The plates of a parallel plate capacitor have an area of `90cm^(2)` and each are separated by 2mm. The capacitor is charged by connecting it to a 400 V supply. Then the energy density of the energy stored (in `Jm^(-3)`) in the capacitor is (take `epsi_(0)=8.8xx10^(-12)Fm^(-1)`)

Options

A. `0.113`
B. `0.117`
C. `0.152`
D. none of these

Explanation

To find the energy density of the energy stored in a parallel plate capacitor, we can follow these steps:

### Step 1: Convert the given area and separation to SI units
- Area \( A = 90 \, \text{cm}^2 = 90 \times 10^{-4} \, \text{m}^2 = 9.0 \times 10^{-2} \, \text{m}^2 \)
- Separation \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)

### Step 2: Calculate the electric field \( E \)
The electric field \( E \) between the plates of a capacitor is given by the formula:
\[ E = \frac{V}{d} \]
Where:
- \( V = 400 \, \text{V} \) (the voltage supply)
- \( d = 2 \times 10^{-3} \, \text{m} \)

Substituting the values:
\[ E = \frac{400 \, \text{V}}{2 \times 10^{-3} \, \text{m}} = \frac{400}{0.002} = 200000 \, \text{V/m} = 2.0 \times 10^5 \, \text{V/m} \]

### Step 3: Calculate the energy density \( u \)
The energy density \( u \) in a capacitor can be calculated using the formula:
\[ u = \frac{1}{2} \epsilon_0 E^2 \]
Where:
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) (the permittivity of free space)

Substituting the values:
\[ u = \frac{1}{2} \times 8.85 \times 10^{-12} \, \text{F/m} \times (2.0 \times 10^5 \, \text{V/m})^2 \]

Calculating \( (2.0 \times 10^5)^2 \):
\[ (2.0 \times 10^5)^2 = 4.0 \times 10^{10} \, \text{(V/m)}^2 \]

Now substituting back into the energy density formula:
\[ u = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4.0 \times 10^{10} \]
\[ u = 0.5 \times 8.85 \times 4.0 \times 10^{-2} \]
\[ u = 0.5 \times 35.4 \times 10^{-2} = 17.7 \times 10^{-2} = 0.177 \, \text{J/m}^3 \]

### Final Answer
The energy density of the energy stored in the capacitor is:
\[ \boxed{0.177 \, \text{J/m}^3} \]