Question

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

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

Answer 1

The correct option is B $\frac{17\rho r_0}{54{ϵ}_0}\text{left}$

Step 1: Electric field due to original sphere.
For a uniformly charged sphere having charge density $\rho$, electric field at the surface is given by:
$=\frac{\rho r_0}{3{ϵ}_0}$
$E_1$ is towards the left since the field is repulsive as shown in figure below.

Step 2: Electric field due to cavity
A cavity can be assumed to be a negatively charged sphere and its charge can be assumed to be concentrated at its center $C_2$.
$\therefore E_2=\frac{4}{4\pi {ϵ}_0}\left[\frac{(-\rho )\times \frac{4}{3}\pi (\frac{r_0}{2}{)}^3}{q{r}_0^2}\right]$



Step 3: Net electric field at point B
$E_B=E_1-E_2$
$=\frac{\rho r_0}{3{ϵ}_0}-\frac{4\rho r_0}{216{ϵ}_0}$
$=\frac{17\rho r_0}{54{ϵ}_0}$
$\Rightarrow E_B\text{is towards the left as}{E}_1>E_2$
Hence, option (𝒂) is correct.