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. none of these

Answer 1

Answer: (A)

$R_1=0$ and $R_2<(R_4-R_3)$

Consider sphere carries positive charge,

potential on the surface is $V_0=\frac{KQ}{R}$

potential expression inside the sphere is $V_{inside}=\frac{KQ}{2R^3}\left(3R^2-r^2\right)$

potential at the center is $V_{centre}=\frac{3KQ}{2R^3}\left(r=0\right)=\frac{3V_0}{2}$

So, radius of equipotential at the center of sphere is $0$

$R_1=0$

Consider the $2nd$ equi-potential surface is $R_2$

Then,

$\begin{array}{c}\frac{5V_0}{4}=\frac{KQ}{2R^3}\left(3R^2-R_2^2\right)\frac{5V_0}{182}\\ =\frac{V_0}{2R^3}\left(3R^2-R_2^2\right)5R^2\\ =6R^2-2R_2^{2}2R_2^2\\ =R^2\Rightarrow R_2=\frac{R}{\sqrt{2}}\end{array}$

Consider the radius of $3rd$ equipotential is

$\frac{V_0}{4}=\frac{KQ}{R_4}\frac{KQ}{4R}=\frac{KQ}{R_4}\Rightarrow R_4=4R{R}_4-R_3=4R-\frac{4R}{3}=\frac{8R}{3}=2.66R{R}_2=\frac{R}{1.414}<R$

So,

$R_2<R_4-R_3$

Hence, the option $A$ is the correct answer