Question

Concentric metallic hollow spheres of radii $R$ and $4R$ hold charges $Q_1$ 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. $\frac{3Q_2}{4{\mathrm{\pi \epsilon }}_0\mathrm{R}}$
• B. $\frac{3Q_1}{4{\mathrm{\pi \epsilon }}_0\mathrm{R}}$
• C. $\frac{3Q_1}{16{\mathrm{\pi \epsilon }}_0\mathrm{R}}$
• D. $\frac{Q_1}{4{\mathrm{\pi \epsilon }}_0\mathrm{R}}$

Answer 1

The correct option is C

$\frac{3Q_1}{16{\mathrm{\pi \epsilon }}_0\mathrm{R}}$

Step 1. Given Data,

Inner spheres of radii $R$,

Outer sphere of radii $4R$,

Inner sphere hold charge$Q_1$,

Outer sphere hold charge $Q_2$,

Surface charge densities $\sigma$ of the concentric spheres are equal.

We have to calculate the potential difference $V\left(R\right)-V\left(4R\right)$ is.

Step 2. Calculate the Charge Distribution,

Since, Surface charge density $\sigma$ of the concentric spheres is equal.

$$\begin{gathered} \sigma =\frac{Q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2} \\ \Rightarrow \sigma =\frac{Q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}_1^2}=\frac{Q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}_2^2} \\ \Rightarrow \sigma =\frac{Q_1}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2}=\frac{Q_2}{4{\mathrm{\pi \epsilon }}_0{\left(4R\right)}^2} \\ \Rightarrow Q_1=\frac{Q_2}{16} \end{gathered}$$

Step 3. Calculate the potential difference $V\left(R\right)-V\left(4R\right)$,

$V\left(R\right)-V\left(4R\right)=k\left[\frac{Q_1}{R}+\frac{Q_2}{4R}\right]-k\left[\frac{Q_1}{4R}+\frac{Q_2}{4R}\right]$ (where, $k=\frac{1}{4{\mathrm{\pi \epsilon }}_0}$)

$$\begin{gathered} \Rightarrow V\left(R\right)-V\left(4R\right)=k\left[\frac{3Q_1}{4R}\right] \\ \Rightarrow V\left(R\right)-V\left(4R\right)=\left[\frac{3Q_1}{16{\mathrm{\pi \epsilon }}_{0}R}\right] \end{gathered}$$

Hence, the correct option is $\text{'}C\text{'}$.