Conduction and displacement current density

tiagobt

Could anyone help me solve the following problem?

Calculate the ratio of the conduction current density to the displacement current density of the electric field $E = E_0 \cos(\omega t)$ in copper, to a frequence of $f = 1 kHz$. (Given: $\epsilon_{Cu} = \epsilon_0$, $\rho_{Cu} = 2 \times 10^{-8} \Omega m$).​

First, I calculated the displacement current density:

$$J_d = \epsilon_{Cu} \frac {dE} {dt} = - \epsilon_0 E_0 \sin(\omega t)\omega$$

I'm not sure if it's correct. Besides, I don't know how to find the conduction current density. I thought about using:

$$\vec{\nabla} \times \vec{B} = \mu_0 (\vec{J} + \epsilon_0 \vec{J_d})$$

But is there a magnetic field? I'm confused...

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