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 the current, potential difference, and capacitance is given by: \[ I = C \frac{dV}{dt} \] Where:
- \( I = 0.15 \, \text{A} \),
- \( \frac{dV}{dt} = 7 \times 10^6 \, \text{V/s} \),
- \( C = \epsilon_0 \frac{A}{d} \), with \( A \) being the area of the plates and \( d \) the distance between them.
The area of the circular plates is: \[ A = \pi r^2 = \pi (0.1 \, \text{m})^2 = 3.14 \times 10^{-2} \, \text{m}^2 \] Substitute \( C \) into the current equation: \[ I = \epsilon_0 \frac{A}{d} \frac{dV}{dt} \]
Now, solving for \( d \): \[ d = \frac{\epsilon_0 \pi r^2 \frac{dV}{dt}}{I} \] Substitute values: \[ d = \frac{(9 \times 10^{-12})(3.14 \times 10^{-2})(7 \times 10^6)}{0.15} \] \[ d = 1.32 \, m = 1320 \, \mu m \] Thus, the distance between the plates is 1320 \(\mu\) m.