A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab of dielectric constant K ,
A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How would (i) the capacitance, (ii) the electric field between the plates and (iii) the energy stored in the capacitor, be affected? Justify your answer.
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(i) The capacitance of capacitor increases to K times (since C = (Kε0A)/d ∝ K)
(ii) The potential difference between the plates becomes 1/K times.
Reason: V = Q/C;Q same, increases to K times
∴ V' = V/K
As E = V/d = and V is decreased; therefore, electric field decreases to 1/K times.
(iii) Energy stored by the capacitor, u =Q2/2C.
As Q = constant, C is increased, and so energy stored by capacitor decreases to 1/K times.
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