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EEJ MAIN Physics QUESTION #6989
Question 1

A parallel plate capacitor of capacitance 90 pF is connected to a 20 V battery. A dielectric slab with dielectric constant $K = \dfrac{5}{3}$ is then inserted between the plates. What is the magnitude of the induced (bound) charge on the dielectric?

  • 2.4 nC
  • 0.9 nC
  • 1.2 nC✔️
  • 0.3 nC
Correct Answer Explanation

Initial charge (without dielectric):

$Q_0 = C_0 V = 90 \times 10^{-12} \times 20 = 1800\ \text{pC} = 1.8\ \text{nC}$

New capacitance with dielectric:

$C = KC_0 = \frac{5}{3} \times 90 = 150\ \text{pF}$

New free charge:

$Q = CV = 150 \times 10^{-12} \times 20 = 3000\ \text{pC} = 3\ \text{nC}$

Induced charge on dielectric:

$Q_{induced} = Q\left(1 - \frac{1}{K}\right) = 3 \times \left(1 - \frac{3}{5}\right) = 3 \times \frac{2}{5} = 1.2\ \text{nC}$