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...

 

[TSny]

A frequency of 1 kHz is pretty small for this situation. So, for the conduction current density, I think you can just use Ohm's law: $J = \sigma E = \frac{1}{\rho} E = \frac{1}{\rho} E_0 \cos(\omega t)$. In this case, the conduction current and the electric field oscillate in phase.