity to point p; where x=∞ to the point P at x=r against the field is given by

$$ W = \int{x = \infty}^{x=0} dW = \int_{x = \infty}^{x=r} – E dx $$

$$ So , \space W = \int_{x = \infty}^{x=r} – \frac {1}{4πϵ0} \frac {q}{x^2} dx $$ $$ = \frac {q}{4πϵ_0} \int{ \infty}^{r} – \frac {1}{x^2} dx $$

$$ = \frac {q}{4πϵ0} \bigg[ \frac {1}{x} \bigg]{\infty}^{r} $$

$$ W = \frac {q}{4πϵ_0} \bigg[ \frac {1}{r} \bigg] $$

$$ = \frac {1}{4πϵ_0} \bigg[ \frac {q}{r} \bigg] $$

Therefore, the total amount of work done in moving a unit positive charge from infinity to point P is the electric potential at P. That is

$$ V = \frac {1}{4πϵ_0} \bigg[ \frac {q}{r} \bigg] $$

If the charge at O is –q, then $ = \frac {1}{4πϵ_0} \bigg[ \frac {-q}{r} \bigg] $

Example 2.1

(a) Calculate the potential at a point P due to a charge of 4 × 10${–7}$C located 9 cm away.

(b) Hence obtain the work done in bringing a charge of 2 × $10^{–9}$ C from infinity to the point P.

Does the answer depend on the path along which the charge is brought?

Solution:

(a) $$ V = \frac {1}{4 \pi \epsilon_o }{Q}{r} = 9 \times Nm^2C^{-2} \times \frac {4 \times 10^{-7}C}{0.09 \space m}$$
= 4 × $10^4$ V

(b) W = qV = 2 × $10^{-9}$ C × 4 × $10^4$V = 8 × $10^{–5}$ J

No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along r and another perpendicular to r. The work done corresponding to the later will be zero.

Questions for section 2.3

1. Derive an equation for the potential due to a point charge.

2.4 Potential due to an electric dipole

An electric dipole moment P=q*2a is considered which acts along the dipole axis. P is a point at a distance r from centre of the dipole. Distance of P from +q and –q be r1 and r¬2 respectively.

Electric potential at P due to +q is,

$V_1 = \frac{1}{4 \pi ϵ_0} \frac{q}{r_1}$

Concept box What is an electric dipole? An electric dipole is a separation of positive and negative charges. The simplest example of this is a pair of electric charges of equal magnitude but opposite sign, separated by some (usually small) distance. The total charge on the dipole is always zero.

Electric potential at P due to +q is,
$$ V_1 = \frac {1}{4πϵ_0} \frac {q}{r_1} $$

Electric potential at P due to -q is,
$$ V_2 = \frac {1}{4πϵ_0} \frac {-q}{r_2} $$

The potential V due to the dipole is sum of potentials due to charges +q and –q
$$ V =V_1 + V_2 = \frac {1}{4πϵ_0} \frac {q}{r_1} – \frac {1}{4πϵ_0} \frac {q}{r_2} = \frac {q}{4πϵ_0} \bigg[ \frac {1}{r_1}- \frac {1}{r_2} \bigg]
$$

From geometry of the figure,

$ r_1^2 = r^2 + a^2 – 2ar \space cosθ = r^2 \bigg [1+ \frac {a^2}{r^2}- \frac {2a}{r} cosθ \bigg] $ and

$ r_2^2 = r^2 + a^2 + 2ar \space cosθ = r^2 \bigg [1+ \frac {a^2}{r^2}+ \frac {2a}{r} cosθ \bigg] $

if r >>a, (a/r) is very small, hence $(a2/r2)$ is very very small, hence can be neglected.

$ r_1^2 = r^2 \bigg [1- \frac {2a}{r} cosθ \bigg] $ ; $ r_1 = r \bigg[1 – \frac {2a}{r} cosθ \bigg ]^{1/2} $

$$ so , \space \frac {1}{r_1} = \frac {1}{r} \bigg [1- \frac {2a}{r} \bigg]^{-1/2}cosθ$$

Using binomial theorem and retaining terms upto first order in (a/r),

$$ \frac {1}{r_1} = \frac {1}{r} \bigg[1- \bigg(- \frac {1}{ 2} \bigg) \frac {2a}{r}cosθ \bigg] \hspace{10mm} \frac{1}{r_1} = \frac {1}{r} \bigg[ 1+\frac {a}{r}cosθ \bigg] \hspace{10mm} \frac {1}{r_1} = \bigg[\frac{1}{r}+ \frac {a}{r^2} cosθ \bigg] $$

Similarly,

$$ \frac {1}{r^2} = \frac{1}{r} \bigg[ 1 + \frac {2a}{r}cosθ \bigg ]^{1/2}= \frac {1}{r} \bigg[ 1 – \frac {a}{r} cosθ \bigg] \ce{ ->} \frac {1}{r_2} = \bigg[ \frac {1}{r} – \frac {a}{r^2} cosθ \bigg] $$

Substituting the values in first equation,

$$ V = \frac {q}{4πϵ_0} \bigg[ \frac {2a}{r^2} cosθ \bigg] = V = \frac {1}{ 4πϵ_0} \frac{P cosθ}{r^2} \space where \space P = q × 2a $$

Questions for section 2.4

1. Derive an equation for the potential due to an electric dipole.

2.5 Potential due to system of charges

A system of charges $q_1$, $q_2$, $q_3$…. $q_n$ with position vectors $r_1$, $r_2$, ….. $r_n$ relative to origin is considered. The potential $V_1$ at P due to charge $q_1$ is given by $ V_1 = \frac {1}{4πϵ0} \frac {q_1}{r{1P}} \space where \space r_{1P} \space is \space distance \space between \space q_1 and \space P $

Similarly, Vn = where rnP is distance from qn to P.

