Question

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 potentials $\frac{3V_0}{2},\frac{5V_0}{4},\frac{3V_0}{4}$ and $\frac{V_0}{4}$ 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 $(R_2-R_1)>(R_4-R_3)$
• C. $2R<R_4$
• D. None of the above

Answer 1

The correct option is C $2R<R_4$

$V\left(R\right)=\frac{kQ}{R}=V_0$

We use results derived for potential.

$V\left(r\right)=\frac{kQ}{r}$ for $r>R$

$V\left(r\right)=\frac{kQ}{2R^3}(3R^2-r^2)$

$V(r=0)=\frac{3kQ}{2R}=\frac{3V_0}{2}$

Hence, $V\left(R_1\right)$ matches with V at center.

$\therefore R_1=0$

Thus, we eliminate option(B).

At $R_2$

$V\left(R_2\right)=\frac{5V_0}{4}=\frac{kQ}{2R^3}(3R^2-R_2^2)$

$\Rightarrow \frac{5V_0}{4}=\frac{1}{2}\left(\frac{kQ}{R}\right)(3-\frac{R_2^2}{R^2})=\frac{1}{2}{V}_0(3-\frac{R_2^2}{R^2})$

$\Rightarrow \frac{5}{2}=3-\frac{R_2^2}{R^2}$

$\Rightarrow \frac{1}{2}=\frac{R_2^2}{R^2}$

$\Rightarrow R_2=\frac{R}{\sqrt{2}}$

For $R_3$

$V\left(R_3\right)=\frac{3V_0}{4}=\frac{kQ}{R_3}$

$\Rightarrow \frac{3V_0}{4}=V_0\frac{R}{R_3}$

$\Rightarrow R_3=\frac{4R}{3}$

For $R_4$

$V\left(R_4\right)=\frac{V_0}{4}=\frac{kQ}{R_4}$

$\Rightarrow R_4=4R$

Thus, $R_4-R_3=4R-\frac{4R}{3}=\frac{8R}{3}$

$R_2=\frac{R}{\sqrt{2}}<\frac{8R}{3}=R_4-R_3$

Also, $2R<4R=R_4$