Question

A spherical shell of radius

R_{1}

with a uniform charge

q

, has a point charge

q_{0}

at its centre. Find the work performed by the electric forces during the shell expansion from radius

R_{1}

to radius

R_{2}

.

Accepted Answer

For the given question, the diagram can be drawn as shown:

As shown, the charge $q_{0}$ is at the centre and the charge $q$ is uniformly distributed over the shell of radius $R_{1}$ . Then, this shell is expanded to radius of $R_{2}$ , as shown in the diagram, and the work done in doing this is given by,

$W = - ∆ U = - (U_{f} - U_{i}) = U_{i} - U_{f} . . . (1)$

Initially, the energy of the system is, $U_{i} = W_{1} + W_{12}$ , where $W_{1}$ is the self energy and $W_{12}$ is the mutual energy.

So, $U_{i} = \frac{1}{2} \frac{q^{2}}{4 π ε_{0} R_{1}} + \frac{q q_{0}}{4 π ε_{0} R_{1}}$

On expansion of the shell to radius $R_{2}$ , the energy of the system is,
$U_{f} = W_{1}' + W_{12}'$

$= \frac{1}{2} \frac{q^{2}}{4 π ε_{0} R_{2}} + \frac{q q_{0}}{4 π ε_{0} R_{2}}$

Now, put the values of $U_{i}$ and $U_{f}$ in equation $(1)$ , we get,

$W = (U_{i} - U_{f}) = \frac{q (q_{0} + \frac{q}{2})}{4 π ε_{0}} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

Alternate Method: The work of electric forces is equal to the decrease in electric energy of the system,

$W = U_{i} - U_{f}$

In order to find the difference $U_{i} - U_{f}$ , we note that upon expansion of the shell, the work is done only in expanding the shell from $R_{1}$ to $R_{2}$ , as shown by shaded portion in the diagram,

$W = U_{i} - U_{f} = \int_{R_{1}}^{R_{2}} \frac{1}{2} (E_{1}^{2} - E_{2}^{2}) \cdot 4 π r^{2} d r$

where, $E_{1}$ and $E_{2}$ are the field intensities in the shaded region, at a distance $r$ ( $r$ is the arbitrary radius in the shaded region), from the centre of the system) before and after the expansion of the shell.

By using Gauss' theorem, we find,

$E_{1} = \frac{1}{4 π ε_{0}} \frac{q + q_{0}}{r^{2}} and E_{2} = \frac{1}{4 π ε_{0}} \frac{q_{0}}{r^{2}}$

As a result of integration, we obtain,
$W = \frac{q (q_{0} + \frac{q}{2})}{4 π ε_{0}} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

I E Irodov Solutions for Chapter: ELECTRODYNAMICS, Exercise 3: ELECTRIC CAPACITANCE. ENERGY OF AN ELECTRIC FIELD

I E Irodov Physics Solutions for Exercise - I E Irodov Solutions for Chapter: ELECTRODYNAMICS, Exercise 3: ELECTRIC CAPACITANCE. ENERGY OF AN ELECTRIC FIELD

Questions from I E Irodov Solutions for Chapter: ELECTRODYNAMICS, Exercise 3: ELECTRIC CAPACITANCE. ENERGY OF AN ELECTRIC FIELD with Hints & Solutions

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