A parallel plate condenser has a unifrom electric field `E (V//m)` in the space between the plates. If the distance between the plates is `d(m)` and area of each plate is `A(m^(2))` the energy (joule) stored in the condenser is

Options

A. `1/2 epsilon_(0)E^(2)`
B. `E^(2)Ad//epsilon_(0)`
C. `1/2 epsilon_(0) E^(2)Ad`
D. `epsilon_(0)EAD`

Explanation

To find the energy stored in a parallel plate capacitor with a uniform electric field \( E \), distance between the plates \( d \), and area of each plate \( A \), we can follow these steps:

### Step-by-step Solution:

1. **Understand the formula for energy stored in a capacitor**:
The energy \( U \) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
where \( C \) is the capacitance and \( V \) is the voltage across the capacitor.

2. **Determine the capacitance \( C \)**:
For a parallel plate capacitor, the capacitance \( C \) is given by:
\[ C = \frac{A \epsilon_0}{d} \]
where \( \epsilon_0 \) is the permittivity of free space.

3. **Calculate the voltage \( V \)**:
The voltage \( V \) across the plates can be expressed in terms of the electric field \( E \) and the distance \( d \):
\[ V = E \cdot d \]

4. **Substitute \( C \) and \( V \) into the energy formula**:
Now, substitute the expressions for \( C \) and \( V \) into the energy formula:
\[ U = \frac{1}{2} \left( \frac{A \epsilon_0}{d} \right) (E \cdot d)^2 \]

5. **Simplify the expression**:
Expanding the equation gives:
\[ U = \frac{1}{2} \left( \frac{A \epsilon_0}{d} \right) (E^2 d^2) \]
This simplifies to:
\[ U = \frac{1}{2} A \epsilon_0 E^2 d \]

6. **Final expression for energy stored**:
Thus, the energy stored in the capacitor is:
\[ U = \frac{1}{2} A \epsilon_0 E^2 d \]

### Final Answer:
The energy stored in the parallel plate capacitor is:
\[ U = \frac{1}{2} A \epsilon_0 E^2 d \text{ (joules)} \]