Question

An ellipsoidal cavity is carved within a perfect conductor. A charge $+q$ is placed at the centre of the cavity. The points $A$ and $B$ are on the cavity surface as shown in the figure. Then

• A. electric field near $A$ in the cavity $=$ electric field near $B$ in the cavity
• B. charge density near $A$ on the conductor surface $=$ charge density near $B$ on the conductor surface
• C. potential at $A=$ potential at $B$
• D. total electric field flux through the surface of the cavity is $\frac{q}{{ϵ}_o}$

Answer 1

Answer: (C), (D)

total electric field flux through the surface of the cavity is $\frac{q}{{ϵ}_o}$

Based on the concept of Gauss's Law, since the electric field inside a conductor is zero, the charge distribution will be as shown in figure.


We know the surface charge density $\left(\sigma \right)$ at a point varies inversely as the radius of curvature $\left(R\right)$.
$\Rightarrow \sigma \propto \frac{1}{R}$
For points near $A$ and $B$, radius of curvature, $R_A>R_B$ $\Rightarrow {\sigma }_B>{\sigma }_A$

Also, electric field just outside the conductor $($in the cavity$)$ is $E=\frac{\sigma }{{ϵ}_o}$.
$\Rightarrow E_B>E_A$

Further, we know that under an electrostatic condition, all points lying on the conductor are at the same potential.
$\Rightarrow V_A=V_B$.

Now, let us draw a Gaussian surface just inside the cavity.


According to Gauss's Law, electric flux through the surface,
$\varphi =\frac{q_{\text{enclosed}}}{{ϵ}_o}=\frac{q}{{ϵ}_o}$.
So, options $\left(c\right)$ and $\left(d\right)$ are correct.