Animated Solution for Physics - Electrostatics: An elliptical cavity is carved within a perfect conductor. A positive 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

Imagine you are exploring the interior of a solid block of metal. This isn't just any metal; it's a perfect conductor. Deep inside, an elliptical cavity has been hollowed out, and right at its geometric center sits a solitary positive point charge, $+q$. On the inner walls of this cavity, we mark two distinct points: $A$ on the minor axis (closer to the center) and $B$ on the major axis (farther away). Our mission is to decode the electrical environment inside this cavity.

The most fundamental rule of electrostatics involving conductors is that they are equipotential volumes. When charges reach electrostatic equilibrium, there is no net movement of charge. If there were a potential difference between any two points on or within the conductor, electrons would flow to neutralize it.

Because points $A$ and $B$ both reside on the inner surface of the conducting cavity, they are part of this equipotential body. Therefore, without any complex calculations, we can confidently state that the electrical potential at $A$ is exactly equal to the potential at $B$:

This elegant property immediately confirms that option (c) is correct.

While the potential is uniform, the electric field and the induced charge density are a different story. The positive charge $+q$ at the center induces a negative charge on the inner wall of the cavity. Because the cavity is elliptical, the distance from the central charge to the wall varies. Point $A$ is physically closer to the charge $+q$ than point $B$ ($r_A<r_B$).

In electrostatics, the electric field just outside a conductor's surface is directly proportional to the local surface charge density, given by $E=\frac{\sigma }{{\epsilon }_0}$. Because point $A$ is closer to the source charge, the electric field lines are denser there, causing a higher concentration of induced negative charge. Consequently, the magnitude of the charge density and the electric field at $A$ will be strictly greater than at $B$:

This asymmetry means options (a) and (b) are incorrect. If the cavity had been perfectly spherical, the symmetry would have guaranteed uniform charge density and electric field, making those options correct as well.

Finally, let's determine the total electric flux passing through the surface of the cavity. Gauss's Law is our ultimate tool here. It states that the total electric flux $\Phi$ through any closed surface is equal to the total enclosed charge divided by the permittivity of free space:

If we construct a Gaussian surface that perfectly traces the inner boundary of the cavity, the only charge enclosed within this volume is the point charge $+q$ at the center. The induced charges reside on the boundary itself and do not contribute to the net enclosed charge of a surface just infinitesimally inside the cavity. Therefore, the total flux is simply:

This confirms that option (d) is also correct. The beauty of Gauss's Law is that it holds true regardless of the cavity's shape—whether it's spherical, elliptical, or completely irregular, the total flux depends solely on the enclosed charge.