Sea water at frequency $v = 4 \times {10}^8 \mathrm{Hz}$ has permittivity $\epsilon \approx 80 {\epsilon }_0$, permeability $\mu \approx {\mu }_0$ and resistivity $\rho = 0.25 \mathrm{\Omega }-\mathrm{m}$. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source $\mathrm{V}\left(\mathrm{t}\right) = {\mathrm{V}}_0 \sin \left(2\pi vt\right)$. The ratio of amplitude of the conduction current density to the displacement current density is:

$\mathrm{V}\left(\mathrm{t}\right)={\mathrm{V}}_0 \sin 2\mathrm{\pi }v\mathrm{t}$

Let distance between the plates $= \mathrm{d}$

Electric field, $\mathrm{E}=\frac{\mathrm{V}\left(\mathrm{t}\right)}{\mathrm{d}}=\frac{{\mathrm{V}}_0}{\mathrm{d}}\sin 2\mathrm{\pi }\nu \mathrm{t}$

Conduction current density

${\mathrm{J}}_{\mathrm{c}}=\frac{\mathrm{E}}{\mathrm{\rho }}=\frac{{\mathrm{V}}_0}{\mathrm{\rho d}}\sin 2\mathrm{\pi \nu t}={\mathrm{J}}_{\mathrm{c},0}\sin \left(2\mathrm{\pi \nu t}\right)$

where ${\mathrm{J}}_{c,0}$ is maximum conduction current density. 

Displacement current density is given by:

$$\begin{gathered} {\mathrm{J}}_{\mathrm{d}}=\mathrm{\epsilon }\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{{\mathrm{V}}_0}{\mathrm{d}}\sin \left(2\mathrm{\pi \nu t}\right)\right]=\frac{2\mathrm{\pi \nu \epsilon }}{\mathrm{d}}{\mathrm{V}}_0 \cos \left(2\mathrm{\pi \nu t}\right) \\ {\mathrm{J}}_{\mathrm{d}}={\mathrm{J}}_{\mathrm{d},0} \cos \left(2\mathrm{\pi \nu t}\right) \end{gathered}$$

where $J_{d,0}$is the maximum displacement current density.

$$\begin{gathered} \therefore \frac{{\mathrm{J}}_{\mathrm{d},0}}{{\mathrm{J}}_{\mathrm{c},0}}=\frac{2{\mathrm{\pi \nu V}}_0\mathrm{\epsilon }}{\mathrm{d}}\times \frac{\mathrm{\rho d}}{{\mathrm{V}}_0}=2\mathrm{\pi \nu \epsilon \rho }=2\mathrm{\pi }\left(4\times {10}^8\right)\left(80 {\mathrm{\epsilon }}_0\right)\left(0.25\right) \\ =\frac{4\times {10}^9}{9\times {10}^9} \left(\because \frac{1}{4{\mathrm{\pi \epsilon }}_0}=9\times {10}^9 \mathrm{N} {\mathrm{m}}^2/{\mathrm{C}}^2\right) \\ \Rightarrow \frac{{\mathrm{J}}_{\mathrm{c},0}}{{\mathrm{J}}_{\mathrm{d},0}}=\frac{9}{4} \end{gathered}$$