A parallel plate capacitor has an electric field of `10^(5)V//m` between the plates. If the charge on the capacitor plate is `1muC`, then force on each capacitor plate is-

Options

A. `0.1 N`
B. `0.05 N`
C. `0.02 N`
D. `0.01 N`

Explanation

To find the force on each plate of a parallel plate capacitor given the electric field and the charge on the plates, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Electric field, \( E = 10^5 \, \text{V/m} \)
- Charge on each plate, \( Q = 1 \, \mu\text{C} = 1 \times 10^{-6} \, \text{C} \)

2. **Understanding the Electric Field Contribution:**
- The electric field \( E \) between the plates of a parallel plate capacitor is uniform. The force on each plate can be calculated using the electric field created by the other plate.
- The electric field due to one plate is given by \( E_{\text{plate}} = \frac{\sigma}{2 \epsilon_0} \), where \( \sigma \) is the surface charge density and \( \epsilon_0 \) is the permittivity of free space.

3. **Calculate the Force on One Plate:**
- The force \( F \) on a charge \( Q \) in an electric field \( E \) is given by the formula:
\[ F = Q \cdot E \]
- However, since the electric field \( E \) is due to both plates, the effective electric field acting on one plate is \( \frac{E}{2} \).
- Therefore, the force on one plate becomes:
\[ F = Q \cdot \left(\frac{E}{2}\right) \]

4. **Substituting the Values:**
- Substitute \( Q \) and \( E \) into the equation:
\[ F = (1 \times 10^{-6} \, \text{C}) \cdot \left(\frac{10^5 \, \text{V/m}}{2}\right) \]
- Simplifying this gives:
\[ F = (1 \times 10^{-6}) \cdot (5 \times 10^4) = 5 \times 10^{-2} \, \text{N} = 0.05 \, \text{N} \]

5. **Conclusion:**
- The force on each capacitor plate is \( 0.05 \, \text{N} \).