Question

A solid conducting sphere having a charge $Q$ is surrounded by an uncharged concentric 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 $-3Q$, the new potential difference between the same two surfaces is :

• A. $V$
• B. $2V$
• C. $-2V$
• D. $4V$

Answer 1

Answer: (A)

The electric field in between the shell and sphere is
$E.\pi x^2=\frac{Q_{en}}{{ϵ}_0}=\frac{Q}{{ϵ}_0}$ (using Gauss's law)
$E=\frac{Q}{4\pi {ϵ}_0{x}^2}$
The potential difference between the shells is $dV=V_r-V_R={\int }_r^{R}Edx={\int }_r^R\frac{Q}{4\pi {ϵ}_0{x}^2}dx=\frac{Q}{4\pi {ϵ}_0}(1/r-1/R)$
Thus, $V=\frac{Q}{4\pi {ϵ}_0}(1/r-1/R)$
As the potential difference between solid sphere and hollow shell depends on the radii of two spheres and charge on the inner sphere, Since the two values have not changed, potential difference does not change. Hence the potential difference remains $V$.