The electromagnetic wave equation in vacuum for the electric field E is which of the following?

The electromagnetic wave equation in vacuum for the electric field E is which of the following?

• A. ∇^2 E = μ0 ε0 ∂^2 E/∂t^2
• B. ∇^2 E = - μ0 ε0 ∂^2 E/∂t^2
• C. ∂^2 E/∂t^2 = μ0 ε0 ∇^2 E
• D. ∇ · E = 0

Answer: (A)

The wave equation for the electric field in vacuum arises when you combine Maxwell’s equations to see how E propagates as a wave. Start with Faraday’s law, ∇ × E = -∂B/∂t, and the Ampère–Maxwell law in vacuum, ∇ × B = μ0 ε0 ∂E/∂t. Take the curl of Faraday’s law: ∇ × (∇ × E) = -∂/∂t (∇ × B). Substitute ∇ × B from Ampère–Maxwell to get ∇ × (∇ × E) = -μ0 ε0 ∂^2 E/∂t^2. Use the vector identity ∇ × (∇ × E) = ∇(∇ · E) - ∇^2 E. In the absence of charges, ∇ · E = 0, so the equation becomes -∇^2 E = -μ0 ε0 ∂^2 E/∂t^2, or ∇^2 E = μ0 ε0 ∂^2 E/∂t^2. This is the electromagnetic wave equation for E in vacuum. Since μ0 ε0 = 1/c^2, it can also be written as ∇^2 E - (1/c^2) ∂^2 E/∂t^2 = 0, showing the wave travels at speed c. Note: ∇ · E = 0 is true in vacuum with no charges, but by itself it does not describe how E varies in time and space like a wave equation does.

The wave equation for the electric field in vacuum arises when you combine Maxwell’s equations to see how E propagates as a wave. Start with Faraday’s law, ∇ × E = -∂B/∂t, and the Ampère–Maxwell law in vacuum, ∇ × B = μ0 ε0 ∂E/∂t. Take the curl of Faraday’s law: ∇ × (∇ × E) = -∂/∂t (∇ × B). Substitute ∇ × B from Ampère–Maxwell to get ∇ × (∇ × E) = -μ0 ε0 ∂^2 E/∂t^2. Use the vector identity ∇ × (∇ × E) = ∇(∇ · E) - ∇^2 E. In the absence of charges, ∇ · E = 0, so the equation becomes -∇^2 E = -μ0 ε0 ∂^2 E/∂t^2, or ∇^2 E = μ0 ε0 ∂^2 E/∂t^2. This is the electromagnetic wave equation for E in vacuum. Since μ0 ε0 = 1/c^2, it can also be written as ∇^2 E - (1/c^2) ∂^2 E/∂t^2 = 0, showing the wave travels at speed c.

Note: ∇ · E = 0 is true in vacuum with no charges, but by itself it does not describe how E varies in time and space like a wave equation does.