I am trying to prove the equation $$ \frac{\bar{P}}{V}=\frac{1}{2}E_0^2\sigma_{AC} $$ which can be rewritten as $$ \begin{align} \frac{\bar{P}}{V} &= \frac{1}{2}E_0^2\sigma_{AC}\\ &=\frac{1}{2}E_0^2\ \omega\ \epsilon_0\ \epsilon^{''}_r\\ &=\frac{1}{2}E_0^2\ \omega\ \epsilon_0\ \epsilon_r^{'}\ \tan(\delta) \end{align} $$ Here $\bar{P}$ stands for the time-averaged power loss which satisfies the equation $$ \bar{P}=\frac{1}{T}\int_0^T U\ I\ dt, $$ where $T=\frac{2\pi}{\omega}$ is the time period, $U=U_0 e^{j\omega t}$ is the complex sinusoidal voltage, and $I=j\omega\epsilon^{'}_rC_0U + \omega\epsilon^{''}_rC_0U$. The instructions say to use $$ \begin{align} U_0 &= E_0h\\ C &= \epsilon_r\epsilon_0\frac{A}{h}\\ V &= A\ h \\ \sigma_{AC}&=\omega\epsilon_0\epsilon^{''}_r=\omega\epsilon_0\epsilon^{'}_r\tan(\delta)\\ \tan(\delta) &= \frac{\epsilon^{''}_r}{\epsilon^{'}_r} \end{align} $$

The problem I face is after solving the main integral part, which is like: $$ \epsilon^{''}_r*(F(T) - F(0)) + j*\epsilon^{'}_r((F(T) - F(0)) $$ where $F(t) = e^{2j\omega t}$ and I neglected all the constants for simplicity. $F(T)$ is equal to $\exp(j*4*\pi)$ which is 1, making $(F(T)-F(0))$ zero.

I thought of root mean squaring both $U$ and $I$ to begin with, but this gives out $\sqrt{\epsilon^{''2}_r+\epsilon^{'2}_r/2}$ term which doesn't seem to lead to the proof result.

I have a hunch that I maybe missing some elementary calculation basics. I would really appreciate any help that you can offer.


Well, this expression is the standard expression for the electric field energy $U$ on a dielectric with permitivity $\epsilon$ inside a volume $V$ but with some replacements.



$$ U_V=\frac{1}{2}\int_V \epsilon E^2 dV $$

The last expression is known from Maxwell's. And proven in Wikipedia.

Normally this expression comes in a volume integral on the dielectric $\epsilon_0 \epsilon_r$ -which you assumed to be a kind of cylinder of area $A$ and heigh $h$?- and for generic electric fields -which on this case you assumed to be constant $E_0$ on that cylinder?-.

Hence. you start with the expression for the energy in a volume, so the integral you must to solve is in the space, not in the time.

After that, you fill in the variables, on the above expression, make an average on a single frequency period $2\pi/\omega$ -again assuming your excitation is sinusoidaly varying with $\omega$ -, and you are trivially done.


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