Question

Consider a sphere of radius $R$ which carries a uniform charge density $\rho$. If a sphere of radius $\frac{R}{2}$ Ls carved out of it, as shown, the ratio $\frac{\left|\overrightarrow{E_{A}}\right|}{\left|\overrightarrow{E_{B}}\right|}$ of magnitude of electric field $\overrightarrow{E_{A}}$ and $\overrightarrow{E}_{B}$ respectively, at points A and B due to the remaining portion is :

• A. $\frac{21}{34}$
• B. $\frac{18}{54}$
• C. $\frac{17}{54}$
• D. $\frac{18}{34}$

Answer 1

Answer: (D)

Fill the empty space with $+\rho$ and $-\rho$ charge density.
$\left|E_{A}\right|=0+\frac{k\rho. \frac{4}{3}\pi\left(\frac{R}{2}\right)^{3}}{\left(\frac{R}{2}\right)^{2}}=k\rho \frac{4}{3}\pi\left(\frac{R}{2}\right)$
$\left|E_{B}\right|=\frac{k\rho. \frac{4}{3}\pi R^{3}}{R^{2}}-\frac{k\rho. \frac{4}{3}\pi\left(\frac{R}{2}\right)^{3}}{\left(\frac{3R}{2}\right)^{2}}$
$=k\rho \frac{4}{3}\pi R-k\rho \frac{4}{3}\pi \frac{R}{18}=k\rho. \frac{4}{3}\pi\left(\frac{17R}{18}\right)$
$\frac{E_{A}}{E_{B}}=\frac{9}{17}=\frac{18}{34}$