12B (Vector Calculus)

The vector identity from Question 3 gives

\nabla \cdot (E \times B) = B \cdot (\nabla \times E) - E \cdot (\nabla \times B) = - B \cdot \frac{\partial B}{\partial t} - E \cdot J - E \cdot \frac{\partial E}{\partial t} = - E \cdot J - \frac{\partial}{\partial t} \frac{∣ E ∣ ^{2} + ∣ B ∣ ^{2}}{2} .

Integrating over

V

\int_{S} P \cdot d S = - \int_{V} E \cdot J d V - \frac{\partial}{\partial t} \int_{V} \frac{∣ E ∣ ^{2} + ∣ B ∣ ^{2}}{2} d V .

b

Both divergences vanish, so

ρ = 0

. Faraday's and Ampere's curl equations respectively require

k E_{0} = ω B_{0}, k B_{0} = ω E_{0} .

For positive constants these imply

ω = k, E_{0} = B_{0}, ρ = 0 .

P = E_{0}^{2} x cos^{2} (k x - ω t) .

Only the two

x

-faces of the box contribute to its outward flux. With

a = π / (2 k)

\int_{S} P \cdot d S = L^{2} E_{0}^{2} [cos^{2} (ka - ω t) - cos^{2} (ω t)] = - L^{2} E_{0}^{2} cos (2 ω t) .

The electromagnetic energy inside the box is

U (t) = L^{2} E_{0}^{2} \int_{0}^{a} cos^{2} (k x - ω t) d x,

and direct differentiation, using

ω = k

, gives

\frac{d U}{d t} = L^{2} E_{0}^{2} cos (2 ω t) .

Thus

\int_{S} P \cdot d S = - \frac{d U}{d t},

confirming the integral identity when

J = 0

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