Question

A metal ball of radius $a$ carries a charge $Q$. A grounded metal shell of radius $b(>a)$ is concentric to the ball. Find the electrostatic potential energy of the system of the two spheres.

• A. $\frac{Q^2}{8\pi {\epsilon }_0}[\frac{1}{a}-\frac{1}{b}]$
• B. $\frac{Q^2}{8\pi {\epsilon }_0}[\frac{1}{a}+\frac{1}{b}]$
• C. $\frac{Q^2}{4\pi {\epsilon }_0}[\frac{1}{a}-\frac{1}{b}]$
• D. $\frac{Q^2}{4\pi {\epsilon }_0}[\frac{1}{a}+\frac{1}{b}]$

Answer 1

The correct option is A $\frac{Q^2}{8\pi {\epsilon }_0}[\frac{1}{a}-\frac{1}{b}]$


Self energy of the inner shell is $U_a=\frac{Q^2}{8\pi {\epsilon }_{0}a}$

Self energy of the outer shell is $U_b=\frac{(-Q{)}^2}{8\pi {\epsilon }_{0}b}$

To write potential energy of interaction of $+Q$ and $-Q$ charge, we can think that $-Q$ charge is located in the field of $+Q$ charge. Therefore,

$U_{ab}=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{b}(-Q)=-\frac{Q^2}{4\pi {\epsilon }_{0}b}$

$\therefore$ Total potential energy of the system is $U=U_a+U_b+U_{ab}$

$=\frac{Q^2}{4\pi {\epsilon }_0}[\frac{1}{2a}+\frac{1}{2b}-\frac{1}{b}]=\frac{Q^2}{8\pi {\epsilon }_0}[\frac{1}{a}-\frac{1}{b}]$

Hence, option (a) is the correct answer.