- #1
BaghDal
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Homework Statement
2. Homework Equations
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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
$$
\begin{align}
U_0 &= E_0h\\
C &= \epsilon_r\epsilon_0\frac{A}{h}\\
V &= A\ h \\
\sigma_{AC}&=\omega\epsilon_0\epsilon^{''}_r=\epsilon_0\epsilon^{'}_r\tan(\delta)\\
\tan(\delta) &= \frac{\epsilon^{''}_r}{\epsilon^{'}_r}
\end{align}
$$
The Attempt at a Solution
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 and thus the whole equation 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.