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 :

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 : (A) 1.2 n C (B) 0.3 n C (C) 2.4 n C (D) 0.9 n C

Tardigrade Admin · Accepted Answer

C'=K C0 Q=K C0 V Q text induced =Q(1-(1/K)) =(5/3) × 90 × 10-12 × 20(1-(3/5)) =1.2 nC

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

1.2 n C

0.3 n C

2.4 n C

0.9 n C

$C^{'} = K C_{0}$
$Q = K C_{0} V$
$Q_{induced} = Q (1 - \frac{1}{K})$
$= \frac{5}{3} \times 90 \times 1 0^{- 12} \times 20 (1 - \frac{3}{5})$
$= 1.2 n C$

Q. A parallel plate capacitor of capacitance 90pF is connected to a battery of emf 20V. If a dielectric material of dielectric constant K=35​ is inserted between the plates, the magnitude of the induced charge will be :