A solid conducting sphere having a charge Q i | vedclass

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(A) Let the radius of the solid sphere be $r_a$ and the radius of the hollow shell be $r_b$.The potential at the surface of the solid sphere is $V_{sp

Vedclass · Accepted Answer

$V$

Vedclass · Answer

(A) Let the radius of the solid sphere be $r_a$ and the radius of the hollow shell be $r_b$.The potential at the surface of the solid sphere is $V_{sphere} = \frac{KQ}{r_a} + \frac{K(0)}{r_b} = \frac{KQ}{r_a}$.The potential at the surface of the hollow shell is $V_{shell} = \frac{KQ}{r_b} + \frac{K(0)}{r_b} = \frac{KQ}{r_b}$.The initial potential difference is $V = V_{sphere} - V_{shell} = K Q \left( \frac{1}{r_a} - \frac{1}{r_b} \right)$.When the shell is given a charge $-3Q$,the new potential at the surface of the solid sphere is $V'_{sphere} = \frac{KQ}{r_a} + \frac{K(-3Q)}{r_b}$.The new potential at the surface of the hollow shell is $V'_{shell} = \frac{KQ}{r_b} + \frac{K(-3Q)}{r_b} = -\frac{2KQ}{r_b}$.The new potential difference is $V' = V'_{sphere} - V'_{shell} = \left( \frac{KQ}{r_a} - \frac{3KQ}{r_b} \right) - \left( -\frac{2KQ}{r_b} \right) = \frac{KQ}{r_a} - \frac{KQ}{r_b} = V$.Thus,the potential difference remains unchanged.

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