Question

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

Answer: (A)

Electric field on surface of a uniformly charged sphere is given by
$\frac{Q}{4\pi {\epsilon }_0{R}^2}=\frac{\rho R}{3{\epsilon }_0}$
Electric field at outside point is given by
$E=\frac{Q}{4\pi {\epsilon }_0{r}^2}=\frac{\rho R^3}{3{\epsilon }_0{r}^2}$
$E_B=E_{\text{whole sphere }}-E_{\text{cavity }}$
$E_B=\frac{\rho r_0}{3{\epsilon }_0}-\frac{\rho {\left(\frac{r_0}{2}\right)}^3}{3{\epsilon }_0{\left(\frac{3r_0}{2}\right)}^2}=\frac{17\rho r_0}{54{\epsilon }_0}$