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 -4Q, the new potential difference between the same two surfaces is:

We have a solid conducting sphere of radius $$R_1$$ carrying a charge $$+Q$$. It is concentrically surrounded by a conducting hollow spherical shell whose inner radius is $$R_2$$ and outer radius is $$R_3$$. Initially the shell is uncharged.

Statement of the electrostatic condition inside a conductor: the electric field inside the material of a conductor must be zero. Therefore any free charge present resides only on its surface(s), and the potential is the same at every point of the metallic body.

Because the inner solid sphere has charge $$+Q$$, a charge $$-Q$$ is induced on the inner surface of the hollow shell (radius $$R_2$$), leaving a compensating charge $$+Q$$ on the outer surface (radius $$R_3$$) so that the shell as a whole remains neutral. Thus the initial charge distribution is

$$\begin{aligned} \text{Sphere (}\,r=R_1\,\text{)} &: +Q,\\ \text{Inner surface of shell (}\,r=R_2\,\text{)} &: -Q,\\ \text{Outer surface of shell (}\,r=R_3\,\text{)} &: +Q. \end{aligned}$$

We now evaluate the initial potential difference between the surface of the sphere ($$r=R_1$$) and the outer surface of the shell ($$r=R_3$$).

Formula stated: the potential at a point due to a point charge is $$V = k\displaystyle\frac{q}{r}$$ with $$k=\dfrac{1}{4\pi\varepsilon_0}$$. A charged spherical shell behaves as though its entire charge were concentrated at the centre for an external point, while the potential is uniform inside the shell.

Initial potential on the sphere’s surface:

$$\begin{aligned} V_{R_1} &= k\frac{(+Q)}{R_1} \;+\; k\frac{(-Q)}{R_2} \;+\; k\frac{(+Q)}{R_3}\\ &= kQ\left(\frac{1}{R_1}-\frac{1}{R_2}+\frac{1}{R_3}\right). \end{aligned}$$

Initial potential on the outer surface of the shell (all three charges are now “inside” the Gaussian surface of radius $$R_3$$):

$$\begin{aligned} V_{R_3} &= k\frac{(+Q)}{R_3} \;+\; k\frac{(-Q)}{R_3} \;+\; k\frac{(+Q)}{R_3}\\ &= kQ\left(\frac{1}{R_3}-\frac{1}{R_3}+\frac{1}{R_3}\right)\\ &= kQ\left(\frac{1}{R_3}\right). \end{aligned}$$

Hence the initial potential difference is

$$\begin{aligned} V &= V_{R_1}-V_{R_3}\\ &= kQ\left(\frac{1}{R_1}-\frac{1}{R_2}+\frac{1}{R_3}-\frac{1}{R_3}\right)\\ &= kQ\left(\frac{1}{R_1}-\frac{1}{R_2}\right). \end{aligned}$$

Now the shell is given an additional charge $$-\,4Q$$. The shell already possessed $$0$$ net charge; after accepting $$-4Q$$ its net charge becomes $$-4Q$$. However, the electrostatic shielding requirement inside the conductor remains: the electric field in the metal of the shell must still be zero. Consequently the charge on the inner surface must stay at exactly $$-Q$$ to cancel the field arising from the central $$+Q$$. Whatever is left, namely $$-4Q-(-Q) = -3Q$$, must appear on the outer surface.

The new charge distribution therefore is

$$\begin{aligned} \text{Sphere (}\,r=R_1\,\text{)} &: +Q,\\ \text{Inner surface of shell (}\,r=R_2\,\text{)} &: -Q,\\ \text{Outer surface of shell (}\,r=R_3\,\text{)} &: -3Q. \end{aligned}$$

We now compute the new potentials with exactly the same method.

New potential on the sphere’s surface:

$$\begin{aligned} V'_{R_1} &= k\frac{(+Q)}{R_1} \;+\; k\frac{(-Q)}{R_2} \;+\; k\frac{(-3Q)}{R_3}\\ &= kQ\left(\frac{1}{R_1}-\frac{1}{R_2}-\frac{3}{R_3}\right). \end{aligned}$$

New potential on the outer surface of the shell:

$$\begin{aligned} V'_{R_3} &= k\frac{(+Q)}{R_3} \;+\; k\frac{(-Q)}{R_3} \;+\; k\frac{(-3Q)}{R_3}\\ &= kQ\left(\frac{1}{R_3}-\frac{1}{R_3}-\frac{3}{R_3}\right)\\ &= kQ\left(-\frac{3}{R_3}\right). \end{aligned}$$

Finally, the new potential difference is

$$\begin{aligned} \Delta V' &= V'_{R_1}-V'_{R_3}\\ &= \left[kQ\left(\frac{1}{R_1}-\frac{1}{R_2}-\frac{3}{R_3}\right)\right] - \left[kQ\left(-\frac{3}{R_3}\right)\right]\\ &= kQ\left(\frac{1}{R_1}-\frac{1}{R_2}-\frac{3}{R_3}+\frac{3}{R_3}\right)\\ &= kQ\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\\ &= V. \end{aligned}$$

The potential difference between the surface of the solid sphere and the outer surface of the hollow shell remains exactly the same as before, namely $$V$$.

Hence, the correct answer is Option C.