Question

Charge on the outer sphere is $q$ and the inner sphere is grounded. Then the charge $q^{\prime }$ on the inner sphere is, for $(r_2>r_1)$

• A. zero
• B. $q^{\prime }=q$
• C. $q^{\prime }=-\frac{r_1}{r_2}q$
• D. $q^{\prime }=\frac{r_1}{r_2}q$

Answer 1

Answer: (C)

Step 1: Given that:

Charge on outer sphere = q

Radius of outer sphere = $r_2$

Radius of inner sphere = $r_1$

The inner sphere is grounded.

Step 2: Calculation of charge(q') on the inner sphere:

As the inner sphere is grounded,

The net potential on the surface of inner sphere = 0

Now,

The potential at the surface of charged sphere is given by;

$V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}$ ..........(1)

Where Q is the charge on the sphere and R is the radius of the sphere.

Now, the electric potential on the inner surface is due to the charge on it as well as the charge on the outer sphere.

Therefore, the net potential on the inner sphere will be;

$V_{inner}=\frac{1}{4\pi {ϵ}_0}\frac{q^{\prime }}{r_1}+\frac{1}{4\pi {ϵ}_0}\frac{q}{r_2}$

But from equation (1), we get;

$\frac{1}{4\pi {ϵ}_0}\frac{q^{\prime }}{r_1}+\frac{1}{4\pi {ϵ}_0}\frac{q}{r_2}=0$

$\frac{1}{4\pi {ϵ}_0}\frac{q^{\prime }}{r_1}=-\frac{1}{4\pi {ϵ}_0}\frac{q}{r_2}$

$\frac{q^{\prime }}{r_1}=-\frac{q}{r_2}$

$q^{\prime }=-\frac{q{r}_1}{r_2}$

$q^{\prime }=-\frac{r_1}{r_2}q$

Hence,

Option C)$-\frac{r_1}{r_2}q$ is the correct option.