Question

Charge $Q,2Q$ and $-Q$ are given to three concentric conducting spherical shells $A,B$ and $C$ respectively as shown in figure. The ratio of charges on the inner and outer surfaces of shell $C$ will be

• A. $+\frac{3}{4}$
• B. $\frac{-3}{4}$
• C. $\frac{3}{2}$
• D. $\frac{-3}{2}$

Answer 1

The correct option is D $\frac{-3}{2}$

For the given conducting spherical shells the charges $Q,2Q,-Q$ on shells $A,B$ and $C$ will reside at their outer surface.
Let $E$ is electric field and $dA$ is elemental Gaussian surface.

For shell $A$
The charge $Q$ will appear at the outer surface of the shell and there will be no charge in the inner surface.

For shell $B$
Draw an imaginary Gaussian surface inside the shell $B$ and apply Gauss law.
$\oint \overrightarrow{E}.\overrightarrow{dA}=\frac{q_{en}}{{ϵ}_0}$
$\Rightarrow \overrightarrow{E}=0,q_{en}=0$

The charge $q_{en}$ will be zero only if the inner surface of shell $B$ will induce $-Q$ charge due to $Q$ charge on the outer surface of shell $A$.
The charge on outer surface $=2Q$

For shell $C$
Draw an imaginary Gaussian surface inside the shell $C$ and apply Gauss law.
$\oint \overrightarrow{E}.\overrightarrow{dA}=\frac{q_{en}}{{ϵ}_0}$
$\Rightarrow \overrightarrow{E}=0,q_{en}=0$

Again the charge $q_{en}$ will be zero only if the inner surface of shell $C$ will induce $-3Q$ charge due to $(2Q+Q)$ charge inside the shell.
The charge on outer surface $=-Q+3Q=2Q$
Now the ratio of charge $=\frac{-3Q}{2Q}=-\frac{3}{2}$