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. `0.01 J//m^(3)`
B. `0.1 J//m^(3)`
C. `1.0 J//m^(3)`
D. `10 J//m^(3)`

Explanation

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:
\[ u = \frac{1}{2} \epsilon_0 E^2 \]
where \( \epsilon_0 \) is the permittivity of free space and \( E \) is the electric field between the plates.

### Step 2: Calculate the electric field (E)
The electric field (E) between the plates of a parallel plate capacitor can be calculated using the formula:
\[ E = \frac{V}{d} \]
where \( V \) is the voltage across the plates and \( d \) is the separation between the plates.

Given:
- \( V = 300 \, \text{V} \)
- \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)

Substituting the values:
\[ E = \frac{300 \, \text{V}}{2 \times 10^{-3} \, \text{m}} = 150000 \, \text{V/m} \]

### Step 3: Substitute E into the energy density formula
Now, substitute the value of \( E \) into the energy density formula:
\[ u = \frac{1}{2} \epsilon_0 E^2 \]
Using \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \):
\[ u = \frac{1}{2} (8.85 \times 10^{-12}) (150000)^2 \]

### Step 4: Calculate \( E^2 \)
Calculating \( E^2 \):
\[ E^2 = (150000)^2 = 22500000000 \, \text{V}^2/\text{m}^2 \]

### Step 5: Calculate the energy density (u)
Now, substituting \( E^2 \) back into the energy density formula:
\[ u = \frac{1}{2} (8.85 \times 10^{-12}) (22500000000) \]
\[ u = \frac{1}{2} (1.98875 \times 10^{-2}) \]
\[ u = 0.00994375 \, \text{J/m}^3 \]
\[ u \approx 0.01 \, \text{J/m}^3 \]

### Final Answer
The energy density is approximately \( 0.01 \, \text{J/m}^3 \).

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