Difference between revisions of "ECE PhD QE FO3 2012 1 Problem1.1" - Rhea

Latest revision as of 09:44, 18 June 2017

1)

a)

$ \frac{\partial}{\partial r}\bigg(r^2\frac{\partial\Phi}{\partial r}\bigg) + \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\bigg(\sin\theta\frac{\partial\Phi}{\partial\theta}\bigg)=0\text{ then} $

$ \Phi(r,\theta) = \begin{cases}Ar\cos\theta\hspace{2cm}r\le a\\-E_0r\cos\theta+\frac{B\cos\theta}{r^2}\hspace{2cm}r\ge a\end{cases} $ \underline{assuming:} n = 1, field goes to $ E_0\hat{z} $ at large r, finite field at r=0.

\underline{BC's}: continuous $ \Phi(r,\theta)<math> at r=a: <math>Aa = -E_0a+\frac{B}{a^2}\to A = -E_0 + \frac{B}{a^3} $

$ \sigma = 0 \to \rho_s = 0\to D_{1n} = D_{2n}\to\epsilon_0E_1 = \epsilon E_2\Rightarrow \text{ otherwise use } \sigma_1E_1 = \sigma_2E_2 $

$ -\epsilon_0A = E_0\epsilon - \frac{2B\epsilon}{a^3}=\epsilon_0E_0 - \frac{B\epsilon_0}{a^3} $

$ B\bigg(\frac{-2\epsilon}{a^3} + \frac{\epsilon_0}{a^3}\bigg) = E_0(\epsilon_0-\epsilon) $

$ B = E_0a^3\bigg(\frac{\epsilon_0- \epsilon}{\epsilon_0-2\epsilon}\bigg)\hspace{1cm}A = E_0\bigg[\bigg(\frac{\epsilon_0-\epsilon}{\epsilon_0-2\epsilon}\bigg)-1\bigg] $

$ \Phi(r,\theta) = \begin{cases}E_0\big(\frac{\epsilon}{\epsilon_0-2\epsilon}\big)r\cos\theta\hspace{1cm} r\le a\\-E_0r\cos\theta + \frac{E_0a^3}{r^2}\cos\theta\big(\frac{\epsilon_0-\epsilon}{\epsilon_0-2\epsilon}\big)\hspace{1cm} r\ge a\end{cases} $

\underline{check:} $ \epsilon_0 = \hspace{1cm} r\ge a $

$ E = E_0 \hat{z} $ everywhere

\underline{outside:}$ \bar{E} = -\nabla\Phi = E_0\hat{z} - \frac{2E_0a^3}{\Gamma^3}\bigg(\frac{\epsilon_0-\epsilon}{\epsilon_0-2\epsilon}\bigg)\hat{z} $

$ J = \sigma E = \bigg[ \sigma E_0 - \frac{2E_0\sigma a^3}{\Gamma^3}\bigg(\frac{\epsilon_0-\epsilon}{\epsilon_0-2\epsilon}\bigg)\bigg]\hat{z} $

b)

Insulator implies $ \sigma = 0 $ inside the sphere, thus $ J = \sigma\bar{E} = 0 $ inside the sphere (couldn't be used as a BC).

2)$ \oint\bar{H}\cdot dl = \int_S(J+\frac{d}{dt}\bar{D})ds\hspace{2cm}\text{ only have } H_z $

a)$ \bar{E} = -\nabla V $

$ \begin{align*} \nabla^2V &=0 \leftarrow \text{ assume } P_v \text{ is small}\\ \frac{d^2V_z}{dz} &= 0\\ V(z) &= Az+B\\ V(z=0) &= B = 0\\ V(z=h) &= V_0\cos\omega t = Ah\to A = \frac{V_0}{h}\cos(\omega t) \\ V &= \frac{V_0}{h}\cos(\omega t)z \end{align*} $

$ \bar{E} = - \nabla V = -\frac{V_0}{n}\cos(\omega t)\hat{z}\to \bar{J} = \sigma\bar{E} = \frac{-V_0\sigma}{h}\cos(\omega t)\hat{z} $

$ \oint\bar{H}\cdot dl = NI_0\cos(\omega t) + \frac{V_0\sigma\omega}{h}\sin(\omega t)[\pi a^2] - \int_0^h H_zdz=-hH_z $

$ \bar{H}= \frac{-NI_0}{n}\cos(\omega t)\hat{z} - \frac{V_0\sigma\omega\pi a^2}{h^2}\sin(\omega t)\hat{z} $

component due to $ \triangle\bar{E} $ in water.

b)

$ P_{ave} = \oint_SP\cdot ds = [\pi a^2]\bigg[\frac{1}{2}Re[E\times H^*]\bigg] = VI $

$ J_{water} = \sigma\bar{E} = \frac{-V_0\sigma}{n}\cos(\omega t)\hat{z} $

$ I = \oint J\cdot d\bar{s} = \frac{-V_0\sigma\pi a^2}{n}\cos(\omega t)\hat{z} $

$ V = \frac{V_0}{n}\cos(\omega t)z $

$ P = IV + \frac{V_0^2\sigma \pi a^2}{n^2} $

c)

$ Z = \frac{V_0}{I_0} = \frac{\frac{V_0}{n}\cancel{\cos(\omega t)}z}{{\frac{-V_0\sigma\pi a^2}{n}\cancel{\cos(\omega t)}}}= \frac{-z}{\sigma\pi a^2} $

a,b,c probably wrong.