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.

Answer 2

Step 1: Given data.
Radius of circular plates, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
Current, \( I = 0.15 \, \text{A} \)
Rate of change of potential difference, \( \frac{dV}{dt} = 7 \times 10^6 \, \text{V/s} \)
We need to find the distance between plates \( d \) (in meters), and then convert it to an integer (in mm).

Step 2: Relation between charging current and changing voltage.
The current through a charging capacitor is related to the rate of change of potential difference as:
\[ I = C \frac{dV}{dt} \] where \( C \) is the capacitance of the capacitor.

Step 3: Expression for capacitance of a parallel plate capacitor.
\[ C = \varepsilon_0 \frac{A}{d} \] where \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \), \( A = \pi r^2 \).

Step 4: Substitute capacitance into the current equation.
\[ I = \varepsilon_0 \frac{A}{d} \frac{dV}{dt} \] \[ d = \varepsilon_0 A \frac{dV}{dt} \div I \]

Step 5: Substitute known values.
\[ A = \pi (0.1)^2 = 3.14 \times 10^{-2} \, \text{m}^2 \] \[ d = \frac{8.85 \times 10^{-12} \times 3.14 \times 10^{-2} \times 7 \times 10^6}{0.15} \] \[ d = \frac{1.946 \times 10^{-6}}{0.15} = 1.296 \times 10^{-5} \, \text{m} \]

Step 6: Convert to millimeters.
\[ d = 1.296 \times 10^{-5} \, \text{m} = 1.296 \times 10^{-2} \, \text{mm} = 1320 \, \text{μm} \] Thus, the integer value of distance between the plates is:

Final Answer:
\[ \boxed{1320} \]