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. $\frac{17 \rho r_{0}}{54 \varepsilon_{0}}$ left
• B. $\frac{\rho r_{0}}{6 \varepsilon_{0}}$ left
• C. $\frac{17 \rho r_{0}}{54 \varepsilon_{0}}$ right
• D. $\frac{\rho r_{0}}{6 \varepsilon_{0}}$ right

Answer 1

Answer: (A)

Electric field on surface of a uniformly charged sphere is given by
$\frac{Q}{4 \pi \varepsilon_{0} R^{2}}=\frac{\rho R}{3 \varepsilon_{0}}$
Electric field at outside point is given by
$E=\frac{Q}{4 \pi \varepsilon_{0} r^{2}}=\frac{\rho R^{3}}{3 \varepsilon_{0} r^{2}}$
$E_{B}=E_{\text {whole sphere }}-E_{\text {cavity }}$
$E_{B}=\frac{\rho r_{0}}{3 \varepsilon_{0}}-\frac{\rho\left(\frac{r_{0}}{2}\right)^{3}}{3 \varepsilon_{0}\left(\frac{3 r_{0}}{2}\right)^{2}}=\frac{17 \rho r_{0}}{54 \varepsilon_{0}}$