Within a spherical charge distribution of charge density $$\rho(r)$$, N equipotential surfaces of potential $$V_0$$, $$V_0 + \Delta V$$, $$V_0 + 2\Delta V$$, ... $$V_0 + N\Delta V$$ $$(\Delta V > 0)$$, are drawn and have increasing radii $$r_0$$, $$r_1$$, $$r_2$$, ... $$r_N$$, respectively. If the difference in the radii of the surfaces is constant for all values of $$V_0$$ and $$\Delta V$$ then:

We consider a spherically symmetric charge distribution whose volume charge density is $$\rho(r)$$. At a distance $$r$$ from the centre the total charge enclosed is written as

$$Q(r)=\int_{0}^{r}\rho(r')\,4\pi r'^2\,dr'. $$

For spherical symmetry the electric field at this point is obtained from Gauss’s law. First we state the formula:

$$E(r)=\frac{1}{4\pi\epsilon_0}\,\frac{Q(r)}{r^2}.$$

The electrostatic potential satisfies the basic relation

$$\frac{dV}{dr}=-E(r).$$

Now the problem tells us that the equipotential surfaces $$V_0,\,V_0+\Delta V,\,V_0+2\Delta V,\ldots$$ appear at radii $$r_0,\,r_1,\,r_2,\ldots$$ with a constant spacing of radii, that is

$$r_{k+1}-r_{k}=\Delta r\quad\text{for every}\;k.$$

Because both $$\Delta V$$ and $$\Delta r$$ are fixed and small intervals, the ratio $$\Delta V/\Delta r$$ represents the magnitude of $$dV/dr$$, and this ratio is the same everywhere. Hence

$$\left|\frac{dV}{dr}\right|=\frac{\Delta V}{\Delta r}= \text{constant}.$$

Calling this positive constant $$C_1$$, we write

$$\frac{dV}{dr}=-C_1.$$

Substituting the expression for $$dV/dr$$ from the electric field formula, we have

$$-\frac{1}{4\pi\epsilon_0}\,\frac{Q(r)}{r^2}=-C_1,$$

or, dropping the negative sign and gathering the constants,

$$\frac{Q(r)}{r^2}=4\pi\epsilon_0 C_1=\text{constant}.$$

So we conclude that

$$Q(r)=k\,r^{2},\quad\text{where}\;k=4\pi\epsilon_0 C_1.$$

Next we relate this to the density. Differentiating the enclosed charge with respect to $$r$$ gives, by the Leibniz rule,

$$\frac{dQ}{dr}=4\pi r^{2}\rho(r).$$

But we also have $$Q(r)=k r^{2}$$, so

$$\frac{dQ}{dr}=2k r.$$

Equating the two expressions for $$dQ/dr$$ we obtain

$$4\pi r^{2}\rho(r)=2k r.$$

Solving for the density,

$$\rho(r)=\frac{2k r}{4\pi r^{2}}=\frac{k}{2\pi r}.$$

The constant factors are unimportant for proportionality, so finally we can state

$$\rho(r)\propto\frac{1}{r}.$$

Hence, the correct answer is Option 3.