Similarly , V_n = $\frac {1}{4πϵ0} \frac {q_n}{r_np}$ where r{nP} is distance from $q_n$ to P

By the principle of superimposition, the potential V at P due to system of charges is algebraic sum of the potentials due to the individual charges.

$$ V = V_1 + V_2 +…….V_n $$

$$ = \frac {1}{4πϵ0} \bigg[ \frac {q_1}{r{1P}} + \frac {q_2}{r_{2P}} + ……..+ \frac {q_n}{r_{3P}} \bigg] $$

$$ V = – \frac {1}{4πϵ0} \Sigma{i=1}^{n} \frac {q_i}{r_{iP}} $$

Example 2.2 Two charges $3 \times 10^{–8}$ C and $–2 \times 10^{–8}$ C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Solution:

Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be the x-axis; the negative charge is taken to be on the right side of the origin (Fig. 2.4).

Let P be the required point on the x-axis where the potential is zero. If x is the x-coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for x < 0.) If x lies between O and A, we have

$$ \frac {1}{4 \pi \epsilon_o} \bigg[ \frac {3 \times10^{-8}}{x \times 10^{-2}}- \frac {2 \times 10^{-8}}{(15-x) \times 10^{-2}} \bigg]=0$$

where x is in cm. That is, $ \frac {3}{x}- \frac {2}{15-x} = 0 $

which gives x = 9 cm.

If x lies on the extended line OA, the required condition is $ \frac {3}{x}- \frac {2}{15-x} = 0 $

which gives x = 45 cm Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.

Example 2.3 Figures 2.5 (a) and (b) show the field lines of a positive and negative point charge respectively.

(a) Give the signs of the potential difference $V_P – V_Q; V_B – V_A$.

(b) Give the sign of the potential energy difference of a small negative charge between the points Q and P; A and B.

(c) Give the sign of the work done by the field in moving a small positive charge from Q to P.

(d) Give the sign of the work done by the external agency in moving a small negative charge from B to A. (e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A?

Solution:

(a) As V ∝ $ \frac {1}{r}$, $V_P > V_Q$. Thus, ($V_P – V_Q$) is positive. Also $V_B$ is less negative than $V_A$. Thus, $V_B > V_A$ or ($V_B – V_A$) is positive.

(b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly, $(P.E.)_A$ > $(P.E.)_B$ and hence sign of potential energy differences is positive.

(c) In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative.

(d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive.

(e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A.

Questions for section 2.5

1. Derive an equation for the potential due to a system of charges.

2.6 Equipotential surfaces

An equipotential surface is a surface with constant value of potential at all points on the surface.

The potential difference between any two points on the surface is zero. So, no work is done in moving a charge from one point to another on the surface.

For a single charge q, the potential is given by
$$ V = \frac {1}{4πϵ_0} \frac {q}{r} $$

The above equation implies that V is constant if r is constant.

Therefore, equipotential surfaces of a single charge are concentric spherical surfaces with charge at the centre.

Now the electric field lines for a single charge q are radial lines starting from or ending at the charge, depending on whether q is positive or negative. Clearly, the electric field at every point is normal to the equipotential surface passing through that point. This is true in general: for any charge configuration, equipotential surface through a point is normal to the electric field at that point. The proof of this statement is simple. If the field were not normal to the equipotential surface, it would have non-zero component along the surface. To move a unit test charge against the direction of the component of the field, work would have to be done. But this is in contradiction to the definition of an equipotential surface: there is no potential difference between any two points on the surface and no work is required to move a test charge on the surface. The electric field must, therefore, be normal to the equipotential surface at every point. Equipotential surfaces offer an alternative visual picture in addition to the picture of electric field lines around a charge configuration.

For a uniform electric field, the equipotential surfaces are planes perpendicular to the direction of electric field.

The equipotential surfaces for a dipole and for two identical positive charges are as shown in the following figure.

Relationship between electric field and potential

Consider two closely spaced equipotential surfaces, A and B with potential value V and (V-dV) respectively, where dV is the decrease in potential in the electric field. Let P and Q be the points on the surface B and A respectively and dx be the perpendicular distance between the points P and Q.

If a unit positive charge is moved along this perpendicular distance dx from the surface B to the surface A against the electric field, the work done is:

$$ dW = (\vec E.)(\vec dx)= E \space dx \space cos180˚ \hspace{10mm} dW = -Edx (against \space the \space field)$$

this work done equals the potential difference between the surfaces A and B.

$$ dW = V_A – V_B = dV $$

$$ dV = – E.dx \space or \space E = – \frac {dV}{dx} $$

The negative sign shows that the direction of the electric field is in the direction of decreasing potential.

Thus, electric field E at a point is equal to the negative potential gradient at that point.

Questions for section 2.6

1. What are equipotential surfaces?
2. Derive a relationship between electric field and electric potential.

2.7 Potential energy of a system of charges

Potential energy of a system of charges is defined as the total amount of work done in bringing the individual charges from their state of rest at infinity to constitute the system.

Two point charges

Two point charges $q_1$ and $q_2$ separated by a distance r.

Let $q_14 be moved from infinity to the point A. no work is done as the charge $q_1$ is moved in the free region. With the charge $q_1$ fixed at A, let the other charge be moved from infinity to the point B against the field due to charge $q_1$. The work done during this process is:

$W = V_1q_2$ where $V_1$ is electric potential at B due to charge $q_1$.

