JEE Main 2019 — Electrostatics Question with Solution

Initially when uncharged shell encloses charge Q, charge distribution due to induction will be as shown,

The potential on surface of inner shell is

$$V_A=\frac{kQ}{a}+\frac{k(-Q)}{b}+\frac{kQ}{b}$$ ..... (i)

where, k = proportionality constant.

Potential on surface of outer shell is

$$V_B=\frac{kQ}{b}+\frac{k(-Q)}{b}+\frac{kQ}{b}$$ ..... (ii)

Then, potential difference is

$$\Delta V_{AB}=V_A-V_B=kQ\left(\frac{1}{a}-\frac{1}{b}\right)$$

Given, $$\Delta V_{AB}=V$$

So, $$kQ\left(\frac{1}{a}-\frac{1}{b}\right)=V$$ ....... (iii)

Finally after giving charge $$-$$ 4Q to outer shell, potential difference will be

$$\Delta V_{AB}=V_A-V_B$$

$$=\left(\frac{kQ}{a}+\frac{k(-4Q)}{b}\right)-\left(\frac{kQ}{b}+\frac{k(-4Q)}{b}\right)$$

$$=kQ\left(\frac{1}{a}-\frac{1}{b}\right)=V$$ [from Eq. (iii)]

Hence, we obtain that potential difference does not depend on the charge of outer sphere, hence potential difference remains same.