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 $\frac{|{\overrightarrow{E}}_A|}{|{\overrightarrow{E}}_B|}$ 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. $\frac{17}{54}$
• B. $\frac{21}{34}$
• C. $\frac{18}{54}$
• D. $\frac{18}{34}$

Answer 1

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

Let us fill the empty space $-\rho$ charge density. In this way we will have two spheres one having charge density $+\rho$ and radius R and other having charge density $-\rho$ with radius $\frac{R}{2}$.

Electric field due to the remaining portion will be equal to the electric field due to these combined charges.



Now, electric field inside the sphere
= $\frac{kq}{r^2}$
q = (Volume charge density)(Volume enclosed) = $\rho$V

Now, electric field inside the sphere
= $\frac{k\rho V}{r^2}$

Where r is the distance from the charge to the point where field is to be calculated and V is the volume enclosed by the charge
$E_A$ = Electric field due to sphere of radius R + Electric field due to cavity
Let $E_1$ = Electric field due to sphere of radius R
and $E_2$ = Electric field due to cavity

If we consider all the charge to be concentrated at the centre then
Since A is at centre electric field due to sphere of radius R would be zero since r = 0, therefore
Electric field at A, $|E_A|$ = 0 + $E_2$

$|E_A|=0+\left(\frac{1}{4\pi {\epsilon }_0}\right)\frac{\rho .\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3}{{\left(\frac{R}{2}\right)}^2}$

$|E_A|=\left(\frac{1}{4\pi {\epsilon }_0}\right)\rho \frac{4}{3}\pi \left(\frac{R}{2}\right)...\left(1\right)$

$E+B$ = Electric field due to sphere of radius R + Electric field due to cavity Again, assuming the charge to be concentrated at the centre
Electric field at B is given by

$|E_B|=E_1-E_2$

$|E_B|=\left(\frac{1}{4\pi {\epsilon }_0}\right)\frac{\rho .\frac{4}{3}\pi R^3}{R^2}$

$-\left(\frac{1}{4\pi {\epsilon }_0}\right)\frac{\rho .\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3}{{\left(\frac{3R}{2}\right)}^2}$

$=\left(\frac{1}{4\pi {\epsilon }_0}\right)\rho \frac{4}{3}\pi R-\left(\frac{1}{4\pi {\epsilon }_0}\right)\rho .\frac{4}{3}\pi \frac{8}{18}$

$|E_B|=\left(\frac{1}{4\pi {\epsilon }_0}\right)\rho .\frac{4}{3}\pi \left(\frac{17R}{18}\right)...\left(2\right)$

[using (1) and (2)]

$\frac{E_A}{E_B}=\frac{9}{17}=\frac{18}{34}$
Hence, option (d) is correct.