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.
$\left|E_A\right|=0+\frac{k\rho .\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3}{{\left(\frac{R}{2}\right)}^2}=k\rho \frac{4}{3}\pi \left(\frac{R}{2}\right)$
$\left|E_B\right|=\frac{k\rho .\frac{4}{3}\pi R^3}{R^2}-\frac{k\rho .\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3}{{\left(\frac{3R}{2}\right)}^2}$
$=k\rho \frac{4}{3}\pi R-k\rho \frac{4}{3}\pi \frac{R}{18}=k\rho .\frac{4}{3}\pi \left(\frac{17R}{18}\right)$
$\frac{E_A}{E_B}=\frac{9}{17}=\frac{18}{34}$