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Step-by-Step Solution
Step 1: Understand the Problem Statement
We have a spherically symmetric charge distribution with charge density $ \rho(r) $. Several equipotential surfaces of potentials
$ V_{0}, \; V_{0} + \Delta V, \; V_{0} + 2\Delta V, \dots, V_{0} + N\Delta V $ are drawn inside the sphere.
The radii of these surfaces are $ r_{0}, \; r_{1}, \; r_{2}, \dots, r_{N} $ respectively,
and the difference in radii for consecutive equipotentials is the same constant for all $\Delta V > 0$.
We need to find how $ \rho(r) $ depends on $ r $ under these conditions.
Step 2: Relate Potential Spacing to Electric Field
For a spherical charge distribution (assuming it starts at the center), the electric field at a distance $ r $ from the center is
$$
E(r) \;=\; \frac{1}{4\pi \varepsilon_0} \,\frac{Q_{\text{enc}}(r)}{r^{2}},
$$
where $ Q_{\text{enc}}(r) $ is the total charge enclosed within radius $ r $. The potential $ V(r) $ is related to the electric field by
$$
E(r) \;=\; -\,\frac{dV}{dr}.
$$
If consecutive equipotential surfaces differ by the same potential interval $ \Delta V $, but their radii differ by a constant
$ \Delta r = r_{n+1} - r_{n} $ as we move outward, then
$ \frac{\Delta V}{\Delta r} $ is constant. In the limit of small intervals,
$ \frac{dV}{dr} $ is constant, implying $ E(r) = -\frac{dV}{dr} $ is a constant inside the region we are considering.
Step 3: Infer the Enclosed Charge with Constant Electric Field
If $ E(r) $ is constant (independent of $ r $), then from Gauss’s law in spherical symmetry,
$$
E(r) = \frac{1}{4\pi \varepsilon_0} \,\frac{Q_{\text{enc}}(r)}{r^2} = \text{constant}.
$$
Hence, for each radius $ r $ inside the sphere,
$$
Q_{\text{enc}}(r) \;\propto\; r^{2}.
$$
This means that the total charge enclosed within radius $ r $ grows proportionally to $ r^2 $.
Step 4: Deduce the Form of $ \rho(r) $
The enclosed charge up to radius $ r $ is given by the integral of the charge density over the volume:
$$
Q_{\text{enc}}(r) \;=\; \int_{0}^{r} 4\pi x^{2}\,\rho(x)\,dx.
$$
Since $ Q_{\text{enc}}(r) \propto r^{2} $, let us denote
$$
Q_{\text{enc}}(r) \;=\; k\,r^{2} \quad (\text{for some constant } k).
$$
Differentiating both sides with respect to $ r $:
$$
\frac{d}{dr}\bigl[\,Q_{\text{enc}}(r)\bigr] = 4\pi r^{2}\,\rho(r) = 2k\,r.
$$
Thus,
$$
4\pi r^{2}\,\rho(r) = 2k\,r.
$$
Rearranging, we get
$$
\rho(r) = \frac{2k\,r}{4\pi r^{2}} = \frac{k'}{r},
$$
where $ k' $ is a constant proportional to $ k $. Therefore,
$$
\rho(r) \;\propto\; \frac{1}{r}.
$$
Step 5: State the Final Answer
The charge density that ensures the same radial spacing of equipotential surfaces for a fixed potential difference
must vary inversely with $ r $. Hence,
$$
\rho(r) \;\alpha\; \frac{1}{r}.
$$