Wkt , $ V_1 = \frac {1}{4\pi ε_0} \frac {q_1}{r} $

Therefore, W $ = \frac {1}{4\pi ε_0} \frac {q_1}{r} q_2$

By definition, $ U =\frac {1}{ 4\pi ε_0} \frac {q_1q_2}{r} $

Three point charges

The three point charge system is considered with 3 point charges $q_1$, $q_2$, $q_3$, located at A, B and C respectively in free space. Let the distances be AB=$r_1 $, BC=$r_2$, CA=$r_3$. Work done in bringing the charge q1 from infinity to A is zero. Work done in bringing the charge q2 from infinity to point B in the field of charge $q_1$ is
$$ W_1=\frac {1}{4\pi ε_0} \frac {q_1q_2}{r} $$

Work done in bringing the charge $q_3$ from infinity to point C in the field of $q_1$ and $q_2$ is $W_2 = (V_1 + V_2) q_3 $

Where $V_1$ and $V_2$ are the potentials at C due to $q_1$ and $q_2$ respectively.

$$ W_2 = \frac {1}{ \bigg[ \frac {q_1}{r_3}+ \frac {q_2}{r_2} \bigg] = \frac {1}{ \bigg[ \frac {q_2q_3}{r_2} + \frac {q_3q_2}{r_3} \bigg] $$

Network done in the formation of system is $ W=W_1+W_2 = \frac {1}{4πϵ_0} \bigg[ \frac{q_1q_2}{r_1} \bigg] +\frac {1}{4πϵ_0} \bigg[ \frac{q_2q_3}{r_2} + \frac {q_3q_1}{r_3} \bigg] $

$$ W = \frac {1}{4πϵ_0} \bigg[ \frac {q_1q_2}{r_1} + \frac {q_2q_3}{r_2}+ \frac {q_3q_1}{r_3} \bigg] $$

Thus electrostatic potential of the system of three point charges is
$$ U = \frac {1}{4πϵ_0} \bigg[ \frac {q_1q_2}{r_1} + \frac {q_2q_3}{r_2}+ \frac {q_3q_1}{r_3} \bigg] $$

Example 2.4 Four charges are arranged at the corners of a square ABCD of side d, as shown in Fig. 2.12.(a) Find the work required to put together this arrangement. (b) A charge $q_0$ is brought to the centre E of the square, the four charges being held fixed at its corners. How much extra work is needed to do this?

Solution (a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A, and then the charges –q, +q, and –q are brought to B, C and D, respectively. The total work needed can be calculated in steps:

(i) Work needed to bring charge +q to A when no charge is present elsewhere: this is zero.

(ii) Work needed to bring –q to B when +q is at A. This is given by (charge at B) × (electrostatic potential at B due to charge +q at A)
$$ = – q \times \bigg( \frac {q}{4 \pi \epsilon_o d } = – \frac {q^2}{4 \pi \epsilon_od} \bigg)$$

(iii) Work needed to bring charge +q to C when +q is at A and –q is at B. This is given by (charge at C) × (potential at C due to charges at A and B)
$$ = + q \times \bigg( \frac {+q}{4 \pi \epsilon_o d \sqrt 2} + \frac {-q}{4 \pi \epsilon_o d } \bigg)$$
$$ \frac {-q^2}{4 \pi \epsilon_o d } \bigg( 2 – \frac {1}{\sqrt 2} \bigg)$$

(iv) Work needed to bring –q to D when +q at A,–q at B, and +q at C. This is given by (charge at D) × (potential at D due to charges at A, B and C)

$$ = – q \times \bigg( \frac {+q}{4 \pi \epsilon_o d } + \frac {-q}{4 \pi \epsilon_o d \sqrt 2} + \frac {q}{4 \pi \epsilon_o d } \bigg)$$
$$ = \frac {-q^2}{4 \pi \epsilon_od} \bigg( 2 – \frac {1}{\sqrt 2} \bigg)$$
Add the work done in steps (i), (ii), (iii) and (iv). The total work required is
$$ = \frac {-q^3}{4 \pi \epsilon_o d} \bigg{ (0) +(1) + \bigg( 1 – \frac {1}{\sqrt 2} \bigg) + \bigg( 2 – \frac {1}{ \sqrt 2} \bigg) \bigg}$$
$$ = \frac {-q^2}{4 \pi \epsilon_o d} (4 – \sqrt 2)$$

The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges. (Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.) (b) The extra work necessary to bring a charge q0 to the point E when the four charges are at A, B, C and D is q0 × (electrostatic potential at E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D. Hence no work is required to bring any charge to point E.

Questions for section 2.7

1. Derive an expression for the potential energy of a system due to two point charges.
2. Derive an expression for the potential energy of a system due to three point charges.
   2.8 Potential energy in an external field

The external field E is not produced by the given charge(s) whose potential energy we wish to calculate. E is produced by sources external to the given charge(s).The external sources may be known, but often they are unknown or unspecified; what is specified is the electric field E or the electrostatic potential V due to the external sources. We assume that the charge q does not significantly affect the sources producing the external field. This is true if q is very small, or the external sources are held fixed by other unspecified forces. Even if q is finite, its influence on the external sources may still be ignored in the situation when very strong sources far away at infinity produce a finite field E in the region of interest.

The external electric field E and the corresponding external potential V may vary from point to point. By definition, V at a point P is the work done in bringing a unit positive charge from infinity to the point P.

Thus, work done in bringing a charge q from infinity to the point P in the external field is qV.

This work is stored in the form of potential energy of q. If the point P has position vector r relative to some origin, we can write: Potential energy of q at r in an external field
= qV(r)

Where V(r) is the external potential at the point r.

If an electron of charge q=1.6×10$^{-19}$ C is accelerated by a potential difference of 1V, the energy gained by it is $ U = qV = 1.6×10^{-19}x1 = 1.6×10^{-19} \space J $

This energy is defined as 1 electron volt or 1eV.

