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Electrostatic Potential And Capacitance

Question

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K= 5/3 is inserted between the plates, the magnitude of the induced charge will be:

0.9 n C

1.2 n C

0.3 n C

2.4 n C

Solution

1.2 n C

Charge on Capacitor, Q₁ =CV

After inserting dielectric of dielectric constant = K

Qf = (kC)V

Induced charges on the dielectric

Q_ind =Q_f-Q_i = KCV-CV

= (K-1)CV = $(\frac{5}{3} - 1) x 90 p F x 2 V = 1.2 n c$