Question

An ellipsoidal cavity is made within a perfect conductor. A positive charge $q$ is placed at the center 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. change density at $A$ $=$ change density at $B$
• C. potential at $A$ $=$ potential at $B$
• D. total electric field flux through the surface of the cavity is $q/{ϵ}_0$

Answer 1

Answer: (C), (D)

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