Question

Two concentric conducting thin spherical shells $A$ and $B$ having radii $r_A$ and $r_B$$(r_B>r_A)$ are charged with charges $Q_A$ and $-Q_B$ $(|Q_B|>|Q_A|)$. The electric field along a line, (passing through the centre) varies with distance $x$ from the centre as

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Given that,
$r_A<r_B$ and $(|Q_B|>|Q_A|)$

Let $x$ is distance from the centre of the shells.

Case I: When $x\le r_A$.
Draw a spherical Gaussian surface through shell $A$.
According to Gauss law
$\overrightarrow{E}.\overrightarrow{S}=\frac{q_{en}}{{ϵ}_0}$
Where $\overrightarrow{E}$ is electric field at Gaussian surface and $\overrightarrow{S}$ is surface area enclosing a charge $q$
Here, $q_{en}=0$
$\overrightarrow{E}=0$

Case II: When $r_A<x<r_B$
By using Gauss law
Charge enclosed by Gaussian surface, $q_{en}=Q_A$
$E\times 4\pi x^2=\frac{Q_A}{{ϵ}_0}$
$E=\frac{Q_A}{4\pi x^2{ϵ}_0}$
We can say that
$E\propto \frac{1}{x^2}$

Case III: When $x=r_B$
The electric field inside the conducting shell $B$ is zero.
The net charge enclosed by the Gaussian surface will be zero when the inner surface of the shell induces $-Q_A$ charge.

Case (IV): When $x>r_B$
Total charge on the surface of the shell $B=(Q_A-Q_B)$
Net charge enclosed by the Gaussian surface = $q_{en}=Q_A-Q_B$ which is negative.

By Gauss law
$E\times 4\pi x^2=\frac{Q_A-Q_B}{{ϵ}_0}$
$E=\frac{Q_A-Q_B}{4\pi x^2{ϵ}_0}$
We can say that
$E\propto -\frac{1}{x^2}$

Correct graph