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 $R_2>\left(R_4{R}_3\right)$
• B. $R_1\ne 0$ and $\left(R_2{R}_1\right)>\left(R_4{R}_3\right)$
• C. R${}_1=0$ and $R_2<\left(R_4{R}_3\right)$
• D. $2R<R_4$

Answer 1

The correct option is C R${}_1=0$ and $R_2<\left(R_4{R}_3\right)$

Radius $=R$

Potential $=V_0$

with potentials $=\frac{{3V}_0}{2},\frac{5V_0}{4},\frac{3V_0}{4},\frac{V_0}{4}$

with radius $=R_1,R_2,R_3,R_4$ respectively.

$\left(C\right)$ option will be correct.

The potential at the surface of the charged sphere,

$V_{\varphi }=\frac{KQ}{R}$

$V=\frac{KQ}{r}\times r\ge R$

$=\frac{KQ}{{2R}^3}({3R}^2-r^2)$ ; $r\le R$

$V_{centre}=V_C=\frac{KQ}{2R^3}\times {3R}^2=\frac{3}{2}\times KQ=\frac{{3V}_0}{2}$

As potential decreases for outside points. Thus, according to the question, we can write

$V_{R_2}=\frac{5V_0}{4}=\frac{KQ}{2R^3}({3R}^2-R_2^2)$

$\frac{{5V}_0}{4}=\frac{V_0}{2R^2}({3R}^2-R_2^2)$

or, $\frac{5}{2}=3-{\left(\frac{R_2}{R}\right)}^2$

or, ${\left(\frac{R_2}{R}\right)}^2=3-\frac{5}{2}=\frac{1}{2}$

or, $R_2=\frac{R}{\sqrt{2}}$,

Similarly,

$V_{R_3}=\frac{{3V}_0}{4}$

or, $\frac{KQ}{R_3}=\frac{3}{4}\times \frac{KQ}{R}$

or, $R_3=\frac{4}{3}R$

or, $V_{R_4}=\frac{KQ}{R_4}\times \frac{V_0}{4}$

or, $\frac{KQ}{R_4}=\frac{1}{4}\times \frac{KQ}{R}$

or, $R_4=4R$

Now, $R_1=0$ and $R_2<\left(R_4{R}_3\right)$