Question

A solid conducting sphere having 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 surface is
a)V
b)2V
c)-2V
d)4V

Answer 1

Hint: In the question it is given to us when the solid sphere is given charge Q, the potential difference between the surface of the sphere and that of the hollow shell is V. First we will determine what is the value of V numerically. Further we will determine the potential difference between the two surfaces by taking the difference in potential on their respective surface when the hollow shell is given a charge -4Q. It is to be noted that the potential on the surface is the same at any point on the potential on the inner side of the sphere or the shell. Further we will compare this value to the numerical value of V and accordingly select the correct alternative.
Formula used:
${{V}_{POINT}}=\dfrac{kQ}{r}$

Complete answer:
In the above figure we can see that a solid sphere of radius a is enclosed within a hollow sphere of radius b Let us say the potential at any point on solid sphere be ${{V}_{A}}$ and that of the shell be ${{V}_{B}}$.
Let us say in general there is either a shell or a sphere of radius R and having charge Q. Hence the potential at any point at a distance r from the centre of the sphere is given by,
${{V}_{POINT}}=\dfrac{kQ}{r}$ where k is constant for a given media. Hence from this expression itself the potential at a point on the sphere of radius R is equal to ${{V}_{R}}=\dfrac{kQ}{R}$.
In the question it is given to us when the solid sphere has a charge Q on it, the potential difference between the surface of the shell and the sphere is V. Hence we can write,
$\begin{align}
  & V={{V}_{A}}-{{V}_{B}} \\
 & \Rightarrow V=\dfrac{kQ}{a}-\dfrac{kQ}{b} \\
 & \Rightarrow V=kQ\left( \dfrac{1}{a}-\dfrac{1}{b} \right)...(1) \\
\end{align}$
When the spherical shell is given a charge -4Q, the potential on the solid sphere becomes equal to the potential due to the sphere plus that of the shell. This can be mathematically be represented as,
${{V}_{A}}=\dfrac{kQ}{a}+\dfrac{k(-4Q)}{b}...(2)$
Similarly the potential on the spherical shell becomes equal to,
${{V}_{B}}=\dfrac{kQ}{b}+\dfrac{k(-4Q)}{b}...(3)$
Taking the difference between equation 1 and 2 we get,
$\begin{align}
  & {{V}_{A}}-{{V}_{B}}=\dfrac{kQ}{a}+\dfrac{k(-4Q)}{b}-\left[ \dfrac{kQ}{b}+\dfrac{k(-4Q)}{b} \right] \\
 & \Rightarrow {{V}_{A}}-{{V}_{B}}=\dfrac{kQ}{a}-\dfrac{k4Q}{b}-\dfrac{kQ}{b}+\dfrac{k4Q}{b} \\
 & \Rightarrow {{V}_{A}}-{{V}_{B}}=\dfrac{kQ}{a}-\dfrac{kQ}{b} \\
 & \Rightarrow {{V}_{A}}-{{V}_{B}}=kQ\left( \dfrac{1}{a}-\dfrac{1}{b} \right)...(4) \\
\end{align}$
If we compare equation 1 and 4 they are the same. Therefore we can imply that the potential remains the same i.e. equal to V.

So, the correct answer is “Option A”.

Note:
K is nothing but the permittivity of free space which is equal to $\dfrac{1}{4\pi {{\in }_{\text{o}}}}$ where ${{\in }_{\text{o}}}$ is the permittivity of free space. It is also to be noted that charge always exists on the surface of the body irrespective whether the given body is solid or hollow. It is also noted that potential at any point inside the surface of a sphere or shell is equal to that of their respective surface. Hence we have added the potential on the solid sphere when the outer shell has acquired some potential.