Consider a parallel plate capcaitor with plates 20 cm by 20 cm and separated by 2 mm. The dielectric constant of the material between the plates is 5. The plates are connected to a voltage source of 500 V. The energy density of the field between the plates will be close to

Options

A. `2.65J//m^(3)`
B. `1.95J//m^(3)`
C. `1.38J//m^(3)`
D. `0.69J//m^(3)`

Explanation

To find the energy density of the electric field between the plates of a parallel plate capacitor, we can use the formula for energy density \( U \):

\[ U = \frac{1}{2} \epsilon E^2 \]

Where:
- \( U \) is the energy density,
- \( \epsilon \) is the permittivity of the dielectric material,
- \( E \) is the electric field between the plates.

### Step 1: 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 plates (500 V),
- \( d \) is the separation between the plates (2 mm = \( 2 \times 10^{-3} \) m).

Substituting the values:

\[ E = \frac{500 \, \text{V}}{2 \times 10^{-3} \, \text{m}} = \frac{500}{0.002} = 250000 \, \text{V/m} \]

### Step 2: Calculate the Permittivity \( \epsilon \)

The permittivity \( \epsilon \) of the dielectric material is given by:

\[ \epsilon = K \epsilon_0 \]

Where:
- \( K \) is the dielectric constant (5),
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.854 \times 10^{-12} \, \text{F/m} \).

Substituting the values:

\[ \epsilon = 5 \times 8.854 \times 10^{-12} \, \text{F/m} = 4.427 \times 10^{-11} \, \text{F/m} \]

### Step 3: Calculate the Energy Density \( U \)

Now, substitute \( E \) and \( \epsilon \) into the energy density formula:

\[ U = \frac{1}{2} \epsilon E^2 \]

Substituting the values:

\[ U = \frac{1}{2} \times 4.427 \times 10^{-11} \, \text{F/m} \times (250000 \, \text{V/m})^2 \]

Calculating \( (250000)^2 \):

\[ (250000)^2 = 62500000000 \, \text{(V/m)}^2 \]

Now substituting this value into the equation for \( U \):

\[ U = \frac{1}{2} \times 4.427 \times 10^{-11} \times 62500000000 \]

Calculating:

\[ U = \frac{1}{2} \times 4.427 \times 10^{-11} \times 6.25 \times 10^{10} \]
\[ U = \frac{1}{2} \times 2.7671875 \approx 1.38359375 \, \text{J/m}^3 \]

Rounding off, we get:

\[ U \approx 1.38 \, \text{J/m}^3 \]

### Final Answer

The energy density of the field between the plates will be close to **1.38 J/m³**.

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