Question

Three concentric conducting spherical shells of radii $R,2R$ and $3R$ carry charges $Q,-2Q$ and $3Q$, respectively. Compute the potential at $r=\frac{5}{2}R$ :

• A. $\frac{3Q}{20\pi {\epsilon }_{0}R}$
• B. $\frac{-3Q}{25\pi {\epsilon }_{0}R}$
• C. $\frac{-6Q}{25\pi {\epsilon }_{0}R}$
• D. $\frac{-Q}{25\pi {\epsilon }_{0}R}$

Answer 1

The correct option is A $\frac{3Q}{20\pi {\epsilon }_{0}R}$

The potential at a distance $r$ from the centre of a spherical shell of charge of radius $a$ is given by

$V=\frac{Q}{4\pi {ϵ}_{0}r}$ if $r>a$ and $V=\frac{Q}{4\pi {ϵ}_{0}a}$ if $r<a$

The potential at $r=\frac{5}{2}R$ for the given system can be written as the sum of potentials due to individual shells.

Potential due to the innermost shell is $V_1=\frac{Q}{4\pi {ϵ}_0\left(5R/2\right)}=\frac{2}{5}\frac{Q}{4\pi {ϵ}_{0}R}$

Potential due to middle shell is $V_2=\frac{-2Q}{4\pi {ϵ}_0\left(5R/2\right)}=-\frac{4}{5}\frac{Q}{4\pi {ϵ}_{0}R}$

Potential due to the outer shell is $V_3=\frac{3Q}{4\pi {ϵ}_0\left(3R\right)}=\frac{Q}{4\pi {ϵ}_{0}R}$

Thus, the potential at $r=\frac{5}{2}R$ due to the system of charges is $V=V_1+V_2+V_3=\frac{Q}{4\pi {ϵ}_{0}R}(\frac{2}{5}-\frac{4}{5}+1)=\frac{3Q}{20\pi {ϵ}_{0}R}$