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Question

(C) The relation between electric field $E$ and potential $V$ is given by $E = -\frac{dV}{dr}$.Given that the difference in radii $\Delta r = r_{i+1}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The relation between electric field $E$ and potential $V$ is given by $E = -\frac{dV}{dr}$.Given that the difference in radii $\Delta r = r_{i+1} - r_i$ is constant for a constant potential difference $\Delta V$,it implies that the electric field $E = -\frac{\Delta V}{\Delta r}$ is constant.For a spherical charge distribution,by Gauss's Law,the electric field at a distance $r$ is $E = \frac{q_{enclosed}}{4\pi\epsilon_0 r^2}$.Since $E$ is constant,we have $q_{enclosed} \propto r^2$.We know that $q_{enclosed} = \int_0^r \rho(r) 4\pi r^2 dr$.Since $q_{enclosed} \propto r^2$,differentiating both sides with respect to $r$ gives $\frac{dq}{dr} \propto 2r$.Thus,$\rho(r) 4\pi r^2 \propto r$,which implies $\rho(r) \propto \frac{1}{r}$.

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