Therefore, $1eV = 1.6 \times 10^{-19}J$

Potential energy of a system of two charges in an external field

The work done in bringing the charge $q_1$ from infinity to $r_1$ is calculated. Work done in this step is $q_1$ V. We then consider the work done in bringing $q_2$ to $r_2$. In this step, work is done not only against the external field E but also against the field due to $q_1$. Work done on $q_2$ against the external field $ = q_2 V (r_2) $

Work done on $q_2$ against the field due to $q_1$
$$ = \frac {1}{4πϵ_0} \frac {q_1q_2}{r} $$

Therefore, total work done on $q_2$ against two electric fields is, $ q_2V_2 + \frac {1}{4πϵ_0} \frac {q_1q_2}{r} $

The potential energy of the system of two charges in an external electric field is equal to the total amount of work done in assembling the configuration of two charges in the external electric field.
$ U = q_1 V_1 + q_2V_2 + \frac {1}{4πϵ_0} \frac {q_1q_2}{r} $

Potential energy of a dipole in an external electric field

Consider a dipole with charges $q_1$= +q and $q_2$= –q placed in a uniform electric field E separated by a distance 2a. The dipole moment is P=qx2a. Let the dipole be placed in an uniform electric field $ (\vec E )$with its dipole moment $ (\vec P )$making an angle θ with the direction of the electric field $ (\vec E )$ .The dipole experiences a $ \tau =PE sinθ $ .This torque will tend to rotate the dipole to align it with the electric field.

Let an external torque of the same magnitude be applied in the opposite direction such that it neutralizes this torque and rotates the dipole through a small angle dθ without angular acceleration. The small amount of work done during this process is
$$ dW = $\tau dθ = PE \space sinθdθ $$.

Therefore, the total amount of work done in rotating the dipole from $θ1$ to $θ_2$ is $W = = \int_{0}^{\frac{\pi}{2}} PE \sin\theta \, d\theta = PE[-cosθ]$ *** application of limits

$$ W = \int{ θ1}^{ θ_2} PE \space sinθdθ = PE[-cosθ]{ θ_1}^{ θ_2} $$

$$ W = -PE [cosθ_2- cosθ_1} $$

This work is stored as the potential energy of the system.

Definition box   The potential energy of a dipole in an electric field is defined as the amount of work done against the electric field in bringing the dipole from infinity and placing it in the desired orientation in the field.

The work so done is zero when the dipole is placed perpendicular to the field. Work is done only in rotating the dipole from the position perpendicular to the field to any other position.

Taking $θ_1$ = π/2 and $θ_2$ = θ,

W = -PE( cosθ – cos π /2)

$W = -PE \space cosθ $, $W=- \vec P. \vec E $

Potential energy of a dipole placed at an angle θ to the direction of the electric field $ \vec E $

$$ U = – PE \space cosθ = – \vec P. \vec E $$

Example 2.5

(a) Determine the electrostatic potential energy of a system consisting of two charges 7 µC and –2 µC (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively.

(b) How much work is required to separate the two charges infinitely away from each other?

(c) Suppose that the same system of charges is now placed in an external electric field E = A (1/r 2 ); A = 9 × 105 C $m^{–2}$.

What would the electrostatic energy of the configuration be?

Solution:

(a) $$ U = \frac {1}{4 \pi \epsilon_o}{q_1q_2}{r} = 9 \times 10^9 \times \frac {7 \times (-2) \times 10^{-12}}{0.18} = -0.7 J$$

(b) W = $U_2 – U_1$ = 0 – U = 0 – (–0.7) = 0.7 J

(c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find, $ q_1V(r_1) + q_2 V(r_2) = A \frac {7µC}{0.09m} +A \frac {-2µC}{0.09m}$ and the net electrostatic energy is $ q_1 V (r_1) + q_2 V(r_2) + \frac {q_1q_2}{4 \pi \epsilon_o r_{12} = A \frac {7µC}{0.09m} + A \frac {-2µC}{0.09m} – 0.7J$ =− 70 20 0.7 49.3 J

Example 2.6 A molecule of a substance has a permanent electric dipole moment of magnitude 10$^{–29}$ C m. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude 106 V m–1. The direction of the field is suddenly changed by an angle of 60º. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarization of the sample.

Solution Here, dipole moment of each molecule = 10$^{–29}$ C m As 1 mole of the substance contains 6 × 10$^{23}$ molecules, total dipole moment of all the molecules, p

= 6 × $10^{23}$ × 10$^{–29}$ C m

= 6 × $10^{–6}$ C m

Initial potential energy, $U_i $= –pE cos θ = –6×$10^{–6}$×$10^6$ cos 0° = –6 J

Final potential energy (when θ = 60°), $U_f$ = –6 × $10^{–6}$ × $10^6$ cos 60° = –3 J

Change in potential energy = –3 J – (–6J) = 3 J So, there is loss in potential energy. This must be the energy released by the substance in the form of heat in aligning its dipoles.

Questions for section 2.8

1. What is 1 electron volt?
2. Derive an expression for Potential energy of a system of two charges in an external field.
3. Derive an expression for Potential energy of a dipole in an external electric field with a neat diagram.

2.9 Electrostatics of conductors

What are conductors?

Conductors are the substances that are used to carry electric charges from one place to another. They contain mobile charge carriers.

In metallic conductors, these charge carriers are electrons. In a metal, the outer (valence) electrons part away from their atoms and are free to move. These electrons are free within the metal but not free to leave the metal. The free electrons form a kind of ‘gas’; they collide with each other and with the ions, and move randomly in different directions. In an external electric field, they drift against the direction of the field.

The positive ions made up of the nuclei and the bound electrons remain held in their fixed positions. In electrolytic conductors, the charge carriers are both positive and negative ions; but the situation in this case is more involved – the movement of the charge carriers is affected both by the external electric field as also by the so-called chemical forces.

Inside the conductor, the electric field is zero

A conductor has free electrons. As long as electric field is not zero, the free charge carriers would experience force and drift. In the static situation, the free charges have so distributed themselves that the electric field is zero everywhere inside. Electrostatic field is zero inside a conductor.

At the surface of a charged conductor, electric field must be normal to the surface at every point

If the electric field is not normal to the surface, it will have a component tangential to the surface which will immediately cause the flow of charges, producing surface currents. But, no such currents can exist under static conditions. Hence, electric field is normal to the surface of conductor at every point.

The interior of a conductor can have no excess charge in the static situation

What is a Gaussian surface?
A Gaussian surface is a closed surface in three-dimensional space through which the flux of a vector field is calculated; usually the gravitational field, the electric field, or magnetic field. It is an arbitrary closed surface used in conjunction with Gauss’s law for the corresponding field by performing a surface integral, in order to calculate the total amount of the source quantity enclosed, i.e. amount of gravitational mass as the source of the gravitational field or amount of electric charge as the source of the electrostatic field, or vice versa.

A neutral conductor has equal amounts of positive and negative charges in every small volume or surface element. When the conductor is charged, the excess charge can reside only on the surface in the static situation. This follows from the Gauss’s law.
We consider a Gaussian surface inside the conductor just near its outer boundary. As the field is zero at all the points inside the conductor, the flux through the Gaussian surface is zero.

According to Gauss theorem, $ \phi = \int \vec E \vec {ds} = \frac {q}{ϵ_0} \hspace{10mm} \phi =0 \space ,q=0 $

Hence, there can be no charge in the interior of the conductor.

Electrostatic potential is constant throughout the volume of the conductor and has the same value as that on its surface

Electric field at any point is equal to the negative of the potential gradient.

$$ E = – \frac {dV}{dx} $$

But, wkt inside the conductor, E=0.

Hence, $ \frac {dV}{dx} =0 $ because V=constant

Hence electric potential is constant through out the volume of the conductor and has the same value on its surface.

Electric field at the surface of a charged conductor $ \vec E = \frac { \sigma}{ϵ_0} \hat n$

σ is the surface charge density and is $ \vec n $ a unit vector normal to the surface in the outward direction.

Consider a pill box (a short cylinder) as the Gaussian surface about any point P on the surface, as shown in figure. The pill box is partly inside and partly outside the surface of the conductor. It has a small area of cross δ S and negligible height.

Just inside the surface, the electrostatic field is zero; just outside, the field is normal to the surface with magnitude E. Thus, the contribution to the total flux through the pill box comes only from the outside (circular) cross- of the pill box. This equals ± EδS (positive for σ > 0, negative for σ < 0), since over the small area δS, E may be considered constant and E and δS are parallel or antiparallel. The charge enclosed by the pill box is σδS.

By Gauss’s law

$$ EδS = \frac {|σ|δS}{ϵ_0} $$ $$

E =\frac {|σ|}{ϵ_0} $$

Including the fact that electric field is normal to the surface, we get the vector relation, which is true for both signs of σ. For σ > 0, electric field is normal to the surface outward; for σ < 0, electric field is normal to the surface inward.

Electrostatic shielding

Let us consider a conductor with a cavity, with no charges inside. A remarkable result is that the electric field inside the cavity is zero, whatever be the size and shape of the cavity and whatever be the charge on the conductor and the external fields in which it might be placed.

We know that the electric field inside a charged spherical shell is zero. The proof of the result for the shell makes use of the spherical symmetry of the shell, but the vanishing of electric field in the cavity of a conductor is a very general result. A related result is that even if the conductor is charged or charges are induced on a neutral conductor by an external field, all charges reside only on the outer surface of a conductor with cavity.

Whatever be the charge and field configuration outside, any cavity in a conductor remains shielded from outside electric influence: the field inside the cavity is always zero. This is known as electrostatic shielding. The effect can be made use of in protecting sensitive instruments from outside electrical influence. Figure gives a summary of the important electrostatic properties of a conductor.

Example 2.7

(a) A comb run through one’s dry hair attracts small bits of paper. Why? What happens if the hair is wet or if it is a rainy day? (Remember, a paper does not conduct electricity.)

(b) Ordinary rubber is an insulator. But special rubber tyres of aircraft are made slightly conducting. Why is this necessary?

(c) Vehicles carrying inflammable materials usually have metallic ropes touching the ground during motion. Why?

(d) A bird perches on a bare high power line, and nothing happens to the bird. A man standing on the ground touches the same line and gets a fatal shock. Why?

Solution (a) This is because the comb gets charged by friction. The molecules in the paper gets polarised by the charged comb, resulting in a net force of attraction. If the hair is wet, or if it is rainy day, friction between hair and the comb reduces. The comb does not get charged and thus it will not attract small bits of paper. (b) To enable them to conduct charge (produced by friction) to the ground; as too much of static electricity accumulated may result in spark and result in fire. (c) Reason similar to (b). (d) Current passes only when there is difference in potential.

Questions for section 2.9

1. What are conductors? What is the state of electric field inside and on the surface of it?
2. What is a Gaussian surface?
3. Derive an expression for Electric field at the surface of a charged conductor.
4. Explain in brief, the electrostatic shielding.

2.10 Dielectrics and polarization

Basically, dielectrics are insulators and hence do not conduct electricity. However, they greatly affect the electric field in which they are placed.

The molecules of dielectric are of two types:

a. Polar molecules

b. Non-polar molecules

A polar molecule is the one in which the centers of positive and negative charge distributions are separated by a small distance. It acts like a tiny electric dipole and possesses permanent electric dipole moment.

