Question

Consider a sphere of radius $R$, which carries a uniform charge density $\rho$. If a sphere of radius $\frac{R}{2}$ is carved out of it, as shown, the ratio $\frac{\left|{\overrightarrow{E}}_A\right|}{\left|{\overrightarrow{E}}_B\right|}$ of the magnitudes of electric field $\overrightarrow{E_A}\text{and}\overrightarrow{E_B}$, respectively, at points $A$ and $B$, due to the remaining portions is:

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

Answer 1

The correct option is B $\frac{18}{34}$


$E\cdot A=\frac{q}{{ϵ}_0}\Rightarrow E=\frac{q}{A{ϵ}_0}$

Electric field at $A$ is, $(R^{\prime }=\frac{R}{2})$
$\Rightarrow E_A=\frac{\rho \times \frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3}{{ϵ}_{0}4\pi {\left(\frac{R}{2}\right)}^2}$

$\Rightarrow E_A=\frac{\rho \left(\frac{R}{2}\right)}{3{ϵ}_0}=\left(\frac{\rho R}{6{ϵ}_0}\right)$

Now, electric field at $B$ is,

$E_B=\frac{k\times \rho \times \frac{4}{3}\pi R^3}{R^2}-\frac{k\times \rho \times \frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3}{{\left(\frac{3R}{2}\right)}^2}$

$\Rightarrow E_B=\frac{\rho R}{3{ϵ}_0}-\left(\frac{1}{4\pi {ϵ}_0}\right)\frac{\left(\rho \right)}{{\left(\frac{3R}{2}\right)}^2}\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3$

$\Rightarrow E_B=\frac{\rho R}{3{ϵ}_0}-\frac{\rho R}{54{ϵ}_0}=\frac{17}{54}\left(\frac{\rho R}{{ϵ}_0}\right)$

$\therefore \left|\frac{E_A}{E_B}\right|=\frac{1\times 54}{6\times 17}=\left(\frac{9}{17}\right)$

$=\frac{9}{17}\times \frac{2}{2}=\frac{18}{34}$

Hence, $\left(B\right)$ is the correct answer.