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. 0.5N
B. 0.05 N
C. 0.005 N
D. None of these

Explanation

To find the force on each plate of a parallel plate capacitor, we can follow these steps:

### Step 1: Understand the given values
- Electric field (E) between the plates = \(10^5 \, \text{V/m}\)
- Charge on each plate (Q) = \(1 \, \mu\text{C} = 1 \times 10^{-6} \, \text{C}\)

### Step 2: Calculate the electric field due to one plate
For a parallel plate capacitor, the electric field (E) between the plates is given by:
\[ E = \frac{\sigma}{\epsilon_0} \]
where \(\sigma\) is the surface charge density and \(\epsilon_0\) is the permittivity of free space.

However, the electric field due to one plate can be expressed as:
\[ E' = \frac{E}{2} \]
Thus, the electric field due to one plate (either positive or negative) is:
\[ E' = \frac{10^5}{2} = 5 \times 10^4 \, \text{V/m} \]

### Step 3: Calculate the force on one plate
The force (F) on a charged plate in an electric field is given by:
\[ F = Q \cdot E' \]
Substituting the values we have:
\[ F = (1 \times 10^{-6} \, \text{C}) \cdot (5 \times 10^4 \, \text{V/m}) \]
Calculating this gives:
\[ F = 5 \times 10^{-2} \, \text{N} = 0.05 \, \text{N} \]

### Conclusion
The force on each capacitor plate is \(0.05 \, \text{N}\).

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