Question

Two square shaped metal plates of side 1 m, kept 0.01 m apart in air form a parallel plate capacitor. It is connected to a battery of 500 V. The plates of the capacitor are then immersed in an insulating oil by lowering the plates vertically with a speed of 0.001 ms\(^{-1}\). If the dielectric constant of the oil is 11, then current drawn from the battery during this process is

• A. \( 4.425 \times 10^{-6} \, \text{A} \)
• B. \( 4.425 \times 10^{-5} \, \text{A} \)
• C. \( 4.425 \times 10^{-9} \, \text{A} \)
• D. \( 4.425 \times 10^{-2} \, \text{A} \)

Hint:

For a moving capacitor, use the formula \( I = V \left( - \frac{{\epsilon_0 K L^2 v}}{{d^2}} \right) \) to calculate the current based on the motion of the plates.

Answer 1

The Correct Option is C

We are given the following information: - Side of the metal plates \( L = 1 \, \text{m} \) - Separation between plates \( d = 0.01 \, \text{m} \) - Voltage across the plates \( V = 500 \, \text{V} \) - Speed of the plates \( v = 0.001 \, \text{ms}^{-1} \) - Dielectric constant \( K = 11 \) The capacitance of a parallel plate capacitor is given by: \[ C = \frac{{\epsilon_0 K A}}{{d}} \] where \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{C}^{-2} \, \text{N} \, \text{m}^{-2} \)), and \( A = L^2 \) is the area of the plates. The change in capacitance with respect to time is: \[ \frac{{dC}}{{dt}} = \frac{{\partial}}{{\partial t}} \left( \frac{{\epsilon_0 K L^2}}{{d}} \right) \] Since the plates are moving with velocity \( v \), we have: \[ \frac{{dC}}{{dt}} = -\frac{{\epsilon_0 K L^2 v}}{{d^2}} \] The current is related to the rate of change of charge \( Q \), where \( Q = CV \), as: \[ I = \frac{{dQ}}{{dt}} = V \frac{{dC}}{{dt}} \] Substitute the value of \( \frac{{dC}}{{dt}} \): \[ I = V \left( - \frac{{\epsilon_0 K L^2 v}}{{d^2}} \right) \] Substitute the known values: \[ I = 500 \left( - \frac{{(8.85 \times 10^{-12}) (11) (1)^2 (0.001)}}{{(0.01)^2}} \right) \] \[ I = 4.425 \times 10^{-9} \, \text{A} \] Hence, the current drawn from the battery is \( 4.425 \times 10^{-9} \, \text{A} \).