Question 215 of 311

A uniformly charged solid sphere of radius R has potential ${V}_{0}$ (measured with respect to $\infty$ ) on its surface. For this sphere the equipotential surfaces with potential $\dfrac{3{V}_{0}}{2},\dfrac{5{V}_{0}}{4},\dfrac{3{V}_{0}}{4}$ and $\dfrac{{V}_{0}}{4}$ have radius ${R}_{1}$ , ${R}_{2},{R}_{3}$ and ${R}_{4}$ respectively. Then
Note : This question had two option correct at the time of examination. Proper corrections are made in the question to avoid it.

A$2R\gt {R}_{4}$B${R}_{1}=0$ and ${R}_{2}\gt \left({R}_{4}-{R}_{3}\right)$C${R}_{1}\neq 0$ and $\left({R}_{2}-{R}_{1}\right)\gt ({R}_{4}-{R}_{3})$D${R}_{1}=0$ and ${R}_{2}\lt \left({R}_{4}-{R}_{3}\right)$

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Answer: (D) ${R}_{1}=0$ and ${R}_{2}\lt \left({R}_{4}-{R}_{3}\right)$

Potential for uniformly charged solid sphere

$\text{v}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{\text{r}}$ outside i.e r > R

$\text{v}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{\text{R}}$ on the surface

$\text{v}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{\text{R}}\left[\dfrac{3}{2}-\dfrac{1}{2}\dfrac{{\text{r}}^{2}}{{\text{R}}^{2}}\right]$ inside i.e. r < R

Clearly

$\dfrac{3{\text{v}}_{0}}{2}\text{,}\dfrac{5{\text{v}}_{0}}{4}$ are inside potentials $\left[∵\gt {\text{v}}_{0}\right]$

$\dfrac{3{\text{v}}_{0}}{4}\text{,}\dfrac{{\text{v}}_{0}}{4}$ are outside potentials $\left[∵\lt {\text{v}}_{0}\right]$

To get ${\text{R}}_{1}$: $\dfrac{3{\text{v}}_{0}}{2}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{\text{R}}\left[\dfrac{3}{2}-\dfrac{1}{2}\dfrac{{\text{R}}_{1}^{2}}{{\text{R}}^{2}}\right]$

${\text{where, v}}_{0}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{\text{R}}$

$\dfrac{3}{2\times 4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{\text{R}}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{\text{R}}\left[\dfrac{3}{2}-\dfrac{1}{2}\dfrac{{\text{R}}_{1}^{2}}{{\text{R}}^{2}}\right]$

$\dfrac{3}{2}=\dfrac{3}{2}-\dfrac{1}{2}\dfrac{{\text{R}}_{1}^{2}}{{\text{R}}^{2}}\Rightarrow {\text{R}}_{1}=0$

To get ${\text{R}}_{2}$: $\dfrac{5}{4}{\text{v}}_{0}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{\text{R}}\left[\dfrac{3}{2}-\dfrac{1}{2}\dfrac{{\text{R}}_{2}^{2}}{{\text{R}}^{2}}\right]$

$\dfrac{5}{4}\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{\text{R}}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{\text{R}}\left[\dfrac{3}{2}-\dfrac{1}{2}\dfrac{{\text{R}}_{2}^{2}}{{\text{R}}^{2}}\right]$

$\dfrac{5}{4}=\dfrac{3}{2}-\dfrac{1}{2}\dfrac{{\text{R}}_{2}^{2}}{{\text{R}}^{2}}$

$\dfrac{1}{2}\dfrac{{\text{R}}_{2}^{2}}{{\text{R}}^{2}}=\dfrac{1}{4}$$\Rightarrow$ ${\text{R}}_{2}=\dfrac{\text{R}}{\sqrt{2}}$

To get ${\text{R}}_{3}$ : $\dfrac{3{\text{v}}_{0}}{4}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{{\text{R}}_{3}}$

$\dfrac{3}{4}\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{\text{R}}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{{\text{R}}_{3}}$

$\dfrac{3}{4\text{R}}=\dfrac{1}{{\text{R}}_{3}}$

${\text{R}}_{3}=\dfrac{4}{3}\text{R}$

To get ${\text{R}}_{4}$: $\dfrac{{\text{v}}_{0}}{4}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{{\text{R}}_{4}}$

$\dfrac{1}{4}\times \dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{\text{R}}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\text{Q}}{{\text{R}}_{4}}$

${\text{R}}_{4}=4\text{R}$

${\text{R}}_{4}-{\text{R}}_{3}=4\text{R}-\dfrac{4\text{R}}{3}=\dfrac{8\text{R}}{3}\gt {\text{R}}_{2}$

${\text{R}}_{1}=0$ and ${\text{R}}_{2}\lt \left({\text{R}}_{4}-{\text{R}}_{3}\right)$