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 $-3Q$, what is the new potential difference between the same two surfaces?

Answer 1

Step1: Given data

1. The total charge in the solid sphere is $q$.
2. The potential difference between the surface of the solid sphere and the outer surface of the solid sphere is $V$.
3. $-3Q$ charge is given on the shell after some time.

Step2: Potential

1. Potential at a point is the total work done to bring a body from infinity to that point
2. The potential due to a charge $q$ at a distance r is given by $V=K\frac{q}{r}$, where, K is the Coulomb constant.

Step2: Diagram

Step4: Finding the potential difference

Let the radius of the solid is a and radius of the hollow sphere be b.

The potential at the surface of the solid sphere is $V_a=K\frac{q}{a}$.

The potential at the surface of the hollow sphere is $V_b=K\frac{q}{b}$.

The potential difference between the surface of the solid sphere and the outer surface of the solid sphere is

$$\begin{gathered} V=V_a-V_b=K\frac{q}{a}-K\frac{q}{b} \\ orV=Kq\left(\frac{1}{a}-\frac{1}{b}\right)...............\left(1\right) \end{gathered}$$.

Step5: Finding the potential difference after giving new charge

After giving the charge $-3Q$ in the hollow sphere,

The potential at the surface of the solid sphere is $V{\text{'}}_a=K\frac{q}{a}+K\frac{-3q}{b}$.

The potential at the surface of the hollow sphere is $V{\text{'}}_b=K\frac{q}{b}+K\frac{-3q}{b}$.

The potential difference between the surface of the solid sphere and the outer surface of the solid sphere is

$$\begin{gathered} V\text{'}=V{\text{'}}_a-V{\text{'}}_b=\left(K\frac{q}{a}-K\frac{3q}{b}\right)-\left(K\frac{q}{b}-K\frac{3q}{b}\right) \\ orV\text{'}=K\frac{q}{a}-\cancel{K\frac{3q}{b}}-K\frac{q}{b}+\cancel{K\frac{3q}{b}} \\ orV\text{'}=K\frac{q}{a}-K\frac{q}{b} \\ orV\text{'}=Kq\left(\frac{1}{a}-\frac{1}{b}\right)...............\left(2\right) \end{gathered}$$

So, equations 1 and 2 are the same.

Therefore, the potential energy is the same as before.