CapacitanceHard

Question

A parallel plate air capacitor is connected to a battery. The quantities charge, 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 the space between the plates with the battery still in connection. 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


Potential difference = V0             Potential difference = V0
Capacitance = C                         Capacitance = KC
                                                       [K is the dielectric constant of Slab K > 1]
Q0 = CV0                                      New charge = KC V0
Potential Energy  = CV02        New potential energy =  KC V02
Correct options are  (A), (D).

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