Question

A positively charged sphere of radius $r_0$ carries a volume charge density $\rho$ as shown in figure. A spherical cavity of radius $\frac{r_0}{2}$ is then scooped out and left empty. $C_1$ is the centre of the sphere and $C_2$ is the centre of the cavity. What is the direction and magnitude of the electric field at point B?

• A. $\frac{17\rho r_0}{54{ϵ}_0}$, towards right
• B. $\frac{\rho r_0}{6{ϵ}_0}$, towards right
• C. $\frac{17\rho r_0}{54{ϵ}_0}$, towards left
• D. $\frac{\rho r_0}{6{ϵ}_0}$, towards left

Answer 1

The correct option is C $\frac{17\rho r_0}{54{ϵ}_0}$, towards left

Given,
Charge density or charge per unit volume$=\rho$
Radius of sphere $=r_0$

If, $Q$ is the total chrage on the sphere of radius $r_0$, then
$\rho =\frac{Q}{\frac{4}{3}\pi r_0^3}$

Electric field on surface of the solid sphere at point $\text{B}$ is given by

$E_{\text{whole sphere}}=\frac{Q}{4\pi {ϵ}_0{r}_0^2}=\frac{\rho r_0}{3{ϵ}_0}...\left(1\right)$

Electric field at a point outside the sphere at distance $r$ is given by

$E_o=\frac{Q}{4\pi {ϵ}_0{r}^2}=\frac{\rho r_0^3}{3{ϵ}_0{r}^2}$

We can assume the spherical cavity to be made of material having charge density $-\rho$.
So, using above formula, the electric field at point $\text{B}$ due to spherical cavity can be calculated by taking, $r_c=\frac{r_0}{2}$ and $r=\frac{3}{2}{r}_0$,

$E_{\text{cavity}}=\frac{\rho {\left(\frac{r_0}{2}\right)}^3}{3{ϵ}_0{\left(\frac{3r_0}{2}\right)}^2}...\left(2\right)$

So the net electric field at point $\text{B}$,

$E_B=E_{\text{whole sphere}}-E_{\text{cavity}}$

Substituting the values from equation $\left(1\right)$ and $\left(2\right)$,
$E_B=\frac{\rho r_0}{3{ϵ}_0}\left(\text{towards left}\right)-\frac{\rho {\left(\frac{r_0}{2}\right)}^3}{3{ϵ}_0{\left(\frac{3r_0}{2}\right)}^2}\left(\text{towards right}\right)$

$\therefore E_B=\frac{17\rho r_0}{54{ϵ}_0}$, towards left.

Hence, option (a) is the correct answer.