Question

A sphere of radius $R$ has a volume density of charge $\rho =kr$, where $r$ is the distance from the centre of the sphere and $k$ is constant. The magnitude of the electric field which exits at the surface of the sphere is given by:

(${\epsilon }_0=$ permittivity of free space)

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

Answer 1

Answer: (D)

$\frac{k{R}^2}{4{\epsilon }_0}$

Given $\rho =K.r$
By Gauss theorem
$E\left(4\pi r^2\right)=\frac{\int \rho \times 4\pi r^{2}dr}{{\epsilon }_0}$
$=\frac{\int Kr\times 4\pi r^{2}dr}{{\epsilon }_0}$
$\Rightarrow E=\frac{K{r}^2}{4{\epsilon }_0}$
Here $r=R$
So, $E=\frac{K{R}^2}{4{\epsilon }_0}$