# A uniformly charged solid sphere of the radius $R$ has potential ${V_0}$ (Measured with respect to $\infty$) on its surface. For this sphere the equipotential surfaces with potentials $\dfrac{{3{V_0}}}{2}$, $\dfrac{{5{V_0}}}{4}$, $\dfrac{{3{V_0}}}{4}$ and (A)$\dfrac{{{V_0}}}{4}$have a 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) $2R < {R_4}$(D) None of the above

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Hint: We have a uniformly charged sphere of a radius $R$, the potential on the surface of the sphere with respect to infinity is also given. We are also given the equipotential surfaces with their respective radii. Now we have to find the relation between the given radii.
Formula used
$V = \dfrac{{KQ}}{r}$(where, $V$ stands for the potential of the sphere, $K$ is a constant, $Q$stands for the charge of the sphere, and $r$ stands for the separation of the charge from infinity.

Complete Step by step solution:
The potential on the surface of the sphere can be written as,
$V = \dfrac{{KQ}}{R} = {V_0}$
The potential at any point outside the sphere can be written as,
$V = \dfrac{{KQ}}{r}$ for $r > R$
The potential at any point inside the sphere is given by,
${V_i} = \dfrac{{kQ}}{{2{R^3}}}\left( {3{R^2} - {r^2}} \right)$ for $r < R$
Where $r$is the radius of the sphere.
At the centre of the sphere, $r = 0$
$\Rightarrow {V_i} = \dfrac{{3KQ}}{{2R}}$
Since the potential on the surface is ${V_0}$, for $r < R$,$V > {V_0}$
For $r > R$,$V < {V_0}$
We know that $\dfrac{{3{V_0}}}{2} > {V_0}$
$\Rightarrow {R_1} < R$
Hence, we can write
$\dfrac{{3{V_0}}}{2} = \dfrac{{KQ}}{{2{R^3}}}\left( {3{R^2} - R_1^2} \right)$
We know that,
$\dfrac{{KQ}}{R} = {V_0}$
Substituting in the above equation,
$\dfrac{{3{V_0}}}{2} = \dfrac{{{V_0}}}{{2{R^2}}}\left( {3{R^2} - R_1^2} \right)$
Canceling common terms on both sides,
$3 = \dfrac{1}{{{R^2}}}\left( {3{R^2} - R_1^2} \right)$
$\Rightarrow 3{R^2} = 3{R^2} - R_1^2$
From this, we get
${R_1} = 0$
For ${R_2}$, $V = \dfrac{{5{V_0}}}{4}$
From this, we know that
$\dfrac{{5{V_0}}}{4} > {V_0}$
$\Rightarrow {R_2} < R$
The potential can be written as,
$\dfrac{{5{V_0}}}{4} = \dfrac{{KQ}}{{2{R^3}}}\left( {3{R^2} - R_2^2} \right)$
Substituting$\dfrac{{KQ}}{R} = {V_0}$
We get,
$\dfrac{{5{V_0}}}{4} = \dfrac{{{V_0}}}{{2{R^2}}}\left( {3{R^2} - R_2^2} \right)$
Canceling the common terms, we get
$\dfrac{5}{2} = \dfrac{1}{{{R^2}}}\left( {3{R^2} - R_2^2} \right)$
Solving, we get
$\dfrac{5}{2}{R^2} = 3{R^2} - R_2^2$
From this, we get
$R_2^2 = \dfrac{{{R^2}}}{2}$
Taking the square root,
${R_2} = \dfrac{R}{{\sqrt 2 }}$
For ${R_3}$
$\dfrac{{3{V_0}}}{4} < {V_0}$
$\Rightarrow {R_3} > R$
The potential can be written as,
$\dfrac{{3{V_0}}}{4} = \dfrac{{kQ}}{{{R_3}}}$
Multiply and divide with $R$on RHS
$\dfrac{{3{V_0}}}{4} = \dfrac{{kQ}}{{{R_3}}} \times \dfrac{R}{R}$
Substituting$\dfrac{{KQ}}{R} = {V_0}$
$\dfrac{{3{V_0}}}{4} = \dfrac{{{V_0}R}}{{{R_3}}}$
Canceling the common terms,
$3{R_3} = 4R$
$\Rightarrow {R_3} = \dfrac{4}{3}R$
For ${R_4}$
$\dfrac{{{V_0}}}{4} < {V_0}$
$\Rightarrow {R_4} > R$
Therefore, we can write the potential as,
$\dfrac{{{V_0}}}{4} = \dfrac{{KQ}}{{{R_4}}}$
Multiply and divide with $R$on RHS
$\dfrac{{{V_0}}}{4} = \dfrac{{KQ}}{{{R_4}}} \times \dfrac{R}{R}$
Substituting$\dfrac{{KQ}}{R} = {V_0}$
$\dfrac{{{V_0}}}{4} = \dfrac{{{V_0}R}}{{{R_4}}}$
Canceling common terms and solving
${R_4} = 4R$
${R_1} = 0$
${R_2} = \dfrac{R}{{\sqrt 2 }}$
${R_3} = \dfrac{4}{3}R$
And, ${R_4} = 4R$
Considering the relations in the options,
${R_1} = 0$ and ${R_2} > \left( {{R_4} - {R_3}} \right)$
${R_4} - {R_3} = 4R - \dfrac{4}{3}R = \dfrac{8}{3}R$
From this, $\dfrac{8}{3}R > \dfrac{R}{{\sqrt 2 }}$
Therefore option A is not correct.
In option B, it is given ${R_1} \ne 0$, therefore option B is also wrong.
In option C, it is given $2R < {R_4}$
We know that ${R_4} = 4R$
$\Rightarrow 2R = \dfrac{{{R_4}}}{2}$
Therefore, option (C) is the correct answer.

The answer is: Option (C): $2R < {R_4}$

Note:
A surface on which every point has the same potential is known as an equipotential surface. The electric field will be perpendicular to the equipotential surface. For moving a charge on an equipotential surface no work is required.