Question

a) Show that the equations below can form a solution to the wave equation: $\mathcal{E}(r,t) = [\mathcal{E}(t)e^{ikx} + \mathcal{E}^(t)e^{-ikx}]\mathbf{u}_z$. $\mathcal{E}(t) = \mathcal{E}(0)e^{-i\omega t}$, b) Using the above information, verify that: $\mathcal{B}(r,t) = \frac{1}{c}[\mathcal{E}(t)e^{ikx} + \mathcal{E}^(t)e^{-ikx}](-\mathbf{u}_y)$. $\frac{d}{dt}X = P.$

          a) Show that the equations below can form a solution to the wave
equation:
$\mathcal{E}(r,t) = [\mathcal{E}(t)e^{ikx} + \mathcal{E}^*(t)e^{-ikx}]\mathbf{u}_z$.
$\mathcal{E}(t) = \mathcal{E}(0)e^{-i\omega t}$,
b) Using the above information, verify that:
$\mathcal{B}(r,t) = \frac{1}{c}[\mathcal{E}(t)e^{ikx} + \mathcal{E}^*(t)e^{-ikx}](-\mathbf{u}_y)$.
$\frac{d}{dt}X = P.$

a) Show that the equations below can form a solution to the wave
equation:
ℰ(r,t) = [ℰ(t)e^ikx + ℰ^*(t)e^-ikx]𝐮z.
ℰ(t) = ℰ(0)e^-iω t,
b) Using the above information, verify that:
ℬ(r,t) = (1)/(c)[ℰ(t)e^ikx + ℰ^*(t)e^-ikx](-𝐮y).
(d)/(dt)X = P.

Added by Jaime K.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Nicholas Mogoi

Step 1

The wave equation is given by: ∇^2E - (1/c^2) ∂^2E/∂t^2 = 0 where ∇^2 is the Laplacian operator, c is the speed of light, and E is the electric field. Substituting E(r,t) = E(t)e^(ikx) into the wave equation, we have: ∇^2(E(t)e^(ikx)) - (1/c^2) Show more…

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a) Show that the equations below can form a solution to the wave equation: E(r,t) = E(t)e^(ikx) + E*^(k) E(t) = E(0)e^(-iot) b) Using the above information, verify that: e^(ikx) + E*^(dX) = P dt