Distribution of Charges In a Conductor and Action at Points

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 r1 and r2 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.

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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.

Distribution Of Charges In A Conductor and Action At Points

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

VA = \(\frac{1}{4 \pi \epsilon} \frac{q_{1}}{r_{1}}\) ……….. (1.110)

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

VB = \(\frac{1}{4 \pi \epsilon_{0}} \frac{q_{2}}{r_{2}}\) ………… (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 \(\frac{q_{1}}{r_{1}}=\frac{q_{2}}{r_{2}}\) ……….. (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 = 4πr12σ1 and q2 = 4πr22. Substituting these values into equation (1.112), we get

σ1r1 = σ2r2 ……….. (1.113)

from which we conclude that

σr = constant ……………. (1.114)

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.

Distribution Of Charges In A Conductor and Action At Points

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.

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(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 VA = \(\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1}}{r_{1}}\)

The electrostatic potential on the surface of the sphere B is VB = \(\frac{1}{4 \pi \epsilon} \frac{q_{2}}{r_{2}}\)

Since VA = VB. We have

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

Q – q2 = (\(\frac{r_{1}}{r_{2}}\))q2

so that q2 = Q(\(\frac{r_{2}}{r_{1}+r_{2}}\))

Distribution Of Charges In A Conductor and Action At Points

Therefore,

q2 = 100 × 10-9 × (\(\frac{2}{10}\)) = 20 nC

and q1 = Q – q2 = 80 nC

The electric charge density on sphere A is σ1 = \(\frac{q_{1}}{4 \pi r_{1}^{2}}\)
The electric charge density on sphere B is σ2 = \(\frac{q_{2}}{4 \pi r_{2}^{2}}\)

Therefore,

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Note that the surface charge density is greater on the smaller sphere compared to the larger sphere (σ2 ~ 4σ1) which confirms the result \(\frac{\sigma_{1}}{\sigma_{2}}=\frac{r_{2}}{r_{1}}\)

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

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Distribution Of Charges In A Conductor and Action At Points

Action of points or Corona discharge

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

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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).

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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.

Distribution Of Charges In A Conductor and Action At Points

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 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.

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Distribution Of Charges In A Conductor and Action At Points

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 = \(\frac{1}{4 \pi \epsilon_{\circ}} \frac{Q}{R^{2}}\)

The potential on the surface of the hollow metallic sphere is given by
V = \(\frac{1}{4 \pi \epsilon_{0}}\) \(\frac{Q}{R}\) = ER

Since Vmax = EmaxR
Here Emax = 3 × 106Vm1. So the maximum potential difference created is given by

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

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