Question

A solid conducting sphere having a charge $Q$ is surrounded by an uncharged concentric conducting 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 surface is:
(A) $V$
(B) $2V$
(C) $4V$
(D) $ - 2V$

Answer 1

Hint:Calculate the potential of the surface and the shell using the formula. Calculate the same after the application of the charge to the shell surrounding the sphere. Substitute the potential before application in it to find the relation between the potential difference.

Useful formula:
The potential at the surface is given by
$V = \dfrac{{kQ}}{r}$
Where $V$ is the potential on the surface, $k$ is the constant, $Q$ is the charge on the surface and $r$ is the distance from the charge.

Complete step by step solution:
It is given that the sphere is surrounded by the shell.
Let us consider that the radius of the solid sphere is ${r_1}$ and the radius of the shell is ${r_2}$ .
Let us use the formula of the potential, to find the potential over the sphere
It is known that the
$V = \dfrac{{kQ}}{r}$
${V_1} = \dfrac{{kQ}}{{{r_1}}}$
Similarly, the potential at the surface of the shell,
${V_2} = \dfrac{{kQ}}{{{r_2}}}$
$V = {V_1} - {V_2}$
By substituting both the potentials at the above equation, we get
$V = \dfrac{{kQ}}{{{r_1}}} - \dfrac{{kQ}}{{{r_2}}}$ ------------(1)
When the charge $ - 3Q$ is added to the shell, the potential at the surface of sphere is
${V_1}^\prime = \dfrac{{kQ}}{{{r_1}}} - \dfrac{{3kQ}}{{{r_2}}}$
The potential at the surface of the shell is ${V_2}^\prime = \dfrac{{kQ}}{{{r_2}}} - \dfrac{{3kQ}}{{{r_2}}}$
Total new potential difference, $V' = {V_1}^\prime - {V_2}^\prime $
Substituting the two potentials,
$V' = \dfrac{{kQ}}{{{r_1}}} - \dfrac{{3kQ}}{{{r_2}}} - \dfrac{{kQ}}{{{r_2}}} + \dfrac{{3kQ}}{{{r_2}}}$
By simplifying the above equation and substituting the equation (1) in the above equation, we get
$V' = V$

Thus the option (A) is correct.

Note:In the above solution, the net potential at a point is due to the sum of the potential due to the individual charges present at that region. If the layer of the surface does not contain any charge, the potential at that point will be due the charge at the other region also.