ElectrostaticsHard

Question

A parallel plate air capacitor is connected to a battery. The quantities charge, voltage electric field and energy associated with this capacitor are given by Q0, V0, E0 and U0 respectively. A dielectric slab is now introduced to fill he sapce between the plates with the battery still in conncetion. The corresponding quantities now given by Q, V, E and U are related to the previous one as

Options

A.Q > Q0
B.V > V0
C.E > E0
D.U > U0

Solution

When dielectric slab is introduced capacity gets increased while potential difference remains unchanged.
∴ V = V0     C > C0
    Q = CV      ∴ Q > Q0
    U = CV2      ∴ U > U0
E = but V and d both are unchanged.
Therefore, E = E0
Therefore, correct options are (a) and (d).

Create a free account to view solution

View Solution Free
Topic: Electrostatics·Practice all Electrostatics questions

More Electrostatics Questions