The energy density in a parallel plate capacitor is given as 2.1 × 10–9 J/m3. The value of the electric field in the region between the plates is :

Options

A. `2.1 NC^(-1)`
B. `21.6 NC^(-1)`
C. `72 NC^(-1)`
D. `8.4 NC^(-1)`

Answer

To find the electric field \( E \) in a parallel plate capacitor given the energy density \( u \), we can use the relationship between energy density and electric field. The energy density \( u \) in an electric field is given by the formula:

\[ u = \frac{1}{2} \varepsilon_0 E^2 \]

where:
- \( u \) is the energy density,
- \( \varepsilon_0 \) is the permittivity of free space (\( \varepsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m} \)),
- \( E \) is the electric field.

Given that the energy density \( u = 2.1 \times 10^{-9} \, \text{J/m}^3 \), we can rearrange the formula to solve for \( E \):

\[ E^2 = \frac{2u}{\varepsilon_0} \]

Now, substituting the values:

1. Calculate \( 2u \):
\[ 2u = 2 \times (2.1 \times 10^{-9}) = 4.2 \times 10^{-9} \, \text{J/m}^3 \]

2. Substitute \( \varepsilon_0 \):
\[ E^2 = \frac{4.2 \times 10^{-9}}{8.85 \times 10^{-12}} \]

3. Calculate \( E^2 \):
\[ E^2 \approx \frac{4.2 \times 10^{-9}}{8.85 \times 10^{-12}} \approx 474.77 \, \text{(V/m)}^2 \]

4. Take the square root to find \( E \):
\[ E \approx \sqrt{474.77} \approx 21.78 \, \text{V/m} \]

Thus, the value of the electric field in the region between the plates is approximately \( 21.78 \, \text{V/m} \).