Question

A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A. If the rate of change of potential difference between the plates is \( 7 \times 10^6 \, \text{V/s} \), then the integer value of the distance between the parallel plates is:

Hint:

Use the relationship between the current, capacitance, and rate of change of potential difference to solve for the distance between the plates in a capacitor.

Answer 1

Correct Answer: 1320

The relationship between current (\(I\)), potential difference rate of change (\(\frac{dV}{dt}\)), and capacitance (\(C\)) is described by the equation: \[ I = C \frac{dV}{dt} \] Given values are: - \( I = 0.15 \, \text{A} \) - \( \frac{dV}{dt} = 7 \times 10^6 \, \text{V/s} \) - Capacitance is defined as \( C = \epsilon_0 \frac{A}{d} \), where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the plate area, and \( d \) is the distance between plates. The area of the circular plates is calculated as: \[ A = \pi r^2 = \pi (0.1 \, \text{m})^2 = 3.14 \times 10^{-2} \, \text{m}^2 \] Substituting the expression for \( C \) into the current equation yields: \[ I = \epsilon_0 \frac{A}{d} \frac{dV}{dt} \] Rearranging to solve for the distance \( d \): \[ d = \frac{\epsilon_0 A \frac{dV}{dt}}{I} \] Substituting the given values and constants: \[ d = \frac{(9 \times 10^{-12})(3.14 \times 10^{-2})(7 \times 10^6)}{0.15} \] The calculated distance is: \[ d = 1.32 \, \text{m} = 1320 \, \mu \text{m} \] Therefore, the separation between the plates is 1320 \(\mu\) m.