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Two plane electromagnetic waves in vacuum have electric fields E1 = E0 j cos(wt - kx) and E2 = E0 k cos(wt - ky). At t = 0 a particle of charge q is at the origin moving with velocity v = 0.8c j (c = speed of light). Find the instantaneous force on the particle at t = 0.

1. E0*q*(-0.8 i + j + k)
2. E0*q*(0.8 i - j + 0.4 k)
3. E0*q*(0.8 i + j + 0.2 k)
4. E0*q*(0.4 i - 3 j + 0.8 k)

Correct answer: E0*q*(0.8 i + j + 0.2 k)

Solution

At t = 0, origin: E1 = E0 j (propagating +x), so B1 = (E0/c) k. E2 = E0 k (propagating +y), so B2 = (1/c)(j x E0 k) = (E0/c) i. Total E = E0(j + k); total B = (E0/c)(i + k). v = 0.8c j. v x B = 0.8c j x (E0/c)(i + k) = 0.8 E0 (j x i + j x k) = 0.8 E0 (-k + i) = 0.8 E0 i - 0.8 E0 k. F = q[E0(j + k) + 0.8 E0 i - 0.8 E0 k] = q E0 [0.8 i + j + (1 - 0.8) k] = q E0 (0.8 i + j + 0.2 k).

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