Question

A parallel-plate capacitor is charged by being connected to a battery and is kept connected to the battery. The separation between the plates is then doubled. How does the electric field, charge and total energy change? Electric field is halved, charge is doubled, total energy is doubled Electric field is doubled, charge is halved, total energy is doubled Electric field is doubled, charge is halved, total energy is halved Electric field is halved, charge is halved, total energy is halved

Answer 1

We know electric field E = v/d E' = v/d' d ' = 2 d E' = v/2 d E' = E/2 The field is halved. We know E = v/d = Q/C d Vertical-bar v = Q/C Now, Q = E e d Q' = C E d' f d ' = 2 d Q ' = 2 e E d Q' = 2 Q change is doubled Total energy U = 1/ 2 E^2 A d and U' = 1/2 E^2 A d ' Vertical-bar d' = 2 d U' = 2/2 E^2 A d = 2 U Total energy is doubled