Physics - Electrostatics Question with Solution | TestHub

The total charge $Q={\int }_0^R\rho (r)(4\pi r^2)dr={\int }_0^{R}Ar(4\pi r^2)dr=4\pi A{\int }_0^R{r}^{3}dr=4\pi A[r^4/4{]}_0^R=\pi A{R}^4$.

Thus, $A=Q/(\pi R^4)$, so P-(1).

For $r<R$, by Gauss's law, $E(4\pi r^2)=(1/{ϵ}_0){\int }_0^r\rho (x)(4\pi x^2)dx=(4\pi A/{ϵ}_0){\int }_0^r{x}^{3}dx=(\pi A{r}^4)/{ϵ}_0$.

So $E=(A{r}^2)/(4{ϵ}_0)=(Q/(\pi R^4))r^2/(4{ϵ}_0)=kQ{r}^2/R^4$. At $r=R/2$, $E_Q=kQ(R/2{)}^2/R^4=kQ/(4R^2)$,

so Q-(2). For $r>R$, $E(4\pi r^2)=Q/{ϵ}_0$,

so $E=kQ/r^2$. At $r=2R$, $E_S=kQ/(2R{)}^2=kQ/(4R^2)$,

so S-(4). The potential at the center is $V(0)=V(\infty )-{\int }_{\infty }^{0}E\cdot dr=0-{\int }_{\infty }^R(kQ/r^2)dr-{\int }_R^0(kQ{r}^2/R^4)dr$.

This simplifies to $V(0)=kQ/R+{\int }_0^R(kQ{r}^2/R^4)dr=kQ/R+(kQ/R^4)[r^3/3{]}_0^R=kQ/R+kQ/(3R)=4kQ/(3R)$. Thus, R-(3).