The volume charge density in a spherical ball | vedclass

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Question

(A) The electric field $E$ at a distance $r$ inside a spherical charge distribution is given by Gauss's Law: $E(r) = \frac{q_{enc}}{4\pi\epsilon_0 r^2

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$\frac{R}{2^{1/3}}$

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(A) The electric field $E$ at a distance $r$ inside a spherical charge distribution is given by Gauss's Law: $E(r) = \frac{q_{enc}}{4\pi\epsilon_0 r^2}$,where $q_{enc}$ is the charge enclosed within a sphere of radius $r$.The enclosed charge is $q(r) = \int_0^r \rho(r') 4\pi r'^2 dr'$.Substituting $\rho(r') = \rho_0 \left[1 - \left(\frac{r'}{R}\right)^3\right]$:$q(r) = 4\pi\rho_0 \int_0^r \left(r'^2 - \frac{r'^5}{R^3}\right) dr' = 4\pi\rho_0 \left[ \frac{r^3}{3} - \frac{r^6}{6R^3} \right]$.Thus,$E(r) = \frac{4\pi\rho_0}{4\pi\epsilon_0 r^2} \left( \frac{r^3}{3} - \frac{r^6}{6R^3} \right) = \frac{\rho_0}{\epsilon_0} \left( \frac{r}{3} - \frac{r^4}{6R^3} \right)$.For maximum field,$\frac{dE}{dr} = 0$:$\frac{d}{dr} \left( \frac{r}{3} - \frac{r^4}{6R^3} \right) = \frac{1}{3} - \frac{4r^3}{6R^3} = 0$.$\frac{1}{3} = \frac{2r^3}{3R^3} \Rightarrow r^3 = \frac{R^3}{2} \Rightarrow r = \frac{R}{2^{1/3}}$.

The volume charge density in a spherical ball of radius $R$ varies with distance $r$ from the centre as $\rho(r)=\rho_0\left[1-\left(\frac{r}{R}\right)^3\right]$,where $\rho_0$ is a constant. The radius at which the electric field would be maximum is

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