Question

The diagram shows a sphere of radius $R$ that has a charge density described by the function $=Ar$.
Which of the following expressions gives the electric field strength at point $P$, a point that lies outside the sphere and is at a distance $r$ from the center of the sphere?

• A. $\frac{A{R}^2}{8\pi {ϵ}_0{r}^2}$
• B. $\frac{A{R}^4}{3{ϵ}_0{r}^2}$
• C. $\frac{A{R}^4}{8{ϵ}_0{r}^2}$
• D. $\frac{A{R}^3}{3{ϵ}_{0}r}$
• E. $\frac{A{R}^4}{4{ϵ}_0{r}^2}$

Answer 1

The correct option is E $\frac{A{R}^4}{4{ϵ}_0{r}^2}$

The total charge on the sphere is $Q={\int }_0^R\rho dV={\int }_0^R\left(Ar\right)4\pi r^{2}dr=\pi A{R}^4$
By Gauss's law, the electric field at distance $r>R$ from centre is
$E.4\pi r^2=\frac{Q_{in}}{{ϵ}_0}=\frac{Q}{{ϵ}_0}$
or $E=\frac{\pi A{R}^4}{4\pi {ϵ}_0{r}^2}=\frac{A{R}^4}{4{ϵ}_0{r}^2}$