Question

A neutral conducting sphere of radius $R$ contains two spherical cavities of radii $a$ and $b$ that contains point charges $q_a$ and $q_b$ placed at the center of each cavity as shown in the figure below . Match the following table:

$\left(\text{i}\right){\sigma }_a$	$\left(\text{a}\right)\frac{q_a+q_b}{4\pi R^2}$
$\left(\text{ii}\right){\sigma }_b$	$\left(\text{b}\right)\frac{-q_a}{4\pi a^2}$
$\left(\text{iii}\right){\sigma }_R$	$\left(\text{c}\right)\frac{-q_b}{4\pi b^2}$

• A. $\left(\text{i}\right)-c;\left(\text{ii}\right)-b;\left(\text{iii}\right)-a$
• B. $\left(\text{i}\right)-a;\left(\text{ii}\right)-c;\left(\text{iii}\right)-b$
• C. $\left(\text{i}\right)-b;\left(\text{ii}\right)-c;\left(\text{iii}\right)-a$
• D. $\left(\text{i}\right)-b;\left(\text{ii}\right)-a;\left(\text{iii}\right)-c$

Answer 1

The correct option is C $\left(\text{i}\right)-b;\left(\text{ii}\right)-c;\left(\text{iii}\right)-a$

Given,

Radius of the conducting sphere be $R$

Radii of two spherical cavities be $a$ and $b$.

A point charge $q_a$ is placed in a spherical cavity of radius $a$ and point charge $q_b$ is placed in the spherical cavity of radius $b$.

Since, the electric field intensity is zero inside the conducting region we can say that,



Induced charge on the surface of the spherical cavity of radius $a$ be$"-q_a"$



Similarly, Induced charge on the surface of the spherical cavity of radius $b$ be $"-q_b"$

We know that, Surface charge density $\sigma =\frac{\text{Total charge}}{\text{Area}}$

$\therefore$ Surface charge density on spherical cavity $a$ be ${\sigma }_a=\frac{-q_a}{4\pi a^2}$

Surface charge density on spherical cavity $b$ be ${\sigma }_b=\frac{-q_b}{4\pi b^2}$

Total charge inside the conducting sphere be $q=q_a+q_b$



So, surface charge density of conducting sphere of radius $R$ be ${\sigma }_R=\frac{q_a+q_b}{4\pi R^2}$

Hence, option (b) is the correct answer.