Question

A parallel plate air capacitor has capacity $C$, distance of separation between plates is $d$ and potential difference $V$ is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is

• A. $ \frac{CV^2}{d} $
• B. $ \frac{C^2V^2}{2d^2} $
• C. $\frac{C^2V^2}{2d} $
• D. $ \frac{CV^2}{2d} $

Answer 1

The Correct Option is D

The question asks for the force of attraction between the plates of a parallel plate air capacitor, given its capacitance, distance between its plates, and the potential difference applied.

1. Given: Capacitance $C$, potential difference $V$, and separation between plates $d$.

2. The energy stored in the capacitor is given by:

   E = \frac{1}{2} C V^2

3. The force of attraction between the plates can be derived from the expression for the energy stored in the capacitor. The force is the derivative of energy with respect to the separation $d$:

   F = -\frac{\partial E}{\partial d}

4. However, the more direct formula for the attractive force between the plates is:

   F = \frac{1}{2} \epsilon_0 \frac{A V^2}{d^2},

   where $A$ is the area of the plate. Here, \epsilon_0 A/d = C.

5. Substitute $C = \epsilon_0 \frac{A}{d}$ into the formula:

   F = \frac{1}{2} C \frac{V^2}{d}

6. This matches the given option of \frac{CV^2}{2d}, confirming it is the correct answer.

7. Conclusion: The correct answer is \frac{CV^2}{2d}. The other options do not match this calculation.