Question

Three concentric conducting spherical shells of radii $R,2R$ and $3R$ carry charges $Q,-2Q$ and $3Q$, respectively. Compute the potential at $r=R$, and the charges on the spheres of radii $R$ and $3R.$

• A. $\frac{Q+2Q_1}{12\pi {\epsilon }_{0}R},Q_1=\frac{Q}{4},Q_2=\frac{7Q}{2}$
• B. $\frac{Q+2Q_1}{12\pi {\epsilon }_{0}R},Q_1=\frac{Q}{2},Q_2=\frac{7Q}{2}$
• C. $\frac{Q+2Q_1}{12\pi {\epsilon }_{0}R},Q_1=\frac{Q}{2},Q_2=\frac{7Q}{4}$
• D. $\frac{Q+2Q_1}{12\pi {\epsilon }_{0}R},Q_1=\frac{Q}{4},Q_2=\frac{7Q}{4}$

Answer 1

The correct option is B $\frac{Q+2Q_1}{12\pi {\epsilon }_{0}R},Q_1=\frac{Q}{2},Q_2=\frac{7Q}{2}$

Let A has charge $Q_1$ and C has charge $Q_2$. After getting connected:

$V_A=V_C$

$\frac{1}{4\pi {ϵ}_0}[\frac{Q_1}{R}+\frac{-2Q}{2R}+\frac{Q_2}{3R}]=\frac{1}{4\pi {ϵ}_0}[\frac{Q_1}{3R}-\frac{2Q}{3R}+\frac{Q_2}{3R}]$

$Q_1-2Q+\frac{Q_2}{3}=\frac{Q_1}{3}-\frac{2Q}{3}+\frac{Q_2}{3}$

$Q_1=\frac{Q}{2}$

Now, $Q_1+Q_2=4Q{Q}_2=\frac{7Q}{2}$

Using $Q_1$ and $Q_2$ potential will be:

$=\frac{1}{4\pi {ϵ}_0}[\frac{Q_1}{R}+\frac{-2Q}{2R}+\frac{Q_2}{3R}]$

$=\frac{Q+2Q_1}{12\pi {ϵ}_{0}R}$