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

### Step 2: Determine the Electric Field
The electric field \( E \) created by one plate (say the positively charged plate) at the location of the other plate can be calculated using the formula for the electric field due to an infinite sheet of charge:
\[ E = \frac{\sigma}{2\epsilon_0} \]
where \( \sigma \) is the surface charge density and \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Calculate the Surface Charge Density
The surface charge density \( \sigma \) can be expressed in terms of the charge \( Q \) and the area \( A \):
\[ \sigma = \frac{Q}{A} \]

### Step 4: Substitute for Electric Field
Substituting \( \sigma \) into the electric field equation gives:
\[ E = \frac{Q}{2\epsilon_0 A} \]

### Step 5: Calculate the Force on One Plate
The force \( F \) on one plate due to the electric field \( E \) created by the other plate can be calculated using the formula:
\[ F = Q \cdot E \]
Substituting the expression for \( E \):
\[ F = Q \cdot \left(\frac{Q}{2\epsilon_0 A}\right) \]
This simplifies to:
\[ F = \frac{Q^2}{2\epsilon_0 A} \]

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

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