Physics - Electrostatics Question with Solution | TestHub

The total charge $Q_{total}={\int }_0^R\rho (r)dV={\int }_0^R(A{r}^2)(4\pi r^{2}dr)=4\pi A{\int }_0^R{r}^{4}dr=4\pi A[\frac{r^5}{5}{]}_0^R=\frac{4\pi A{R}^5}{5}$. (Option A is correct) For the electric field inside the sphere ($r<R$), apply Gauss's Law: $\oint \vec{E}\cdot d\vec{S}=\frac{Q_{enclosed}}{{ϵ}_0}$.$E(r)(4\pi r^2)=\frac{1}{{ϵ}_0}{\int }_0^r(A{r}^{\prime 2})(4\pi r^{\prime 2}d{r}^{\prime })=\frac{4\pi A}{{ϵ}_0}{\int }_0^r{r}^{\prime 4}d{r}^{\prime }=\frac{4\pi A}{{ϵ}_0}[\frac{r^{\prime 5}}{5}{]}_0^r=\frac{4\pi A{r}^5}{5{ϵ}_0}$. Thus, $E(r)=\frac{A{r}^3}{5{ϵ}_0}$. (Option B is correct) For the electric field outside the sphere ($r>R$), $E(r)(4\pi r^2)=\frac{Q_{total}}{{ϵ}_0}=\frac{4\pi A{R}^5}{5{ϵ}_0}$. So, $E(r)=\frac{A{R}^5}{5{ϵ}_0{r}^2}$. This is proportional to $r^{-2}$, not $r^{-3}$. (Option D is incorrect)

To find the potential at the center, we first find the potential at the surface $V(R)$ and then integrate inwards.$V(\infty )=0$.$V(R)=-{\int }_{\infty }^{R}E(r^{\prime })d{r}^{\prime }=-{\int }_{\infty }^R\frac{A{R}^5}{5{ϵ}_0{r}^{\prime 2}}d{r}^{\prime }=-\frac{A{R}^5}{5{ϵ}_0}[-\frac{1}{r^{\prime }}{]}_{\infty }^R=\frac{A{R}^5}{5{ϵ}_{0}R}=\frac{A{R}^4}{5{ϵ}_0}$.$V(0)=V(R)-{\int }_R^{0}E(r^{\prime })d{r}^{\prime }=\frac{A{R}^4}{5{ϵ}_0}-{\int }_R^0\frac{A{r}^{\prime 3}}{5{ϵ}_0}d{r}^{\prime }=\frac{A{R}^4}{5{ϵ}_0}-\frac{A}{5{ϵ}_0}[\frac{r^{\prime 4}}{4}{]}_R^0$.$V(0)=\frac{A{R}^4}{5{ϵ}_0}-\frac{A}{20{ϵ}_0}(0-R^4)=\frac{A{R}^4}{5{ϵ}_0}+\frac{A{R}^4}{20{ϵ}_0}=\frac{4A{R}^4+A{R}^4}{20{ϵ}_0}=\frac{5A{R}^4}{20{ϵ}_0}=\frac{A{R}^4}{4{ϵ}_0}$. (Option C is correct)