An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, $$\sigma$$ = 10mho/m. The ratio of maximum conduction current density to maximum displacement current
density is ___________.
[Take $$\frac{1}{4\pi \epsilon_{0}} = 9 \times 10^{9} Nm^{2}/C^{2}$$]

We need to find the ratio of maximum conduction current density to maximum displacement current density for an electromagnetic wave propagating through a conducting medium.

First, recall the relevant formulas for conduction and displacement current densities. The conduction current density is given by $$J_c = \sigma E$$ (Ohm's law in differential form), where $$\sigma$$ is the conductivity, and the displacement current density is $$J_d = \epsilon_0\frac{\partial E}{\partial t}$$ (from Maxwell's equations).

Next, we determine their maximum values. For an electromagnetic wave with electric field $$E = E_0\sin(\omega t - kx)$$ propagating through a medium, the maximum conduction current density is $$J_{c,max} = \sigma E_0$$. Since $$\frac{\partial E}{\partial t} = \omega E_0\cos(\omega t - kx)$$, the maximum displacement current density is $$J_{d,max} = \epsilon_0 \omega E_0$$.

From these we obtain the ratio of the maximum conduction current density to the maximum displacement current density: $$ \frac{J_{c,max}}{J_{d,max}} = \frac{\sigma E_0}{\epsilon_0 \omega E_0} = \frac{\sigma}{\epsilon_0 \omega} $$.

Now we substitute the given values. The frequency is $$f = 100$$ MHz $$= 10^8$$ Hz and the conductivity is $$\sigma = 10$$ mho/m. The angular frequency is $$\omega = 2\pi f = 2\pi \times 10^8$$ rad/s.

The permittivity of free space follows from the given relation $$\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$$, giving $$\epsilon_0 = \frac{1}{4\pi \times 9 \times 10^9} = \frac{1}{36\pi \times 10^9}$$.

Then we compute $$\epsilon_0 \omega = \frac{1}{36\pi \times 10^9} \times 2\pi \times 10^8 = \frac{2\pi}{36\pi \times 10^9} \times 10^8 = \frac{2}{36 \times 10} = \frac{2}{360} = \frac{1}{180}$$.

Finally, substituting these values into the ratio gives $$ \frac{\sigma}{\epsilon_0 \omega} = \frac{10}{\frac{1}{180}} = 10 \times 180 = 1800 $$, so the ratio of maximum conduction current density to maximum displacement current density is $$\boxed{1800}$$.