A parallel plate capacitors of plate area A has a total surface charge density `+sigma` on one plate and a total surface charge density `-sigma` on the other plate. The force one plate due to the other plate is given by :

Options

A. `(sigma^(2)A^(2))/(2epsi_(0))`
B. `(sigma^(2)A^(2))/(epsi_(0))`
C. `(sigma^(2)A)/(2epsi_(0))`
D. `(sigma^(2)A)/(epsi_(0))`

Explanation

To find the force on one plate of a parallel plate capacitor due to the other plate, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Charge Distribution**:
- We have two plates of a parallel plate capacitor. One plate has a surface charge density of \( +\sigma \) and the other has a surface charge density of \( -\sigma \). The area of each plate is \( A \).

2. **Electric Field Due to One Plate**:
- The electric field \( E \) created by a single infinite plate with surface charge density \( \sigma \) is given by the formula:
\[ E = \frac{\sigma}{2\epsilon_0} \]
- For the plate with charge density \( -\sigma \), the electric field at the location of the positively charged plate (due to the negatively charged plate) will also be:
\[ E = \frac{\sigma}{2\epsilon_0} \]
- The direction of the electric field due to the negative plate is towards the plate, which means it points towards the negative plate.

3. **Force on One Plate**:
- The force \( F \) on a charged plate in an electric field is given by:
\[ F = qE \]
- Here, \( q \) is the charge on the positively charged plate. The charge \( q \) can be expressed in terms of surface charge density \( \sigma \) and area \( A \):
\[ q = \sigma A \]

4. **Substituting Values**:
- Now substituting \( q \) into the force equation:
\[ F = (\sigma A) \left(\frac{\sigma}{2\epsilon_0}\right) \]
- This simplifies to:
\[ F = \frac{\sigma^2 A}{2\epsilon_0} \]

5. **Final Result**:
- Therefore, the force on one plate due to the other plate is:
\[ F = \frac{\sigma^2 A}{2\epsilon_0} \]