Question

There are four concentric shells A, B, C and D of radii $a$, $2a$, $3a$ and $4a$ respectively. Shell B and D are given charges $+q$ and $-q$ respectively. Shell C is now earthed. The potential difference $V_A-V_C$ is $k={\displaystyle \frac{1}{4\pi {\epsilon }_o}}$:
(A) ${\displaystyle \frac{kq}{2a}}$
(B) ${\displaystyle \frac{kq}{3a}}$
(C) ${\displaystyle \frac{kq}{4a}}$
(D) ${\displaystyle \frac{kq}{6a}}$

Answer 1

Hint
We need to find the potential at shell C and then equate that value to zero. From there we can find the charge induced on C. Therefore, by calculating the value of the potential at the shells A and C and we can find the difference in their potential.

Formula Used: In this solution, we will be using the following formula
$V={\displaystyle \frac{kq}{d}}$
where $V$ is the potential
$k={\displaystyle \frac{1}{4\pi {\epsilon }_o}}$ where ${\epsilon }_o$ is the permittivity in free space, $q$ is the charge and $d$ is the distance.

Complete step by step answer
In this case there are 4 concentric shells A, B, C, and D of radius $a$, $2a$, $3a$ and $4a$. A charge $q$ is placed on shell B and $-q$ is placed on the shell D. So we can draw the figure as,

The electric potential due to a charge can be given by the formula,
$V={\displaystyle \frac{kq}{d}}$
Now the potential on the shell C due to the charges placed on the shells D end B is,
$V_c={\displaystyle \frac{kq}{3a}}+{\displaystyle \frac{k{q}^{\prime }}{3a}}-{\displaystyle \frac{kq}{4a}}$
where $q^{\prime }$ is the charge induced on C.
And the potential on the surface of A is,
$V_A={\displaystyle \frac{kq}{2a}}+{\displaystyle \frac{k{q}^{\prime }}{3a}}-{\displaystyle \frac{kq}{4a}}$
Now the shell C is grounded. So the potential on C will be 0. That is, $V_c=0$.
So equating the equation we get,
$0={\displaystyle \frac{kq}{3a}}+{\displaystyle \frac{k{q}^{\prime }}{3a}}-{\displaystyle \frac{kq}{4a}}$
We can cancel the $k$ and $a$ from the numerator and denominator of all the terms.
So we get,
$0={\displaystyle \frac{q}{3}}+{\displaystyle \frac{q^{\prime }}{3}}-{\displaystyle \frac{q}{4}}$
Therefore taking the term containing $q^{\prime }$ to one side,
${\displaystyle \frac{q^{\prime }}{3}}={\displaystyle \frac{q}{4}}-{\displaystyle \frac{q}{3}}$
On the R.H.S taking LCM, we find $q^{\prime }$ as,
$q^{\prime }={\displaystyle \frac{\left(3-4\right)}{12}}3q$
On doing the calculation we get
$\Rightarrow q^{\prime }=-{\displaystyle \frac{q}{4}}$
Now putting this value of $q^{\prime }$ in the equation for $V_A$ we get
$V_A={\displaystyle \frac{kq}{2a}}-{\displaystyle \frac{k{\displaystyle \frac{q}{4}}}{3a}}-{\displaystyle \frac{kq}{4a}}$
$\Rightarrow V_A={\displaystyle \frac{kq}{2a}}-{\displaystyle \frac{kq}{12a}}-{\displaystyle \frac{kq}{4a}}$
On doing the LCM and calculating further we get3
$V_A={\displaystyle \frac{6kq-kq-3kq}{12a}}={\displaystyle \frac{2kq}{12a}}$
Hence we get the value of the potential at the surface of A as,
$V_A={\displaystyle \frac{kq}{6a}}$
Since $V_C=0$
Therefore $V_A-V_C={\displaystyle \frac{kq}{6a}}$
This is the difference in potential. So the correct option is D.

Note
The electric potential at a point in an electric field is the amount of work that is done in bringing a unit positive charge from infinity to that point. And when a body is charged it can attract and repulse an oppositely charged body. This shows the ability of a charged body to do work. This ability is called the potential of that body.