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Question: A point charge q is brought from infinity (slowly so that heat developed in the shell is negligible)...

A point charge q is brought from infinity (slowly so that heat developed in the shell is negligible) and is placed at the centre of a conducting neutral spherical shell of inner radius a and outer radius b, then work done by external agent is:

A

0

B

kq22b\frac{kq^2}{2b}

C

kq22b−kq22a\frac{kq^2}{2b} - \frac{kq^2}{2a}

D

kq22a−kq22b\frac{kq^2}{2a} - \frac{kq^2}{2b}

Answer

kq22b−kq22a\frac{kq^2}{2b} - \frac{kq^2}{2a}

Explanation

Solution

The work done by the external agent to bring a charge qq from infinity and place it at the center of a conducting neutral spherical shell is equal to the change in the potential energy of the system.

When the charge qq is placed at the center of the conducting shell, a charge −q-q is induced on the inner surface (radius aa) and a charge +q+q is induced on the outer surface (radius bb).

The potential at the center due to the induced charge −q-q on the inner surface is k(−q)a\frac{k(-q)}{a}.
The potential at the center due to the induced charge +q+q on the outer surface is kqb\frac{kq}{b}.
So, the potential at the center due to the induced charges is V=−kqa+kqbV = \frac{-kq}{a} + \frac{kq}{b}.
The work done by the external agent to bring the charge qq from infinity to the center in the potential of the induced charges is qV=q(kqb−kqa)=kq2b−kq2aqV = q \left( \frac{kq}{b} - \frac{kq}{a} \right) = \frac{kq^2}{b} - \frac{kq^2}{a}.

However, the work done by the external agent is the change in the total energy of the system. The total energy of the system includes the energy of the point charge and the shell.

The total energy can be calculated as the sum of the potential energy of the point charge at the center in the potential created by the induced charges, plus the potential energy stored in the induced charges.

Potential energy of qq in the field of induced charges: q(kqb−kqa)=kq2b−kq2aq \left( \frac{kq}{b} - \frac{kq}{a} \right) = \frac{kq^2}{b} - \frac{kq^2}{a}.
Potential energy of induced charges: self-energy of −q-q on inner shell + self-energy of +q+q on outer shell + interaction energy between −q-q and +q+q.
Self-energy of −q-q on radius aa is k(−q)22a=kq22a\frac{k(-q)^2}{2a} = \frac{kq^2}{2a}.
Self-energy of +q+q on radius bb is k(+q)22b=kq22b\frac{k(+q)^2}{2b} = \frac{kq^2}{2b}.
Interaction energy between −q-q on inner shell and +q+q on outer shell is −q×(potential at r=a due to +q on outer shell)=−q×kqb=−kq2b-q \times (\text{potential at } r=a \text{ due to } +q \text{ on outer shell}) = -q \times \frac{kq}{b} = -\frac{kq^2}{b}.
Total potential energy of induced charges = kq22a+kq22b−kq2b=kq22a−kq22b\frac{kq^2}{2a} + \frac{kq^2}{2b} - \frac{kq^2}{b} = \frac{kq^2}{2a} - \frac{kq^2}{2b}.
Total potential energy of the system = (Potential energy of qq in the field of induced charges) + (Potential energy of induced charges)
Utotal=(kq2b−kq2a)+(kq22a−kq22b)=kq2b−kq22b−kq2a+kq22a=kq22b−kq22aU_{total} = \left( \frac{kq^2}{b} - \frac{kq^2}{a} \right) + \left( \frac{kq^2}{2a} - \frac{kq^2}{2b} \right) = \frac{kq^2}{b} - \frac{kq^2}{2b} - \frac{kq^2}{a} + \frac{kq^2}{2a} = \frac{kq^2}{2b} - \frac{kq^2}{2a}.

The work done by the external agent is equal to the total potential energy of the system in the final configuration.
W=Ufinal−Uinitial=(kq22b−kq22a)−0=kq22b−kq22aW = U_{final} - U_{initial} = \left( \frac{kq^2}{2b} - \frac{kq^2}{2a} \right) - 0 = \frac{kq^2}{2b} - \frac{kq^2}{2a}.