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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:

1. 1.2 n C
2. 0.3 n C
3. 2.4 n C
4. 0.9 n C

Correct answer: 1.2 n C

Solution

The induced charge in a capacitor with a dielectric is calculated using the formula Q = C * V, where C is the capacitance with the dielectric inserted (C' = K * C). Here, C' = (5/3) * 90 pF = 150 pF, and with V = 20 V, the induced charge Q = 150 pF * 20 V = 3 nC. However, the charge on the original capacitor without the dielectric is 1.8 nC, and the difference gives the induced charge of 1.2 nC.

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