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 solve the problem, we need to calculate the energy density of the energy stored in a parallel plate capacitor. Here's a step-by-step breakdown of the solution:

### Step 1: Convert Given Values to SI Units
- The area of the plates \( A = 90 \, \text{cm}^2 \).
- Convert to square meters:
\[ A = 90 \, \text{cm}^2 = 90 \times 10^{-4} \, \text{m}^2 = 0.009 \, \text{m}^2 \]

- The separation between the plates \( d = 2 \, \text{mm} \).
- Convert to meters:
\[ d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \]

### 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 \( V \) is the voltage across the capacitor.

Given \( V = 400 \, \text{V} \) and \( d = 2 \times 10^{-3} \, \text{m} \):
\[ E = \frac{400 \, \text{V}}{2 \times 10^{-3} \, \text{m}} = 200000 \, \text{V/m} = 2 \times 10^5 \, \text{V/m} \]

### Step 3: Calculate the Energy Density \( u \)
The energy density \( u \) stored in the electric field of the capacitor is given by the formula:
\[ u = \frac{1}{2} \epsilon_0 E^2 \]
where \( \epsilon_0 \) is the permittivity of free space, given as \( \epsilon_0 = 8.8 \times 10^{-12} \, \text{F/m} \).

Substituting the values:
\[ u = \frac{1}{2} \times (8.8 \times 10^{-12} \, \text{F/m}) \times (2 \times 10^5 \, \text{V/m})^2 \]

Calculating \( (2 \times 10^5)^2 \):
\[ (2 \times 10^5)^2 = 4 \times 10^{10} \]

Now substituting back into the energy density formula:
\[ u = \frac{1}{2} \times (8.8 \times 10^{-12}) \times (4 \times 10^{10}) \]
\[ u = \frac{1}{2} \times 35.2 \times 10^{-2} \, \text{J/m}^3 \]
\[ u = 17.6 \times 10^{-2} \, \text{J/m}^3 = 0.176 \, \text{J/m}^3 \]

### Final Answer
Thus, the energy density of the energy stored in the capacitor is:
\[ \boxed{0.176 \, \text{J/m}^3} \]