Question

A spherically symmetric charge distribution is characterised by a charge density having the following variation :
$p\left(r\right)=p_o(1-\frac{r}{R})$ for $r<R$
$p\left(r\right)=0$ for $r⩾R$
Where r is the distance from the centre of the charge distribution and $p_o$ is a constant. The electric field at an internal point (r<R) is :

• A. $\frac{p_o}{4{ϵ}_o}(\frac{r}{3}-\frac{r^2}{4R})$
• B. $\frac{p_o}{{ϵ}_o}(\frac{r}{3}-\frac{r^2}{4R})$
• C. $\frac{p_o}{3{ϵ}_o}(\frac{r}{3}-\frac{r^2}{4R})$
• D. $\frac{p_o}{12{ϵ}_o}(\frac{r}{3}-\frac{r^2}{4R})$

Answer 1

Answer: (B)

$\frac{p_o}{{ϵ}_o}(\frac{r}{3}-\frac{r^2}{4R})$

Answer is B.

Electric field at the internal point r is the sum of fields due to outer and inner spheres. However contribution due to outer part is zero.

For the inner part, consider a shell of thickness dr' at a distance of r' from the center of sphere.

Therefore, charge contained in it

$dq=4\pi r^{\prime 2}d{r}^{\prime }{p}_o(1-\frac{r^{\prime }}{R})$

Or, Electric field dE is

$dE= \dfrac{1}{4\pi \epsilon_or^2}

4\pi r'^{2} dr' p_o(1-\dfrac{r'}{R})$

Integrating above, we get

$E=\frac{p_o}{{ϵ}_o{r}^2}{\int }_0^r(r^{\prime 2}-\frac{r^{\prime 3}}{R})dr$

$E=\frac{p_o}{{ϵ}_o}(\frac{r}{3}-\frac{r^3}{4R})$