Question:
The energy density of the electric field \( 2 \, \text{V/m} \) in a capacitor \( C \) is \( \varepsilon_0 \) is the permittivity of free space
The energy density of the electric field \( 2 \, \text{V/m} \) in a capacitor \( C \) is \( \varepsilon_0 \) is the permittivity of free space
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To calculate the energy density, use the formula \( u = \frac{1}{2} \varepsilon_0 E^2 \). It tells you how much energy is stored in a given electric field.
Updated On: Apr 7, 2026
- \( 3 \varepsilon_0 \)
- \( \frac{\varepsilon_0}{2} \)
- \( 4 \varepsilon_0 \)
- \( \frac{\varepsilon_0}{4} \)
- \( 2 \varepsilon_0 \)
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The Correct Option is
Solution and Explanation
The energy density of 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
- \( E \) is the electric field
Given:
- The electric field \( E = 2 \, \text{V/m} \)
Now substituting the value of \( E \) into the formula:
\[
u = \frac{1}{2} \varepsilon_0 (2)^2
\]
\[
u = \frac{1}{2} \varepsilon_0 \times 4 = 2 \varepsilon_0
\]
Thus, the energy density is \( 2 \varepsilon_0 \).
Hence, the correct answer is (E) \( 2 \varepsilon_0 \).
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