Question

A sphere of radius $r$ has a volume density of charge $\rho =kr$. Find the electric field intensity at the surface of the sphere.

• A. $\frac{4\pi k{R}^4}{3{ϵ}_0}$
• B. $\frac{kR}{3{ϵ}_0}$
• C. $\frac{4\pi kR}{{ϵ}_0}$
• D. $\frac{k{R}^2}{4{ϵ}_0}$

Answer 1

Answer: (D)

$\frac{k{R}^2}{4{ϵ}_0}$

Given : Charge per unit volume of sphere $\rho =kr$

Total charge enclosed by the sphere $Q_{enc}={\int }_o^R\rho \left(4\pi r^2\right)dr$

$\therefore$ $Q_{enc}=4\pi k{\int }_o^R{r}^{4}dr$

OR $Q_{enc}=4\pi k\times \frac{r^4}{4}{|}_o^R$

OR $Q_{enc}=4\pi k\times \frac{R^4-0}{4}$ $⟹Q_{enc}=\pi k{R}^4$

Let the surface of the sphere is the Gaussian surface.

Using Gauss's law, $\int E.dS=\frac{Q_{enc}}{{ϵ}_o}$

$\therefore$ $E_{surface}\int dS=\frac{Q_{enc}}{{ϵ}_o}$

OR $E_{surface}\times 4\pi R^2=\frac{\pi k{R}^4}{{ϵ}_o}$ $⟹E_{surface}=\frac{k{R}^2}{4{ϵ}_o}$