Question

An ellipsoidal cavity is made 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

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

Answer 1

Answer: (C), (D)

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