Question

As shown in the figure below, an ellipsoidal cavity is carved within a perfect conductor. A positive charge $Q$ is placed at the centre of the cavity. If points $A$ and $b$ are shown on the cavity surface, then which among the following choices is correct?

• A. $V_A=V_b$
• B. $E_A=E_b$
• C. Charge density at $A$ = Charge density at $b$
  
• D. Total electric flux through the surface of the cavity is zero

Answer 1

The correct option is A $V_A=V_b$

Under electrostatic condition, all points lying on one conductor are in same potential. Therefore, potential at $A=$ potential at $B$

Using Gauss's theorem total flux through the surface of the cavity will be $\frac{a}{{ϵ}_0}$