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Step-by-Step Solution
Step 1: Understand the problem
We have a uniformly charged solid sphere of radius $R$. Its potential on the surface (with respect to infinity) is given as $V_0$. We are asked to compare the radii of equipotential surfaces corresponding to different potential values:
$ \tfrac{3V_0}{2},\, \tfrac{5V_0}{4},\, \tfrac{3V_0}{4}, \text{ and } \tfrac{V_0}{4}
$.
We denote these equipotential surface radii by $R_1,\, R_2,\, R_3, \text{ and } R_4$, respectively. We need to identify the correct statement among the options given, and the correct one provided is $2R < R_4$.
Step 2: Recall the potential expressions for a uniformly charged sphere
Outside the sphere ($r \geq R$):
$V(r) = \dfrac{K\,q}{r}$,
where $K = \dfrac{1}{4\pi \epsilon_0}$.
On the surface ($r = R$):
$V_0 = \dfrac{K\,q}{R}$.
Inside the sphere ($r < R$):
$V(r) = \dfrac{K\,q}{2R^3}\Bigl(3R^2 - r^2\Bigr)$.
Here, $q$ is the total charge on the sphere. The result
$V(0) = \dfrac{3}{2}\,\dfrac{Kq}{R}$
is the potential at the center ($r=0$), which is also
$\tfrac{3}{2}V_0$ since $V_0 = \dfrac{K\,q}{R}$.
Step 3: Analyze the given equipotential values
$\tfrac{3V_0}{2}$ surface:
This is $\tfrac{3}{2} V_0$, which equals the potential at $r=0$.
Therefore, $R_1 = 0.$
(This indicates the center of the sphere is at potential $\tfrac{3V_0}{2}$.)
$\tfrac{5V_0}{4}$ surface:
Since $\tfrac{5V_0}{4} > V_0$, this potential must lie inside the sphere (because the potential at the surface is $V_0$, and at the center is $\tfrac{3V_0}{2}$ which is higher). We can set up:
$
\dfrac{5V_0}{4}
= \dfrac{Kq}{2R^3}(3R^2 - r^2),
$
then solve for that radius, often found as
$
R_2 = \frac{R}{\sqrt{2}}.
$
(Exact numeric details or further derivation can follow from the above expression.)
$\tfrac{3V_0}{4}$ surface:
Since $\tfrac{3V_0}{4} < V_0$, its equipotential can lie either inside or outside the sphere depending on the exact radius; it turns out it will be inside the sphere if it does not drop below the value at the surface, or could be just on or just inside. A more precise formula can be used to find $R_3$. It is often found from:
$
\dfrac{3V_0}{4}
= \dfrac{Kq}{2R^3}(3R^2 - r^2)\quad (\text{if inside}),
$
or
$
\dfrac{3V_0}{4} = \dfrac{Kq}{r} \quad (\text{if } r > R).
$\tfrac{V_0}{4}$ surface:
Because this potential is significantly smaller than $V_0$, the point at which $V(r) = \tfrac{V_0}{4}$ may lie outside the sphere. Using $V(r) = \dfrac{Kq}{r} = \tfrac{V_0}{4}$, we get:
$
\dfrac{Kq}{r} = \dfrac{1}{4}\dfrac{Kq}{R} \
\Longrightarrow \
r = 4R.
$
Hence,
$
R_4 = 4R.
$
Step 4: Conclude the correct statement
From the above, since $R_4 = 4R$, we immediately see that $2R < 4R$. Hence, the statement $2R < R_4$ is correct.
This matches the given correct answer in the options.
