Question

A point charge $ q $ is a distance $ r $ from the centre O of an uncharged conducting layer whose inner and outer radii equal to $ a $ and $ b $ respectively. The potential at the point O if $ a < r $ is $ \dfrac{q}{{4\pi {\varepsilon _0}}} $ times
(A) $ \left( {\dfrac{1}{r} - \dfrac{1}{a} + \dfrac{1}{b}} \right) $
(B) $ \left( {\dfrac{1}{a} - \dfrac{1}{r} + \dfrac{1}{b}} \right) $
(C) $ \left( {\dfrac{1}{b} - \dfrac{1}{a} - \dfrac{1}{r}} \right) $
(D) $ \left( {\dfrac{1}{a} - \dfrac{1}{b} - \dfrac{1}{r}} \right) $

Answer 1

Hint : Since, the conducting sphere is uncharged the total charge in the sphere is equal to zero. The charge inside the layer will induce a negative charge at the inner layer, and hence induce a positive charge on the outer layer.

Formula used: In this solution we will be using the following formula;
 $ V = \dfrac{q}{{4\pi {\varepsilon _0}r}} $ where $ V $ is the electric potential, $ q $ is the charge creating an electric field, $ {\varepsilon _0} $ is the permittivity of free space and $ r $ is the distance of interest from the charge.

Complete step by step answer:
Now, a charge $ q $ was said to be placed inside an uncharged spherical layer at a distance $ r $ away from the centre. Hence, due to the charge $ q $ only at the centre is equal to
 $ V = \dfrac{q}{{4\pi {\varepsilon _0}r}} $
However, due to the charge, the inner layer will be induced with charge $ - q $ and since the layer is in general uncharged the outer layer will have a charge $ q $ . Hence, the total electric potential will be
 $ {V_t} = \dfrac{q}{{4\pi {\varepsilon _0}r}} + \dfrac{{ - q}}{{4\pi {\varepsilon _0}a}} + \dfrac{q}{{4\pi {\varepsilon _0}b}} $ where $ a $ is the inner radius of the sphere and $ b $ is the outer radius.
Hence, by factoring out all common quantities and constants, we have
 $ {V_t} = \dfrac{q}{{4\pi {\varepsilon _0}}}\left( {\dfrac{1}{r} - \dfrac{1}{a} + \dfrac{1}{b}} \right) $
Which means that $ \dfrac{q}{{4\pi {\varepsilon _0}}} $ multiplies $ \left( {\dfrac{1}{r} - \dfrac{1}{a} + \dfrac{1}{b}} \right) $ .
Thus, the correct option is A.

Note:
For clarity, the reason why the inner radius gets a negative charge is because of the attraction of the negative charges in the conducting layer to the positive charge close to the centre. Recall that a conducting layer, although may be generally neutral in charge, contains free electrons. The migration of the negative charge to the inner part of the sphere allows the outer layer to build up positive charges.