A parallel plate capacitor having a plate separation of `2mm` is charged by connecting it to a `300v` supply. The energy density is

Options

A. `0.01j//m^(2)`
B. `0.01j//m^(3)`
C. `0.1j//m^(3)`
D. `10j//m^(3)`

Answer

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

### Step 1: Understand the formula for energy density
The energy density (u) in a capacitor is given by the formula:
\[ u = \frac{1}{2} \epsilon E^2 \]
where:
- \( \epsilon \) is the permittivity of free space (\( \epsilon_0 \approx 8.854 \times 10^{-12} \, \text{F/m} \))
- \( E \) is the electric field between the plates.

### Step 2: Relate electric field to voltage and distance
The electric field (E) between the plates of a capacitor can be expressed as:
\[ E = \frac{V}{d} \]
where:
- \( V \) is the voltage across the plates (300 V in this case)
- \( d \) is the separation between the plates (2 mm = \( 2 \times 10^{-3} \) m).

### Step 3: Substitute E into the energy density formula
Substituting \( E \) into the energy density formula, we get:
\[ u = \frac{1}{2} \epsilon \left(\frac{V}{d}\right)^2 \]

### Step 4: Plug in the values
Now we can substitute the known values into the equation:
- \( \epsilon = 8.854 \times 10^{-12} \, \text{F/m} \)
- \( V = 300 \, \text{V} \)
- \( d = 2 \times 10^{-3} \, \text{m} \)

Thus, we have:
\[ u = \frac{1}{2} \times 8.854 \times 10^{-12} \times \left(\frac{300}{2 \times 10^{-3}}\right)^2 \]

### Step 5: Calculate the electric field
Calculating the electric field:
\[ E = \frac{300}{2 \times 10^{-3}} = 150000 \, \text{V/m} \]

### Step 6: Substitute E back into the energy density formula
Now substituting \( E \) back into the energy density formula:
\[ u = \frac{1}{2} \times 8.854 \times 10^{-12} \times (150000)^2 \]

### Step 7: Calculate the energy density
Calculating \( (150000)^2 \):
\[ (150000)^2 = 22500000000 \, \text{V}^2/\text{m}^2 \]

Now substituting this value into the energy density equation:
\[ u = \frac{1}{2} \times 8.854 \times 10^{-12} \times 22500000000 \]
\[ u = \frac{1}{2} \times 8.854 \times 10^{-12} \times 2.25 \times 10^{10} \]
\[ u = 0.0997 \, \text{J/m}^3 \approx 0.1 \, \text{J/m}^3 \]

### Final Answer
The energy density is approximately:
\[ u \approx 0.1 \, \text{J/m}^3 \]