1. Interpret the condition

For a spherically symmetric charge distribution, equipotential surfaces are spheres of radius $r$.

Given potentials $V_0,\; V_0+\Delta V,\; V_0+2\Delta V,\; \dots$ at radii $r_0,\; r_1,\; r_2,\; \dots$ with $r_{n+1}-r_n=\text{constant}$ for all $V_0$ and $\Delta V$.

This means that equal changes in potential correspond to equal changes in radius. So, $\frac{dr}{dV}=\text{constant}$ or equivalently, $\frac{dV}{dr}=\text{constant}.$

But for electrostatics, $E(r)=-\frac{dV}{dr}.$ Hence the electric field magnitude must be constant inside the distribution: $E(r)=\text{constant}.$

2. Use Gauss's law for spherical symmetry

For a spherical Gaussian surface of radius $r$, $E(r)\cdot 4\pi r^2=\frac{Q_{\text{enc}}(r)}{\varepsilon_0}.$ Thus, $Q_{\text{enc}}(r)=4\pi \varepsilon_0 E r^2.$

Since $E$ is constant, $Q_{\text{enc}}(r)\propto r^2.$

Now, $\rho(r)=\frac{1}{4\pi r^2}\frac{dQ_{\text{enc}}}{dr}.$ Differentiating $Q_{\text{enc}}\propto r^2$, $\frac{dQ_{\text{enc}}}{dr}\propto r.$ Therefore, $\rho(r)\propto \frac{r}{r^2}=\frac{1}{r}.$

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

3. Check options

• A: $\rho(r)\propto r$
  Then $Q_{\text{enc}}\propto r^4$, so $E\propto r^2$; not constant. ❌

• B: $\rho(r)=\text{constant}$
  Then $Q_{\text{enc}}\propto r^3$, so $E\propto r$; not constant. ❌

• C: $\rho(r)\propto \dfrac{1}{r}$
  Then $Q_{\text{enc}}\propto r^2$, so $E=\text{constant}$. ✅

• D: $\rho(r)\propto \dfrac{1}{r^2}$
  Then $Q_{\text{enc}}\propto r$, so $E\propto \dfrac{1}{r}$; not constant. ❌

4. Final answer

The correct option is $\boxed{\text{C: } \rho(r)\propto \frac{1}{r}}.$