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I am working on a practice problem involving maxwells equations. We have an infinite sheet of charge density $\sigma$ in the x-y plane and it is oscillating as x= $\Re [x_0e^{-i\omega t}]$. I want to use maxwells equations to show that the time dependent E field satisfies: $$\frac{\partial^2E}{\partial z^2}+\alpha E=\beta\delta(z)$$ And find $\alpha, \beta$.

My thoughts were to take the curl of Maxwells Curl(E) equation: $$\nabla\times(\nabla\times E=-\frac{1}{c^2}\frac{\partial B}{\partial t})$$ $$\nabla(\nabla\cdot E)-\nabla^2E=\frac{-\partial}{c^2\partial t}(\nabla\times B)$$ $$\nabla^2E=\frac{4\pi}{c^2}\frac{\partial J}{\partial t}+\frac{1}{c^2}\frac{\partial^2 E}{\partial t^2}+\nabla(4\pi \rho)$$ This is where i get confused a bit. My guess is that because the sheet is oscillating as $\exp(-i\omega t)$, then I assume E does as well. This means that the second derivative of E in time is just $-\omega^2 E$. This gives me something regarding the equation I need to find. Just not sure how to find the delta function part. Is it in the dJ/dt part or in the $\nabla(\rho)$ for this geometry

Additionally, I am unsure of how to get from $\nabla^2E-\partial J/\partial t$ to $\partial^2E/\partial z^2$

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