The orientation of molecules becomes random in the absence of external electric field, so the average dipole moment per unit volume of the dielectric is zero.

Eg. $N_2O$, $NH_3$, $H_2O$, HCl, CO etc.

In a non-polar molecule, the centre of positive and negative charge distributions coincide; it has no permanent dipole moment. However, in the presence of an electric field, the positive and negative charges are pulled in opposite directions till the forces acting on them due to external field are balanced by the internal forces.

Eg. $O_2$, $N_2$, $H_2$, $CO_2$ etc.

When a dielectric slab is placed in a uniform electric field, permanent or induced dipole moment orients itself in the direction of field; the alignment is due to the torque acting on each dipole. The degree of alignment depends upon the temperature and strength of the electric field.

The alignment of dipole moments results in a layer of positive charges on one side and negative on another side of the dielectric slab. These are induced or polarized charges. Under this condition, the dielectric slab is said to be polarized. The phenomenon is called electric polarization.

Due to the polarization, an electric field is set up inside the slab. The internal field opposes the applied electric field. The reduced net electric field in the region is given by $ E = E_a – E_i $

Questions for section 2.10

1. What are the types of dielectric molecules?
2. What are polar and non-polar molecules?
3. What happens when a dielectric slab is placed in uniform electric field?

2.11 Capacitors and capacitances

A capacitor is a system of two conductors separated by an insulator. Capacitance is the ability of the system to store charges. The conductors have charges; say $Q_1$ and $Q_2$, and potentials $V_1$ and $V_2$. Generally the two conductors have charges +Q and – Q, with potential difference V=$V_1 – V_2$ between them.
We shall consider only this kind of charge configuration of the capacitor. The conductors may be charged by connecting them to the two terminals of a battery. Q is called the charge of the capacitor, though this, in fact, is the charge on one of the conductors – the total charge of the capacitor is zero.

The electric field in the region between the conductors is proportional to the charge Q. Now, potential difference V is the work done per unit positive charge in taking a small test charge from the conductor 2 to 1 against the field.

Consequently, V is also proportional to Q, and the ratio Q/V is a constant.

$$ C = \frac {Q}{V} $$

The term C is called the capacitance of the capacitor. It depends only on the geometrical configuration (shape, size, separation) of the system of two conductors.

The maximum electric field that a dielectric medium can withstand without break-down) is called its dielectric strength.

The dielectric strength of air medium is $3 \times 10^6 \space Vm^{-1}$. Separation between conductors of the order of 1 cm or so, this field corresponds to a potential difference of $3 \times 10^4$ V between the conductors. Thus, for a capacitor to store a large amount of charge without leaking, its capacitance should be high enough so that the potential difference and hence the electric field do not exceed the break-down limits.

Questions for section 2.11

1. Define capacitor and capacitance. Give an equation for the same.
2. Define dielectric strength.

2.12 The parallel plate capacitor

A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance. Let A be the area of each plate and d the separation between them. The two plates have charges Q and –Q. Since d is much smaller than the linear dimension of the plates ($d_2$<< A), we can use the result on electric field by an infinite plane sheet of uniform surface charge density. Plate 1 has surface charge density σ = Q/A and plate 2 has a surface charge density –σ. The electric field in different regions is

For outer region 1,
$$ E = \frac {σ}{2ϵ_0 } – \frac {σ}{2ϵ_0 } = 0 $$

For outer region 2,
$$ E = \frac {σ}{2ϵ_0 } – \frac {σ}{2ϵ_0 } = 0 $$

In the inner region between the plates 1 and 2, the electric fields due to the two charged plates add up, giving
$$ E = \frac {σ}{2ϵ_0 }+ \frac {σ}{2ϵ_0} = \frac {σ}{ϵ_0} = \frac {Q}{ϵ_0 A} $$

The direction of electric field is from the positive to the negative plate. Thus, the electric field is localized between the two plates and is uniform throughout.

For plates with finite area, this will not be true near the outer boundaries of the plates. The field lines bend outward at the edges – an effect called ‘fringing of the field’. By the same token, σ will not be strictly uniform on the entire plate. However, for $d_2$ << A, these effects can be ignored in the regions sufficiently far from the edges, and the field there. For uniform electric field, potential difference is simply the electric field times the distance between the plates, that is,
$$ V = Ed = \frac {1}{ϵ_0} \frac {Qd}{A} $$

The capacitance C of the parallel plate capacitor is then
$$ C = \frac {Q}{V} = \frac {ϵ_0 A}{d} $$

Thus, the capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to to the distance between two plates.

Questions for section 2.12

1. What is a parallel plate capacitor?
2. Derive an equation for the capacitance of a parallel plate capacitor.

2.13 Effect of dielectrics on capacitance

A parallel plate capacitor having two large plates, each of area A, separated by a small distance d is considered. Let the charge on the two plates be +Q and –Q , while surface densities be +σ and –σ respectively.

If medium between the two plates is air,

Electric field is $E_0 = σ/ϵ_o$

Potential difference is $V_o = E_0d = σd/E_o$
$$ C = \frac {Q}{V}= \frac {ϵ_0 A}{d} $$

For linear dielectrics, $σ_P$ is proportional to $E_o$ and hence proportional to σ. Thus, (σ – $σ_P$ ) is proportional to σ.

σ – $σ_P$ = σ/K

Then, $ V = \frac {σd}{ϵ_0 K} = \frac {Qd}{ϵ_0 KA} $

The capacitance C with dielectric between two plates is $ C = \frac {Q}{V} = \frac {Q}{ \frac {Qd}{ϵ_o KA}} = \frac {Kϵ_0 A}{d} $

Where $ϵ_oK$ is called permittivity of the medium and is denoted by ϵ.

