QJEEM19Phd51c: JEE Main 2019 Physics - Class 12 JEEM19Phd51c Solution: A solid conducting sphere, having a charge Q, is surrounded by…

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

The potential on surface of inner shell is

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

where, k = proportionality constant.

Potential on surface of outer shell is

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

Then, potential difference is

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

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

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

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

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

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

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

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