Question

The potential energy of the system of two identically charged spheres as shown in the figure is equal to (Assume the charge distribution to be uniform)

• A. $\frac{q^2}{4\pi {\epsilon }_0}[\frac{1}{R}+\frac{1}{r}]$
• B. $\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r}$
• C. $\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{(R+r)}$
• D. None of these

Answer 1

The correct option is A $\frac{q^2}{4\pi {\epsilon }_0}[\frac{1}{R}+\frac{1}{r}]$

Each of the charged sphere will have potential energy to its own charge

$U_1=\frac{1}{8\pi {\epsilon }_0}\frac{q^2}{r}$

The mutual potential energy stored in one sphere is the electric field of the other is

$U_2=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{R}$

therefore the $P.E.$ strored in the system of spheres in

$U=[2U_1+U_2]=\frac{q^2}{4\pi {\epsilon }_0}[\frac{1}{R}+\frac{1}{r}]$