Question 705057

SCQHARD

In a non-magnetic material with no free charges and no free currents, the permittivity ϵ\epsilon is a function of position. If E⃗\vec{E} represents the electric field and μ0,ϵ0\mu_0, \epsilon_0 are free space permeability and permittivity respectively, which one of the following expressions is correct?

(A)

∇2E⃗−μ0∂2(ϵE⃗)∂t2−1ϵ0∇(E⃗⋅∇ϵ)=0\nabla^2 \vec{E} - \mu_0 \frac{\partial^2(\epsilon\vec{E})}{\partial t^2} - \frac{1}{\epsilon_0} \nabla(\vec{E} \cdot \nabla\epsilon) = 0

(B)

∇2E⃗−μ0∂2(ϵE⃗)∂t2+1ϵ0∇(E⃗⋅∇ϵ)=0\nabla^2 \vec{E} - \mu_0 \frac{\partial^2(\epsilon\vec{E})}{\partial t^2} + \frac{1}{\epsilon_0} \nabla(\vec{E} \cdot \nabla\epsilon) = 0

(C)

∇2E⃗−μ0∂2(ϵE⃗)∂t2+∇(1ϵE⃗⋅∇ϵ)=0\nabla^2 \vec{E} - \mu_0 \frac{\partial^2(\epsilon\vec{E})}{\partial t^2} + \nabla(\frac{1}{\epsilon} \vec{E} \cdot \nabla\epsilon) = 0

(D)

∇2E⃗−μ0∂2(ϵE⃗)∂t2−∇(1ϵE⃗⋅∇ϵ)=0\nabla^2 \vec{E} - \mu_0 \frac{\partial^2(\epsilon\vec{E})}{\partial t^2} - \nabla(\frac{1}{\epsilon} \vec{E} \cdot \nabla\epsilon) = 0

Detailed Solution

Using Maxwell's equations with ρf=0\rho_f = 0 and Jf=0J_f = 0, Faraday's law ∇×E⃗=−∂B⃗∂t\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} and Ampere-Maxwell law ∇×B⃗=μ0ϵ∂E⃗∂t\nabla \times \vec{B} = \mu_0 \epsilon \frac{\partial \vec{E}}{\partial t}. Taking the curl and using ∇×(∇×E⃗)=∇(∇⋅E⃗)−∇2E⃗\nabla \times (\nabla \times \vec{E}) = \nabla(\nabla \cdot \vec{E}) - \nabla^2 \vec{E}, along with ∇⋅(ϵE⃗)=0  ⟹  ∇⋅E⃗=−1ϵE⃗⋅∇ϵ\nabla \cdot (\epsilon \vec{E}) = 0 \implies \nabla \cdot \vec{E} = -\frac{1}{\epsilon} \vec{E} \cdot \nabla \epsilon, yields the wave equation with the gradient term.

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