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 \[\dfrac{{3{V_0}}}{2}\], \[\dfrac{{5{V_0}}}{4}\], \[\dfrac{{3{V_0}}}{4}\] and \[\dfrac{{{V_0}}}{4}\] 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) \[2R < {R_4}\]
(c) \[{R_1} = 0\] and \[{R_2} > \left( {{R_4} - {R_3}} \right)\]
(d) \[{R_1} \ne 0\] and \[\left( {{R_2} - {R_1}} \right) > \left( {{R_4} - {R_3}} \right)\]

Answer 1

Hint: We will write the expressions for the potential of equipotential surfaces inside and outside the sphere which gives us the relation between the potential at the surface of the sphere, radius of sphere and distance from the centre. We will find the values of the radius at given equipotential surfaces to find the relation between their radii.

Complete step by step answer:
Given:
The potential of an equipotential surface at radius \[{R_1}\] is \[{V_1} = \dfrac{{3{V_0}}}{2}\].
The potential of an equipotential surface at radius \[{R_2}\] is \[{V_2} = \dfrac{{5{V_0}}}{4}\].
The potential of an equipotential surface at radius \[{R_3}\] is \[{V_3} = \dfrac{{3{V_0}}}{4}\].
The potential of an equipotential surface at radius \[{R_4}\] is \[{V_4} = \dfrac{{{V_0}}}{4}\].

We are required to establish the relationship between \[{R_1},{R_2},{R_3}\] and \[{R_4}\].

We can write the expression for the potential for a radius which is more than that of the sphere that means potential outside the sphere.
\[{V_{out}} = \dfrac{{kQ}}{r}\]……(1)
Here r \[\left( {r > R} \right)\] is the radius of the sphere from the centre, k is proportionality constant, and Q is the charge on the sphere.

We can write the value of potential on the surface as below:
\[{V_0} = \dfrac{{kQ}}{R}\]

We can also write the expression for potential at a radius which is less than the radius of the sphere that is the expression for potential inside the sphere.
\[{V_{in}} = \dfrac{{kQ}}{{2{R^3}}}\left( {3{R^2} - {r^2}} \right)\]……(2)
Here r is less than the radius of the sphere \[\left( {r < R} \right)\].

We are given that the potential at the surface of the sphere is \[{V_0}\]. We know that the given values of potential having a value greater than that on the surface are valid inside the sphere and potential having values less than that on the surface are valid outside the sphere.

For \[{R_1}\] radius, the value of potential is greater than that on the surface, using equation (2) we can write:
\[\dfrac{{3{V_0}}}{2} = \dfrac{{kQ}}{{2{R^3}}}\left( {3{R^2} - R_1^2} \right)\]

We will substitute \[{V_0}\] for \[\dfrac{{kQ}}{R}\] in the above expression to find the value of radius \[{R_1}\].
$
\dfrac{{3{V_0}}}{2} = \dfrac{{{V_0}}}{{2{R^2}}}\left( {3{R^2} - R_1^2} \right)\\
\implies 3{R^2} = 3{R^2} - R_1^2\\
\implies {R_1} = 0
$

For \[{R_2}\] radius, the value of potential is greater than that on the surface, using equation (2) we can write:
\[\dfrac{{5{V_0}}}{4} = \dfrac{{kQ}}{{2{R^3}}}\left( {3{R^2} - R_2^2} \right)\]

We will substitute \[{V_0}\] for \[\dfrac{{kQ}}{R}\] in the above expression to find the value of radius \[{R_2}\].
$
\dfrac{{5{V_0}}}{4} = \dfrac{{{V_0}}}{{2{R^2}}}\left( {3{R^2} - R_2^2} \right)\\
\implies {R_2} = \dfrac{R}{{\sqrt 2 }}
$


For \[{R_3}\] radius, the value of potential is less than that on the surface, using equation (1) we can write:
$
\dfrac{{3{V_0}}}{4} = \dfrac{{kQ}}{{{R_3}}}\\
\implies \dfrac{{3{V_0}}}{4} = \dfrac{{kQR}}{{R{R_3}}}
$

We will substitute \[{V_0}\] for \[\dfrac{{kQ}}{R}\] in the above expression to find the value of radius \[{R_3}\].
$
\dfrac{{3{V_0}}}{4} = \dfrac{{R{V_0}}}{{{R_3}}}\\
\implies {R_3} = \dfrac{{4R}}{3}
$

For \[{R_4}\] radius, the value of potential is less than that on the surface, using equation (1) we can write:
$
\dfrac{{{V_0}}}{4} = \dfrac{{kQ}}{{{R_4}}}\\
\implies \dfrac{{{V_0}}}{4} = \dfrac{{kQR}}{{R{R_3}}}
$

We will substitute \[{V_0}\] for \[\dfrac{{kQ}}{R}\] in the above expression to find the value of radius \[{R_4}\].
$
\dfrac{{{V_0}}}{4} = \dfrac{{R{V_0}}}{{{R_4}}}\\
\implies {R_4} = 4R
$
Let us find the value \[{R_4} - {R_3}\] to find the correct answer:
\[ {R_4} - {R_3}\]

We will substitute \[\dfrac{{4R}}{3}\] for \[{R_3}\] and 4R for \[{R_4}\] in the above expression.
$
= 4R - \dfrac{{4R}}{3}\\
= \dfrac{{8R}}{3}
$

From the above calculations, we can say that \[{R_4} - {R_3}\] is greater than \[{R_2}\], 2R is less than \[{R_4}\] and \[{R_1}\] is equal to zero.

So, the correct answer is “Option C”.

Note:
 To find the exact answer, we have to find the values of various expressions given in options, but on closely observing all the options we find that option (B) and (D) is not correct as \[{R_1}\] is zero. Therefore, we only checked for option (A) and option (C).