State the energy density in a vacuum and in a linear dielectric in terms of E and D.

State the energy density in a vacuum and in a linear dielectric in terms of E and D.

• A. Vacuum: u = (1/2) ε0 E^2; Dielectric: u = (1/2) E · D = (1/2) ε E^2
• B. Vacuum: u = ε0 E^2; Dielectric: u = (1/2) E · D
• C. Vacuum: u = (1/2) D^2 / ε0; Dielectric: u = (1/2) ε E^2
• D. Vacuum: u = (1/2) ε0 E^2; Dielectric: u = (1/2) ε0 E^2

Answer: (A)

The main idea is how electric-field energy is stored in different media. In vacuum there’s no polarization, so the energy per volume depends on the field as u = (1/2) ε0 E^2. In a dielectric, it’s natural to use the general expression u = (1/2) E·D, which already accounts for the medium’s response. For a linear dielectric, the relation D = ε E holds (with ε = ε0 εr), so u = (1/2) E·D becomes (1/2) ε E^2, which can also be written as (1/2) ε0 εr E^2. So the correct statements are: vacuum energy density is (1/2) ε0 E^2, and dielectric energy density is (1/2) E·D (equivalently (1/2) ε E^2). This matches the standard definitions and shows how E and D relate in different media.

The main idea is how electric-field energy is stored in different media. In vacuum there’s no polarization, so the energy per volume depends on the field as u = (1/2) ε0 E^2. In a dielectric, it’s natural to use the general expression u = (1/2) E·D, which already accounts for the medium’s response. For a linear dielectric, the relation D = ε E holds (with ε = ε0 εr), so u = (1/2) E·D becomes (1/2) ε E^2, which can also be written as (1/2) ε0 εr E^2.

So the correct statements are: vacuum energy density is (1/2) ε0 E^2, and dielectric energy density is (1/2) E·D (equivalently (1/2) ε E^2). This matches the standard definitions and shows how E and D relate in different media.