In a parallel plate capacitor with plate area A and charge Q, the force on one plate because of the charge on the other is equal to

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understand the Configuration**:
A parallel plate capacitor consists of two plates, one positively charged (+Q) and the other negatively charged (-Q). The area of each plate is denoted as A.

2. **Calculate the Electric Field (E)**:
The electric field (E) between the plates of a parallel plate capacitor can be calculated using the formula:
\[ E = \frac{\sigma}{\epsilon_0} \]
where \(\sigma\) is the surface charge density and \(\epsilon_0\) is the permittivity of free space. The surface charge density \(\sigma\) is given by:
\[ \sigma = \frac{Q}{A} \]
Therefore, the electric field becomes:
\[ E = \frac{Q}{A \epsilon_0} \]

3. **Determine the Electric Field Due to One Plate**:
Each plate creates an electric field. The electric field due to one plate (say the positively charged plate) is:
\[ E_1 = \frac{Q}{2A\epsilon_0} \]
The factor of 1/2 arises because each plate contributes half of the total electric field in the region between them.

4. **Calculate the Force on One Plate**:
The force (F) on one plate due to the electric field created by the other plate can be calculated using the formula:
\[ F = qE \]
Here, \(q\) is the charge on the plate experiencing the force, and \(E\) is the electric field due to the other plate. For the negatively charged plate, the force will be:
\[ F = Q \cdot E_1 = Q \cdot \frac{Q}{2A\epsilon_0} \]
Simplifying this gives:
\[ F = \frac{Q^2}{2A\epsilon_0} \]

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