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).
∴ V = V0 C > C0
Q = CV ∴ Q > Q0
U =
CV2 ∴ U > U0E =
but V and d both are unchanged. Therefore, E = E0
Therefore, correct options are (a) and (d).
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