Question

An elliptical cavity is carved out in a perfect conductor. A positive charge q is placed at the centre of the cavity. The points A and B are shown in figure. Then :

• A. electric field near A in the cavity = electric field near B in the cavity
• B. charge density at A = charge density at B
• C. potential at A = potential at B
• D. total electric field flux through the surface of the cavity = $\frac{q}{{\epsilon }_0}$

Answer 1

Answer: (C), (D)

The correct options are
C total electric field flux through the surface of the cavity = $\frac{q}{{\epsilon }_0}$
D potential at A = potential at B

Here the distances from charge q to A and B is different. So the charge density is different at A and B.
The field is the effect due to charge and the field is proportional to inverse square of distance between source point and measure point. As distance is different so field at A $\ne$ field at B.
Since, potential inside a conductor is constant so $V_A=V_B$
By Gauss's law , the flux $\varphi =\frac{q_{enclosed}}{{ϵ}_0}=\frac{q}{{ϵ}_0}$