Dielectric constant of a substance is a factor by which the capacitor increases from its vacuum value.

Example 2.8 A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?
Solution Let $E_0 = V_0/d$ be the electric field between the plates when there is no dielectric and the potential difference is $V_0$. If the dielectric is now inserted, the electric field in the dielectric will be E = $E_0$/K. The potential difference will then be

$$ V = E_o ( \frac {1}{4}d) + \frac {E_o}{K} ( \frac {3}{4} d)$$

$$ = E_od ( \frac {1}{4} + \frac {3}{4K}) = V_o \frac {K+3}{4K}$$

The potential difference decreases by the factor (K + 3)/4K while the free charge $Q_0$ on the plates remains unchanged. The capacitance thus increases

$$ C = \frac {Q_o}{V} = \frac {4K}{K+3} \frac {Q_o}V_o} = \frac {4K}{K+3}C_o$$

Questions for section 2.13

1. Derive an equation for the effect of dielectrics on capacitance.

2.14 Combination of capacitors

We can combine several capacitors of capacitance $C_1$, $C_2$ …… $C_n$ obtain a system with some effective capacitance C. The effective capacitance depends on the way the individual capacitors are combined. The two possible combinations are

1. Capacitors in series

The left plate of $C_1$ and the right plate of $C_2$ are connected to two terminals of a battery and have charges Q and –Q , respectively. It then follows that the right plate of $C_1$ has charge –Q and the left plate of $C_2$ has charge Q. If this was not so, the net charge on each capacitor would not be zero. This would result in an electric field in the conductor connecting $C_1$ and $C_2$. Charge would flow until the net charge on both $C_1$ and $C_2$ is zero and there is no electric field in the conductor connecting $C_1$ and $C_2$. Thus, in the series combination, charges on the two plates (±Q) are the same on each capacitor. The total potential drop V across the combination is the sum of the potential drops $V_1$ and $V_2$ across $C_1$ and $C_2$, respectively.

$$ V = V_1 + V_2 = \frac{Q}{C_1} + \frac {Q}{C_2} $$

$$ \frac {V}{Q} = \frac {1}{C_1} + \frac {1}{C_2} $$

$$ \frac {1}{C} = \frac {1}{C_1} + \frac { 1}{C_2} $$

The proof clearly goes through for any number of capacitors arranged in a similar way. Equation for n capacitors arranged in series, generalizes to

$$ V = V_1 + V_2 + …… + V_n = \frac {Q}{C_1} + \frac {Q}{C_2} + ……. + \frac {Q}{C_n} $$

$$ \frac {1}{C} = \frac {1}{C_1} + \frac {1}{C_2} + ……. + \frac {1}{C_n} $$

2. Capacitors in parallel

Figure shows two capacitors arranged in parallel. In this case, the same potential difference is applied across both the capacitors. It is seen that the voltage across both the capacitors are the same as they are in parallel. But the plate charges (±$Q_1$) on capacitor 1 and the plate charges (±$Q_2$) on the capacitor 2 are not necessarily the same: $ $Q_1 =C_1V$, $ Q_2 = C_2V

The equivalent capacitor is one with charge $ Q = Q_1 + Q_2 $ and potential difference V.

$$ Q = CV = C_1V + C_2V $$

The effective capacitance C is, $ C = C_1 + C_2 $

The general formula for effective capacitance C for parallel combination of n capacitors is given by
$$ Q = Q_1 + Q_2+ … + Q_n $$

i.e., $ CV = C_1V + C_2V + …… + C_nV $

which gives $C = C_1+ C_2 + …… + C_ n $4

The equivalent capacitance of a number of capacitors in parallel is equal to sum of their individual capacitances.

Example 2.9 A network of four 10 µF capacitors is connected to a 500 V supply, as shown in Fig. 2.29. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor. (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.)

(a) In the given network, $C_1, C_2$ and $C_3$ are connected in series. The effective capacitance C′ of these three capacitors is given by
$$ \frac {1}{C’} = \frac {1}{C_1} +\frac {1}{C_2} +\frac {1}{C_3}$$

For $C_1 = C_2 = C_3$ = 10 µF, C′ = (10/3) µF. The network has C′ and $C_4$ connected in parallel. Thus, the equivalent capacitance C of the network is

$$ C = C’ + C_4 = \bigg( \frac {10}{3} + 10 \bigg) µF = 13.3 µF $$

(b) Clearly, from the figure, the charge on each of the capacitors, $C_1, C_2$ and $C_3$ is the same, say Q. Let the charge on $C_4$ be Q′. Now, since the potential difference across AB is Q/$C_1$, across BC is Q/$C_2$, across CD is Q/$C_3$ , we have $ \frac {Q}{C_1} +\frac {Q}{C_2} +\frac {Q}{C_3} = 500 \space V $

Also, Q′/$C_4$ = 500 V.

This gives for the given value of the capacitances,
$ Q = 500 \space V \times \frac {10}{3} µF = 1.7 \times 10^{-3}C $ and $ Q’ = 500 \space V \times 10 µF = 5.0 \times 10^{-3}C $

Questions for section 2.14

1. Derive an expression for the equivalent capacitance when the capacitors are connected in series.
2. Derive an expression for the equivalent capacitance when the capacitors are connected in parallel.

2.15 Energy stored in a capacitor

Consider a capacitor with plates A and B connected to a battery of volt V and a plug key K. When the key is closed, free electrons in the conductor A are pulled out of A and transferred to B through the battery. So, plate A gets positively charged, whereas plate B gets negatively charged to the same extent. The flow of electrons from A to B through the battery will continue indefinitely. The potential difference between the plates becomes equal to the battery potential V. The capacitor is said to be fully charged. During the process of charging, work is done. This energy is stored as electrostatic potential energy in the capacitor.

