Question

A solid sphere having a radius $R$ and uniform charge density $\rho$. If a sphere of radius $R/2$ is carved out of it as shown in the figure. Find the ratio of the magnitude of electric field at point $A$and $B$

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

Answer 1

The correct option is C

$\frac{18}{34}$

Step 1 : Given data

Radius of the solid sphere $=R$

Uniform charge density $=\rho$

$E=$Electric field

$B=$Magnetic field

Step 2 : To find the magnitude of electric field at point $A$ and $B$

For a solid sphere,

Field inside the sphere $E_{inside}=\frac{\rho r}{3{\epsilon }_0}$

And field outside the sphere ,$E_{outside}=\frac{\rho R^3}{3{\left(r\right)}^2{\epsilon }_0}$, (where, $r$ is distance from center and is $R$ radius of the sphere)

The electric field at A due to sphere of radius $R$ is zero and therefore, net electric field will be because of the sphere of radius $\frac{R}{2}$ has charge $\left(-\rho \right)$

$$\begin{gathered} E_A=\frac{-\rho R}{2\left(3{\epsilon }_0\right)} \\ \left|E_A\right|=\frac{\rho R}{6{\epsilon }_0} \end{gathered}$$

$B=E_B=E_{1B}+E_{2B}$

Where,

Electric field due to solid sphere of radius $R,E_{1B}=\frac{\rho r}{3{\epsilon }_0}=\frac{\rho R}{3{\epsilon }_0}$

Electric field due to solid sphere having charge density $\left(-\rho \right)$ and radius $\frac{R}{2},E_{1B}=\left[\frac{-\rho R^3}{3{\left(r\right)}^2{\epsilon }_0}\right]=\left[\frac{-\rho \left({\displaystyle \frac{R}{2}}\right)}{3{\left({\displaystyle \frac{3R}{2}}\right)}^2{\epsilon }_0}\right]$

$\Rightarrow E_{1B}=-\frac{\rho R}{54{\epsilon }_0}$

$$\begin{gathered} E_B=E_{1A}+E_{2A} \\ \Rightarrow E_B=\left(\frac{\rho R}{3{\epsilon }_0}\right)-\left(\frac{\rho R}{54{\epsilon }_0}\right) \\ \Rightarrow E_B=\frac{17\rho R}{54{\epsilon }_0} \\ \left|\frac{E_A}{E_B}\right|=\frac{9}{17}=\frac{18}{34} \end{gathered}$$

Hence the ratio of magnitude of electric field at point $A$ and $B$ is $\frac{18}{34}$.

Therefore, the correct answer is Option $\text{'}C\text{'}$.