Seawater at a frequency $$f = 9 \times 10^2$$ Hz, has permittivity $$\varepsilon = 80\varepsilon_0$$ and resistivity $$\rho = 0.25$$ $$\Omega$$ m. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source $$V(t) = V_0 \sin(2\pi ft)$$. Then the conduction current density becomes $$10^x$$ times the displacement current density after time $$t = \frac{1}{800}$$ s. The value of $$x$$ is ________. (Given: $$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$$ N m$$^2$$ C$$^{-2}$$)

The conduction current density is $$J_c = \frac{E}{\rho}$$ and the displacement current density is $$J_d = \varepsilon \frac{\partial E}{\partial t}$$. For a parallel plate capacitor driven by $$V(t) = V_0 \sin(2\pi f t)$$, the electric field between the plates is proportional to $$V(t)$$, so $$E = E_0 \sin(2\pi f t)$$ and $$\frac{\partial E}{\partial t} = 2\pi f E_0 \cos(2\pi f t)$$.

The ratio of conduction to displacement current density is $$\frac{J_c}{J_d} = \frac{E_0 \sin(2\pi ft)/\rho}{\varepsilon \cdot 2\pi f E_0 \cos(2\pi ft)} = \frac{\tan(2\pi ft)}{2\pi f \varepsilon \rho}$$.

At $$t = \frac{1}{800}$$ s, we compute $$2\pi f t = 2\pi \times 9 \times 10^2 \times \frac{1}{800} = \frac{9\pi}{4}$$. Since $$\frac{9\pi}{4} = 2\pi + \frac{\pi}{4}$$, we get $$\tan\left(\frac{9\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1$$.

Now we need $$2\pi f \varepsilon \rho$$. Here $$\varepsilon = 80\varepsilon_0$$ and $$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$$, so $$\varepsilon_0 = \frac{1}{4\pi \times 9 \times 10^9}$$. Thus $$\varepsilon = \frac{80}{4\pi \times 9 \times 10^9} = \frac{20}{9\pi \times 10^9}$$.

Therefore $$2\pi f \varepsilon \rho = 2\pi \times 900 \times \frac{20}{9\pi \times 10^9} \times 0.25$$. The numerator is $$2\pi \times 900 \times 20 \times 0.25 = 9000\pi$$. The denominator is $$9\pi \times 10^9$$. So the ratio is $$\frac{9000\pi}{9\pi \times 10^9} = \frac{1000}{10^9} = 10^{-6}$$.

Hence $$\frac{J_c}{J_d} = \frac{1}{10^{-6}} = 10^6$$. Since this equals $$10^x$$, we get $$x = 6$$.