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 potential $$\frac{3V_0}{2}$$, $$\frac{5V_0}{4}$$, $$\frac{3V_0}{4}$$ and $$\frac{V_0}{4}$$ have radius $$R_1$$, $$R_2$$, $$R_3$$ and $$R_4$$ respectively. Then
Note: This question had two options correct at the time of examination. Proper corrections are made in the question to avoid it.

For a uniformly charged solid sphere whose total charge is $$Q$$ and radius is $$R$$, the electro-static potential at any point is obtained from the well-known results of electrostatics. We first state these results.

• At an external point, that is for $$r \ge R$$, the sphere behaves like a point charge at its centre, so

$$V_{\text{out}}(r)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}.$$

• At an internal point, that is for $$r \le R$$, the potential is

$$V_{\text{in}}(r)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{2R^3}\left(3R^2-r^2\right).$$

On the surface $$r = R$$ these two expressions must give the same value, which is the given surface potential $$V_0$$. Putting $$r = R$$ in either formula we get

$$V_0=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}.$$

For convenience we now write every subsequent potential in terms of $$V_0$$.

1. Potential inside the sphere

Substituting $$\dfrac{Q}{4\pi\varepsilon_0} = V_0R$$ in the internal formula gives

$$V_{\text{in}}(r)=\dfrac{V_0R}{2R^3}\left(3R^2-r^2\right)=V_0\left[\dfrac{3}{2}-\dfrac{r^2}{2R^2}\right].$$

2. Potential outside the sphere

Similarly, substituting in the external formula yields

$$V_{\text{out}}(r)=V_0\dfrac{R}{r}.$$

We now locate the radii of the four equipotential surfaces mentioned in the problem.

(a) Surface with potential $$\dfrac{3V_0}{2}$$

This value of potential exceeds $$V_0$$, so the point must lie inside the sphere. We therefore equate it to the internal potential:

$$V_0\left[\dfrac{3}{2}-\dfrac{r^2}{2R^2}\right]=\dfrac{3V_0}{2}.$$

Cancelling $$V_0$$ from both sides,

$$\dfrac{3}{2}-\dfrac{r^2}{2R^2}=\dfrac{3}{2}\;\;\Longrightarrow\;\;-\dfrac{r^2}{2R^2}=0 \;\;\Longrightarrow\;\;r^2=0\;\;\Longrightarrow\;\;r=0.$$

Hence $$R_1 = 0.$$ This is only a single point—the centre of the sphere.

(b) Surface with potential $$\dfrac{5V_0}{4}$$

Again this value is larger than $$V_0$$, so we use the internal expression:

$$V_0\left[\dfrac{3}{2}-\dfrac{r^2}{2R^2}\right]=\dfrac{5V_0}{4}.$$

Dividing by $$V_0$$ and rewriting the left side with a common denominator 4,

$$\dfrac{6}{4}-\dfrac{r^2}{2R^2}=\dfrac{5}{4}.$$

Taking the difference,

$$\dfrac{6}{4}-\dfrac{5}{4}=\dfrac{r^2}{2R^2} \;\;\Longrightarrow\;\; \dfrac{1}{4}=\dfrac{r^2}{2R^2}.$$

Multiplying both sides by $$2R^2$$,

$$r^2=\dfrac{R^2}{2}\;\;\Longrightarrow\;\;r=\dfrac{R}{\sqrt{2}}.$$

Thus $$R_2=\dfrac{R}{\sqrt{2}}\;(\approx0.707\,R).$$

(c) Surface with potential $$\dfrac{3V_0}{4}$$

Now the potential is below $$V_0$$, so we are outside the sphere. Using the external formula,

$$V_0\dfrac{R}{r}=\dfrac{3V_0}{4}.$$

Eliminating $$V_0$$ gives

$$\dfrac{R}{r}=\dfrac{3}{4}\;\;\Longrightarrow\;\;r=\dfrac{4R}{3}.$$

Hence $$R_3=\dfrac{4R}{3}\;(\approx1.333\,R).$$

(d) Surface with potential $$\dfrac{V_0}{4}$$

Again we are outside, so

$$V_0\dfrac{R}{r}=\dfrac{V_0}{4}\;\;\Longrightarrow\;\;\dfrac{R}{r}=\dfrac{1}{4}\;\;\Longrightarrow\;\;r=4R.$$

Thus $$R_4=4R.$$

We have obtained

$$R_1=0,\qquad R_2=\dfrac{R}{\sqrt{2}},\qquad R_3=\dfrac{4R}{3},\qquad R_4=4R.$$

Let us now examine each option.

Option A states $$2R > R_4.$$ Substituting $$R_4=4R$$ gives $$2R > 4R,$$ which is clearly false.

Option B asserts two points: $$R_1 = 0$$ (true) and $$R_2 > (R_4 - R_3).$$ Compute $$R_4-R_3=4R-\dfrac{4R}{3}=\dfrac{12R-4R}{3}=\dfrac{8R}{3}\;( \approx 2.667\,R).$$ But $$R_2=\dfrac{R}{\sqrt{2}}\;( \approx 0.707\,R).$$ Since $$0.707\,R > 2.667\,R$$ is false, the whole option is false.

Option C says $$R_1 \neq 0$$ (false, we have $$R_1=0$$), so the option is already ruled out.

Option D combines $$R_1 = 0$$ (true) with $$R_2 < (R_4 - R_3).$$ We have $$R_2 \approx 0.707\,R$$ and $$R_4-R_3 \approx 2.667\,R$$, so indeed $$0.707\,R < 2.667\,R.$$ Both parts are satisfied, therefore Option D is correct.

Hence, the correct answer is Option D.