Electric Circuits – Understanding Complex Permittivity

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I have been trying to solve the following question:

$U(t)$ has the frequency $\omega$. Without an electric, the Voltage drop around the resistance is $U_R$. When half of the capacitor is filled with a dielectric (b), the voltage drop around the resistance becomes $2U_R$. What is the dielectric constant $\epsilon$ as a function of $\omega, R, C$?
1

My solution so far:
$$\frac{U_R}{R}=I_1=\frac{U_0}{R+\frac{1}{i\omega C}}\\ \frac{2U_R}{R}=I_2=\frac{U_0}{R+\frac{2}{i\omega C(\epsilon +1)}} \\ \frac{I_2}{I_1}=2=\frac{R+\frac{1}{i\omega C}}{R+\frac{2}{i\omega C(\epsilon +1)}} \\ \Rightarrow \epsilon=\frac{4+i\omega CR}{\omega^2C^2R^2+1}-1$$

I can't see my error. If it is correct, what does a complex permittivity mean?

Best Answer

When you take the ratio of $I_2/I_1$, you are getting a complex value because of disagreements in phase. Instead, take the ratio of their magnitudes.

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