E and V respectively represent magnitudes of electrostatic field and potential inside a charged metallic shell. We have:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $E=0, V≠0$
For a charged, metallic shell, the electrostatic field is zero due to the symmetry of charge distribution.
$∴E=0$
and, The electric potential V inside the shell, however, is constant and non-zero. This is because the shell forms a equipotential surface.
$∴V=K≠0$ [K = Constant]