Question

Concentric metallic hollow spheres of radii $R$ and $4R$ hold charges $Q_1\text{and}{Q}_2$ respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference $V\left(R\right)-V\left(4R\right)$ is:

• A. $\frac{3Q_1}{16\pi {\epsilon }_{0}R}$
• B. $\frac{3Q_2}{4\pi {\epsilon }_{0}R}$
• C. $\frac{Q_2}{4\pi {\epsilon }_{0}R}$
• D. $\frac{3Q_1}{4\pi {\epsilon }_{0}R}$

Answer 1

The correct option is A $\frac{3Q_1}{16\pi {\epsilon }_{0}R}$

Given that, the two metallic hollow spheres of radii $R$ and $4R$ having charges $Q_1\text{and}{Q}_2$ respectively.


Given that, the surface charge densities are the same,

$\Rightarrow \frac{Q_1}{4\pi R^2}=\frac{Q_2}{4\pi 16R^2}$

$\Rightarrow Q_2=16Q_1$

Potential on the surface of inner sphere (at A)

$V_A=\frac{k{Q}_1}{R}+\frac{k{Q}_2}{4R}=\frac{5k{Q}_1}{R}$

Potential on the surface of outer sphere (at B)

$V_B=\frac{k{Q}_1}{4R}+\frac{k{Q}_2}{4R}=\frac{17k{Q}_1}{4R}$

Potential difference,
$\Delta V=V_A-V_B=\frac{3}{4}\frac{k{Q}_1}{R}$

$\Rightarrow \Delta V=\frac{3}{16\pi {ϵ}_0}\frac{Q_1}{R}$

Hence, option $\left(A\right)$ is correct.