Question 581: Electrostats & Capacitance

Question

A solid conducting sphere, having a charge Q,
is surrounded by an uncharged conducting
hollow spherical shell. Let the potential
difference between the surface of the solid
sphere and that of the outer surface of the
hollow shell be V. If the shell is now given a
charge of –4 Q, the new potential difference
between the same two surfaces is :

Accepted Answer

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.

Answer

V

Answer

2V

Answer

–2V

Answer

4V

Electrostats & Capacitance

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