Suppose that a right-moving EM wave overlaps with a left-moving EM wave so that, in a certain region of space, the total electric field in the y direction and magnetic field in the z direction are given by Eᵧ = E₀ sin(kx - ωt) + E₀ sin(kx + ωt) and Bz = B₀ sin(kx - ωt) - B₀ sin(kx + ωt) (b) Determine the Poynting vector and find the x locations at which it is zero at all times.

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1. First, recall the definition of the Poynting vector, \( \vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B} \), where \( \vec{E} \) is the electric field and \( \vec{B} \) is the magnetic field. Here, \( \mu_0 \) is the permeability of free space.
2. Simplify the expressions for \( E_\gamma \) and \( B_z \) using trigonometric identities. The identity for sum and difference of sines, \( \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \) and \( \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \), can be applied here.
3. Apply the identities to get \( E_\gamma = 2E_0 \cos(\omega t) \sin(kx) \) and \( B_z = -2B_0 \sin(\omega t) \cos(kx) \).
4. Calculate the cross product \( \vec{E} \times \vec{B} \). Since \( \vec{E} \) is in the y-direction and \( \vec{B} \) is in the z-direction, the cross product will be in the x-direction. Thus, \( \vec{S} = \frac{1}{\mu_0} (2E_0 \cos(\omega t) \sin(kx))(-2B_0 \sin(\omega t) \cos(kx)) \hat{x} \).
5. Simplify the expression for the Poynting vector and find the x locations where it is zero. The expression simplifies to \( \vec{S} = -\frac{4E_0 B_0}{\mu_0} \cos(\omega t) \sin(\omega t) \sin(kx) \cos(kx) \hat{x} \). The Poynting vector is zero when \( \sin(kx) \cos(kx) = 0 \), which occurs at \( kx = n\pi/2 \) for integer n. Thus, the x locations are \( x = n\pi/(2k) \).