Sea water at frequency $v = 4 \times 10^8\, Hz$ has permittivity $\epsilon \approx 80\epsilon_0$,permeability $\mu = \mu_0$ and resistivity $\rho = 0.25\,\Omega m$. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source $V(t) = V_0 \sin(2\pi vt)$. What fraction of the conduction current density is the displacement current density?

(4/9) Let the distance between the two plates of the capacitor be $d$ and the voltage applied be $V(t) = V_0 \sin(2\pi vt)$.
The electric field is given by $E = \frac{V_0}{d} \sin(2\pi vt)$.
By Ohm's law,the conduction current density is $J_c = \sigma E = \frac{E}{\rho} = \frac{V_0}{\rho d} \sin(2\pi vt)$.
Let $J_0^c = \frac{V_0}{\rho d}$,so $J_c = J_0^c \sin(2\pi vt)$.
The displacement current density is $J_d = \epsilon \frac{\partial E}{\partial t} = \epsilon \frac{\partial}{\partial t} \left[ \frac{V_0}{d} \sin(2\pi vt) \right] = \frac{\epsilon (2\pi v) V_0}{d} \cos(2\pi vt)$.
Let $J_0^d = \frac{2\pi v \epsilon V_0}{d}$,so $J_d = J_0^d \cos(2\pi vt)$.
The ratio of the amplitudes is $\frac{J_0^d}{J_0^c} = \frac{2\pi v \epsilon V_0}{d} \times \frac{\rho d}{V_0} = 2\pi v \epsilon \rho$.
Substituting the given values: $\epsilon = 80\epsilon_0$,$\rho = 0.25\,\Omega m$,and $v = 4 \times 10^8\, Hz$.
$\frac{J_0^d}{J_0^c} = 2\pi v (80\epsilon_0) \rho = 160\pi \epsilon_0 v \rho$.
Using $4\pi\epsilon_0 = \frac{1}{9 \times 10^9}$,we have $160\pi\epsilon_0 = 40 \times (4\pi\epsilon_0) = \frac{40}{9 \times 10^9}$.
$\frac{J_0^d}{J_0^c} = \frac{40}{9 \times 10^9} \times (4 \times 10^8) \times 0.25 = \frac{40 \times 4 \times 10^8 \times 0.25}{9 \times 10^9} = \frac{40}{90} = \frac{4}{9}$.