a) $ \nabla^2\bar{E} - \mu\epsilon\big(\frac{d^2E}{dt^2}\big)\\ \cancelto{0}{\nabla^2(E_o\sin(\omega t)\hat{z})} - \mu\epsilon\frac{d^2}{dt^2}[E_o\sin(\omega t)] = 0\\ (\omega^2)\mu\epsilon E_o \sin(\omega t) = 0\\ sin(\omega t) \ne 0\\ \text{WE not satisfied \to ME not satisfied in region between plates.}\\ \text{b) retardation of the time-dependant fields is ignored in this approximation; causes \bar{E} to depend on z between plates.}\\ \text{c) } \bar{E}= f\big(t-frac{Z}{\mu_p}\big) \; where \; \mu_p is velocity. (see next page)\\ d) \nabla \times \bar{E} = -\frac{\bar{\partial B}}{\partial t}\hspace{0.5cm}\nabla \times \bar{H} = \bar{J} +\frac{\partial \bar{D}}{\partial t}\hspace{0.5cm} \nabla\cdot\bar{D} =0\hspace{0.5cm}\nabla\cdot\bar{B} = 0\\ \nabla\times\nabla\times\bar{E} = - \nabla^2\bar{E} = -\frac{\partial}{\partial t}(\nabla\times\bar{B}) = -\mu\frac{\partial}{\partial t}\big(\cancelto{0}{\bar{J}}+\frac{\partial D}{\partial t}\big)\\ \nabla^2\bar{E} = \nabla(\cancelto{0}{\nabla\cdot\bar{E}}) - \nabla\times\nabla\times\bar{E}\\ -\nabla^2\bar{E} = -\mu\epsilon\frac{\delta^2}{\delta t^2}\bar{E}\\ \nabla^2\bar{E} -\mu\epsilon\frac{\delta^2}{\delta t^2}\bar{E} = 0 \\ \\ \frac{\partial^2E_x}{\partial x^2} + \frac{\partial^2E_x}{\partial y^2} + \frac{\partial^2E_x}{\partial z^2} - \mu\epsilon\frac{\delta^2}{\delta t^2}\bar{E}_x = 0\\ \frac{\partial^2E_y}{\partial x^2} + \frac{\partial^2E_y}{\partial y^2} + \frac{\partial^2E_y}{\partial z^2} - \mu\epsilon\frac{\delta^2}{\delta t^2}\bar{E}_y = 0\\ \frac{\partial^2E_z}{\partial x^2} + \frac{\partial^2E_z}{\partial y^2} + \frac{\partial^2E_z}{\partial z^2} - \mu\epsilon\frac{\delta^2}{\delta t^2}\bar{E}_z = 0 e) \bar{E} = E(z)\sin(\omega t)\hat{z}\\ = E(z)e^{j(\omega t + \frac{\pi}{2})}\\ \frac{\delta^2}{\delta z^2}E(z) - \omega^2\mu\epsilon E(z) = 0\\ E(z) = E_oe^{-j[\omega\sqrt{\mu\epsilon}]z}\\ \bar{E} = E_o\cos(\omega t +\frac{\pi}{2} - \omega\sqrt{\mu\epsilon} z ) $

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009