Question

A uniformly charged solid sphere or radius $R$ has potential $V_0$ (measured with respect to $\infty$) on its surface. For this sphere, the equipotential surfaces with potentials $\frac{3V_0}{2},\frac{{5V}_0}{4},\frac{3V_0}{4},\frac{V_0}{4}$ have radius $R_1,R_2,R_3$ and $R_4$ respectively. Then

• A. $R_1=0$ and $R_2<(R_4-R_3)$
• B. $R_1\ne 0$ and $(R_1-R_1)>(R_4-R_3)$
• C. $R_1=0$ and $R_2>(R_4-R_3)$
• D. $2R>R_4$

Answer 1

The correct option is A $R_1=0$ and $R_2<(R_4-R_3)$

The potential at the centre $(for{R}_1=0)$$=k\frac{Q}{\frac{4}{3}\pi r^3}{\int }_0^R\frac{4\pi r^{2}dr}{r}=\frac{3}{2}\frac{kQ}{R}=\frac{3}{2}{V}_{0}wherek=\frac{1}{4\pi {\epsilon }_0}$
Potential at $R_2=\frac{5V_0}{4}$
$\Rightarrow \frac{kQ}{2R^3}(3R^2-r^2)=\frac{5kQ}{4R}$
$\Rightarrow R_2=\frac{R}{\sqrt{2}}$
Similarily,
Potential at $R_3$, $\frac{kQ}{R_3}=\frac{3V_0}{4}=\frac{3kQ}{4R}$
$\Rightarrow R_3=\frac{4R}{3}$
Potential at $R_4=\frac{V_0}{4}$
$\Rightarrow R_4=4R$

Option (a) is correct.