Energy stored in the capacitor = Amount of work done in charging the capacitor

At any time t, let q be the charge on the plates and V be the potential difference between the plates then,
q = CV
Where C – capacitance of the capacitor

The amount of work done in transferring an additional quantity of charge dq against the potential difference V is $ dW = Vdq = \bigg (\frac {q}{C} \bigg)dq $

The total amount of work done in transferring charge Q is given by
$$ W = \int_{0}^{Q} \frac {q}{C} dq = \frac {1}{C} \int_{0}^{Q} q \space dq $$
$$ = \frac {1}{C} \bigg[ \frac {q^2}{2} \bigg]_{0}^{Q} = \frac {Q^2}{2C} $$

The amount of work done is stored as electrostatic potential energy U in the capacitor. Thus,
$$ U = \frac {Q^2}{2C} $$

As Q=CV,
$$ U = \frac {CV^2}{2} $$
$$ U = \frac {Q^2}{2C} $$
$$ = \frac {QV}{2} $$

Example 2.10

(a) A 900 pF capacitor is charged by 100 V battery [Fig. 2.21(a)]. How much electrostatic energy is stored by the capacitor?

(b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor [Fig. 2.21(b)]. What is the electrostatic energy stored by the system?

Solution:

(a) The charge on the capacitor is
Q = CV = 900 × $10^{–12}$ F × 100 V = 9 × $10^{–8}$ C
The energy stored by the capacitor is
$$ = (1/2) CV^2 = (1/2) QV$$
$$ = (1/2) × 9 × 10^{–8}C × 100 V = 4.5 × 10^{–6} J$$

(b) In the steady situation, the two capacitors have their positive plates at the same potential, and their negative plates at the same potential. Let the common potential difference be V′. The charge on each capacitor is then Q′ = CV′. By charge conservation, Q′ = Q/2. This implies V′ = V/2. The total energy of the system is
$$ = 2 \times \frac {1}{2} Q’V’ = \frac {1}{4}QV = 2.25 \times 10^{-6}J$$

Thus in going from (a) to (b), though no charge is lost; the final energy is only half the initial energy. Where has the remaining energy gone? There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in the form of heat and electromagnetic radiation.

Questions for section 2.15

1. Derive an expression for the energy stored in a capacitor.

2.16 Van De Graff Generator

This is a machine that can build up high voltages of the order of a few million volts. The resulting large electric fields are used to accelerate charged particles (electrons, protons, ions) to high energies needed for experiments to probe the small scale structure of matter.

The principle underlying the machine is as follows: Suppose we have a large spherical conducting shell radius R, on which we place a charge Q. This charge spreads itself uniformly all over the sphere. The field outside the sphere is just that of a point charge, Q at the centre; while the field inside the sphere vanishes. So the potential outside is that of a point charge; and inside it is constant, namely the value at the radius R. We thus have:
Potential inside conducting spherical shell of radius R carrying charge Q = constant = $ \frac {1}{4 \pi ϵ_0} \frac {Q}{R}$

Let us suppose that in some way we introduce a small sphere of radius r, carrying some charge q, into the large one, and place it at the centre. The potential due to this new charge clearly has the following values at the radii indicated: Potential due to small sphere of radius r carrying charge

$q = \frac {1}{4πϵ_0} \frac {q}{r} $at surface of small spheres

$q = \frac {1}{4πϵ_0} \frac {q}{R} $at large shall of radius R
Taking both charges q and Q into account we have for the total potential V and the potential difference the values

$$ V ( R ) =\frac {1}{4πϵ_0} \bigg( \frac {Q}{R} + \frac {q}{R} \bigg) $$

$$ V ( r ) =\frac {1}{4πϵ_0} \bigg( \frac {Q}{R} + \frac {q}{r} \bigg) $$

$$ V( r) – V(R) =\frac {1}{4πϵ_0} \bigg( \frac {1}{r} – \frac {1}{R} \bigg) $$

Assume now that q is positive. We see that, independent of the amount of charge Q that may have accumulated on the larger sphere and even if it is positive, the inner sphere is always at a higher potential: the difference V(r )–V(R) is positive.

The potential due to Q is constant upto radius R and so cancels out in the difference. This means that if we now connect the smaller and larger sphere by a wire, the charge q on the former will immediately flow onto the matter, even though the charge Q may be quite large. The natural tendency is for positive charge to move from higher to lower potential.

Thus, provided we are somehow able to introduce the small charged sphere into the larger one, we can in this way keep piling up larger and larger amount of charge on the latter. The potential at the outer sphere would also keep rising, at least until we reach the breakdown field of air.

This is the principle of the van de Graaff generator. It is a machine capable of building up potential difference of a few million volts, and fields close to the breakdown field of air which is about 3 × 106 V/m.

A schematic diagram of the van de Graaff generator is given in the figure above. A large spherical conducting shell (of few metres radius) is supported at a height several meters above the ground on an insulating column. A long narrow endless belt insulating material, like rubber or silk, is wound around two pulleys – one at ground level, one at the centre of the shell. This belt is kept continuously moving by a motor driving the lower pulley. It continuously carries positive charge, sprayed on to it by a brush at ground level, to the top. There it transfers its positive charge to another conducting brush connected to the large shell. Thus positive charge is transferred to the shell, where it spreads out uniformly on the outer surface. In this way, voltage differences of as much as 6 or 8 million volts (with respect to ground) can be built up.

Questions for section 2.16

1. With a neat diagram, explain the working of a Van de gaff generator.
2. Where is the van de gaff generator used?