Question

The potential of outer shell is

• A. $\frac{q}{32\pi {ϵ}_{0}a}$
• B. $\frac{q}{16\pi {ϵ}_{0}a}$
• C. $\frac{q}{8\pi {ϵ}_{0}a}$
• D. $\frac{q}{4\pi {ϵ}_{0}a}$

Answer 1

The correct option is A $\frac{q}{32\pi {ϵ}_{0}a}$

Let the charge on the inner sphere is $q^{\prime }$.

As shell is uncharged so its net charge is zero.

The inner sphere is grounded, hence its potential is zero.

Now potential at the center of inner sphere $=$ potential due to inner sphere $+$ potential due to uncharged shell $+$ potential due to charge q .

$\Rightarrow 0=\frac{k{q}^{\prime }}{a}+\frac{k\left(0\right)}{2a}+\frac{kq}{4a}$

$\therefore q^{\prime }=-\frac{q}{4}$

As inner sphere is conducting so $(-q/4)$ reside on the surface of the sphere. As charge inside a shell is zero so charge $(+q/4)$ induce on inner surface of shell.

The electric field at b is $E=\frac{(-q/4)}{4\pi {ϵ}_0{x}^2}$

Magnitude of filed is $|E|=\frac{q}{16\pi {ϵ}_0{x}^2}$ and direction is towards centre.

Potential of outer shell is $V_c$

so $V_c-V_a=-{\int }_{2a}^{a}Edx=-\frac{q}{16\pi {ϵ}_0}{\int }_{2a}^{a}dx/x^2$

or $V_c-0=\frac{q}{16\pi {ϵ}_0}[1/a-1/2a]$

$V_c=\frac{q}{32\pi {ϵ}_{0}a}$