Question

A positively charged sphere of radius $r_0$ carries a volume charge density $\rho$. A spherical cavity of radius $r_0/2$ is then scooped out and left empty. $C_1$ is the center of the sphere and $C_2$ that of the cavity. What is the direction and magnitude of the electric field at a point B?

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

Answer 1

The correct option is A $\frac{17\rho r_0}{54{\epsilon }_0}left$

The cavity in the sphere can be replaced by superposition of a sphere with density $+\rho$ and a sphere of density $-\rho$.

Thus, the electric field at $B$ due to the large sphere of density $+\rho$ is

$E_1=\frac{1}{4\pi {ϵ}_0}\frac{Q_1}{r_0^2}$ towards left and $Q_1=\frac{4}{3}\pi r_0^3\rho$

Thus, $E_1=\frac{1}{4\pi {ϵ}_0{r}_0^2}\frac{4\pi r_0^3\rho }{3}=\frac{\rho r_0}{3{ϵ}_0}$ towards left.

Similarly, the field at $B$ due to negatively charged sphere is

$E_2=\frac{1}{4\pi {ϵ}_0}\frac{Q_2}{(3r_0/2{)}^2}$ towards right and $Q_2=\frac{4}{3}\pi (\frac{r_0}{2}{)}^3\rho$

Thus, $E_2=\frac{1}{9\pi {ϵ}_0{r}_0^2}\frac{\pi r_0^3\rho }{6}=\frac{\rho r_0}{54{ϵ}_0}$ towards right.

Thus, the net electric field at $B$ is $|E_1|-|E_2|=\frac{\rho r_0}{3{ϵ}_0}-\frac{\rho r_0}{54{ϵ}_0}=\frac{17\rho r_0}{54{ϵ}_0}$ leftwards.