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

• A. \( 3 \varepsilon_0 \)
• B. \( \frac{\varepsilon_0}{2} \)
• C. \( 4 \varepsilon_0 \)
• D. \( \frac{\varepsilon_0}{4} \)
• E. \( 2 \varepsilon_0 \)

Hint:

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.

Answer 1

Answer: (E)

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) \( 2 \varepsilon_0 \).