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Consider a sphere of radius $\mathrm{R}$ which carries a uniform charge density $\rho .$ If a sphere of radius $\frac{\mathrm{R}}{2}$ is carved out of it, as shown, the ratio $\frac{\left|\overrightarrow{\mathrm{E}}_{\mathrm{A}}\right|}{\left|\overrightarrow{\mathrm{E}}_{\mathrm{B}}\right|}$ of magnitude of electric field $\overrightarrow{\mathrm{E}}_{\mathrm{A}}$ and $\overrightarrow{\mathrm{E}}_{\mathrm{B}}$ respectively, at points $\mathrm{A}$ and $\mathrm{B}$ due to the remaining portion is
$\frac{18}{54}$
$\frac{21}{34}$
$\frac{17}{54}$
$\frac{18}{34}$
Solution
Fill the empty space with $+\rho$ and $-\rho$ charge density.
$\left|\mathrm{E}_{\mathrm{A}}\right|=0+\frac{\operatorname{k\rho} \cdot \frac{4}{3} \pi\left(\frac{\mathrm{R}}{2}\right)^{3}}{\left(\frac{\mathrm{R}}{2}\right)^{2}}=\operatorname{k\rho} \frac{4}{3} \pi\left(\frac{\mathrm{R}}{2}\right)$
$\left|\mathrm{E}_{\mathrm{B}}\right|=\frac{\mathrm{k} \rho \cdot \frac{4}{3} \pi \mathrm{R}^{3}}{\mathrm{R}^{2}}-\frac{\mathrm{k} \rho \cdot \frac{4}{3} \pi\left(\frac{\mathrm{R}}{2}\right)^{3}}{\left(\frac{3 \mathrm{R}}{2}\right)^{2}}$
$=\operatorname{k\rho} \frac{4}{3} \pi \mathrm{R}-\mathrm{k} \rho \frac{4}{3} \pi \frac{\mathrm{R}}{18}=\mathrm{k} \rho \cdot \frac{4}{3} \pi\left(\frac{17 \mathrm{R}}{18}\right)$
$\frac{E_{A}}{E_{B}}=\frac{9}{17}=\frac{18}{34}$
