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 $\dfrac{{3{V_0}}}{2},\dfrac{{5{V_0}}}{4},\dfrac{{3{V_0}}}{4}\,{\text{and}}\,\dfrac{{{V_0}}}{4}$​​ have radius ${R_1},{R_2},{R_3}{\text{ and}}\;{R_4}$ 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

In this solution, we will use the potential due to the solid charged sphere and the potentials provided at different surfaces to determine the values of ${R_1},{R_2},{R_3} ,{R_4}$​ respect.
Formula used: In this solution, we will use the following formula:
Potential inside a solid charged sphere: $V(r) = \dfrac{{kQ}}{{2{R^3}}}(3{R^2} - {r^2})$

Complete step by step answer:
We’ve been told that the potential of the sphere at its surface is ${V_0}$. So, we can write that
${V_0} = \dfrac{{kQ}}{R}$.
Let us start by finding the potential of the charged sphere at the centre of the sphere. Substituting $r = 0$in $V(r) = \dfrac{{kQ}}{{2{R^3}}}(3{R^2} - {r^2})$, we get
$V(0) = \dfrac{{3kQ}}{{2R}} = \dfrac{{3{V_0}}}{2}$
This is similar to the equation given for the potential at ${R_1}$. So, ${R_1} = 0$.
Let us find the distance from the centre of the sphere such that
$V({R_2}) = \dfrac{{5{V_0}}}{4}$
So, we can equate
$V({R_2}) = \dfrac{{5{V_0}}}{4} = \dfrac{{kQ}}{{2{R^3}}}(3{R^2} - {R_2}^2)$
Diving the right side of the equation by ${R^2}$ and substituting ${V_0} = \dfrac{{kQ}}{R}$, we get
$\dfrac{{5{V_0}}}{4} = \dfrac{1}{2}{V_0}\left( {3 - \dfrac{{{R_2}^2}}{{{R^2}}}} \right)$
$\Rightarrow \dfrac{5}{2} = \left( {3 - \dfrac{{{R_2}^2}}{{{R^2}}}} \right)$
Solving for ${R_2}$, we get
$\dfrac{1}{2} = \dfrac{{{R_2}^2}}{{{R^2}}}$
$\Rightarrow {R_2} = R/\sqrt 2$
Similarly, for ${R_3}$, the point will be outside the sphere
$V({R_3}) = \dfrac{{3{V_0}}}{4} = \dfrac{{kQ}}{{{R_3}}}$
Multiplying and dividing by $R$ on the right side, we get
$\dfrac{{3{V_0}}}{4} = {V_0}\dfrac{R}{{{R_3}}}$
$\Rightarrow {R_3} = \dfrac{{4R}}{3}$
For point ${R_4}$, we have
$V({R_4}) = \dfrac{{{V_0}}}{4}\, = \dfrac{{kQ}}{{{r_4}}}$
$\Rightarrow {R_4} = 4R$
So, we can write that
${R_4} - {R_3} = 4R - \dfrac{{4R}}{3} = \dfrac{{8R}}{3}$
And also
${R_2} = \dfrac{R}{{\sqrt 2 }} < {R_4} - {R_3}$ so, option (A) is incorrect.

${R_4} = 4R > 2R$ so option (C) is only correct.

Note: To answer such questions, we must be aware of the potential of the solid charged sphere inside and outside the surface of the sphere. The potential of the sphere will be higher than the surface inside it and lower than the surface outside it. This will help us in deciding whether a point is situated inside or outside the sphere based on whether the potential is higher or lower than the potential on the surface.