Question

Three concentric conducting spherical shells of radius R, 2R & 3R carry charges Q, -2Q and 3Q respectively, then

• A. Electric potential at $r=R$ is $\frac{Q}{4\pi {ϵ}_{0}R}$
• B. Electric field at $r=\frac{5}{2}R$ is $\frac{-Q}{25\pi {ϵ}_0{R}^2}$
• C. Electrostatic energy of the system is $\frac{Q^2}{4\pi {ϵ}_{0}R}$
• D. Electrostatic energy of the system is $\frac{-Q^2}{2\pi {ϵ}_{0}R}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Electric potential at $r=R$ is $\frac{Q}{4\pi {ϵ}_{0}R}$
B Electric field at $r=\frac{5}{2}R$ is $\frac{-Q}{25\pi {ϵ}_0{R}^2}$
C Electrostatic energy of the system is $\frac{Q^2}{4\pi {ϵ}_{0}R}$

$V_R=\frac{KQ}{R}-\frac{2KQ}{2R}+\frac{3KQ}{3R}=\frac{KQ}{R}=\frac{Q}{4\pi {ϵ}_{0}R}$

By Gauss law, $\oint E.ds=\frac{q_{en}}{{ϵ}_0}$
at, $r=\frac{5}{2}R,q_{enclosed}=-2Q+Q=-Q$
so field $E=\frac{q_{encosed}/{ϵ}_0}{4\pi r^2}=\frac{-Q}{25\pi {ϵ}_0{R}^2}$

Energy $E=$$U_{12}+U_{13}+U_{23}+U_1+U_2+U_3$
$=\frac{KQ}{2R}(-2Q)+\frac{KQ}{3R}\left(3Q\right)-\frac{2KQ}{3R}3Q+\frac{K}{2}\frac{Q^2}{R}+\frac{K}{2}\frac{4Q^2}{2R}+\frac{K}{2}\frac{9Q^2}{3R}$
Energy = $\frac{K{Q}^2}{R}$