The electric field in free space is

Question

Examoogle · Accepted Answer

Option A: $\frac{\mathbf{D}}{\epsilon_0}$

Explanation & Concepts:
In electromagnetics, the electric displacement field $\mathbf{D}$ is related to the electric field intensity $\mathbf{E}$ by the constitutive relation $\mathbf{D} = \epsilon \mathbf{E}$. For free space (vacuum), the permittivity is the vacuum permittivity $\epsilon_0$, hence $\mathbf{E} = \frac{\mathbf{D}}{\epsilon_0}$.

Working Principle: The electric displacement field $\mathbf{D}$ accounts for the effects of free charges within a material, whereas $\mathbf{E}$ represents the total force field. In free space, there are no bound charges or polarization effects, simplifying the medium to its permittivity $\epsilon_0$. Thus, $\mathbf{D}$ is directly proportional to $\mathbf{E}$ scaled by the constant $\epsilon_0$.

Key Formulas: $\mathbf{D} = \epsilon_0 \mathbf{E}$ — Constitutive relation in free space; $\mathbf{E} = \frac{\mathbf{D}}{\epsilon_0}$ — Electric field intensity in terms of displacement field

Key Points: $\mathbf{E}$ is the electric field intensity measured in $V/m$.; $\mathbf{D}$ is the electric flux density measured in $C/m^2$.; $\epsilon_0$ is the permittivity of free space, approximately $8.854 	imes 10^{-12} F/m$.

Advantages: Allows simplification of Maxwell's equations in non-homogeneous media.; Decouples the effect of source charges from the effect of polarization.

Disadvantages: Requires knowledge of medium properties to relate $\mathbf{D}$ and $\mathbf{E}$ in non-linear media.

Applications: Solving boundary value problems in electrostatics.; Analysis of transmission lines and wave propagation in vacuum.

Additional Info: Option D ($\sigma/\epsilon_0$) represents the electric field at the surface of a charged conductor, not the general relation in space.. Option B introduces $\mu_0$ (permeability), which relates magnetic field intensity $\mathbf{H}$ and flux density $\mathbf{B}$.

Answer: A is correct — The electric field in free space is defined as the electric displacement field divided by the vacuum permittivity, expressed as $\mathbf{E} = \frac{\mathbf{D}}{\epsilon_0}$.

Exam Tip: Remember that $\mathbf{D}$ depends on free charges and $\mathbf{E}$ depends on both free and bound charges; in free space, they are strictly proportional.

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Option B: $\frac{\mathbf{D}}{\mu_0}$

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Option C: $\epsilon_0 \mathbf{D}$

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Option D: $\frac{\sigma}{\epsilon_0}$

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