Revision as of 20:26, 3 June 2017 by Oelgendy (Talk | contribs)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

a) $\nabla^2\bar{E} - \mu\epsilon\big(\frac{d^2E}{dt^2}\big)$\\ $\cancelto{0}{\nabla^2(E_o\sin(\omega t)\hat{z})} - \mu\epsilon\frac{d^2}{dt^2}[E_o\sin(\omega t)] = 0$\\ $(\omega^2)\mu\epsilon E_o \sin(\omega t) = 0$\\ $sin(\omega t) \ne 0$\\ WE not satisfied $\to$ ME not satisfied in region between plates.

b) retardation of the time-dependant fields is ignored in this approximation; causes $\bar{E}$ to depend on $z$ between plates.

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

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett