Question

How can the capacity of a conductor be increased? Radius of the plates of a parallel plate air capacitor is \( 3 \times 10^{-2} \, \text{m} \) and the capacitance is equal to the capacitance of a charged sphere of radius 1 m. Find the distance between the plates of the capacitor.

Hint:

The capacitance of a parallel plate capacitor depends on the area of the plates and the distance between them. Increasing the area or reducing the distance increases the capacitance.

Answer 1

Step 1: Formula for the Capacitance of a Parallel Plate Capacitor.
The capacitance \( C \) of a parallel plate capacitor is given by the formula: \[ C = \frac{\epsilon_0 A}{d} \] where: - \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{C}^2 / \text{N} \cdot \text{m}^2 \)), - \( A \) is the area of the plates, - \( d \) is the distance between the plates.
Step 2: Formula for the Capacitance of a Charged Sphere.
The capacitance of a charged sphere of radius \( r \) is given by: \[ C = 4 \pi \epsilon_0 r \] where \( r = 1 \, \text{m} \).
Step 3: Setting Capacitances Equal.
We are given that the capacitance of the parallel plate capacitor is equal to the capacitance of the charged sphere. So, we set the capacitance formulas equal: \[ \frac{\epsilon_0 A}{d} = 4 \pi \epsilon_0 r \]
Step 4: Solving for Distance.
The area \( A \) of the plates of the parallel plate capacitor is given by \( A = \pi R^2 \), where \( R = 3 \times 10^{-2} \, \text{m} \) is the radius of the plates. Substituting into the equation: \[ \frac{\epsilon_0 \pi R^2}{d} = 4 \pi \epsilon_0 r \] Simplifying and solving for \( d \): \[ \frac{R^2}{d} = 4r \] \[ d = \frac{R^2}{4r} \] Substituting the given values: \[ d = \frac{(3 \times 10^{-2})^2}{4 \times 1} = \frac{9 \times 10^{-4}}{4} = 2.25 \times 10^{-4} \, \text{m} \]
Step 5: Conclusion.
Thus, the distance between the plates of the capacitor is \( 2.25 \times 10^{-4} \, \text{m} \).