Sea water at frequency `v=4 xx 10^(8)` Hz has permittivity `epsilon ~~ 80 epsilon_(0)` permeability `mu= mu_(0)` and resistivity `rho= 0.25`M. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source `V(t) =V_(0) " sin "(2pivt).` What fraction of the conduction current density is the displacement current density ?

Explanation

Suppose distance between the parallel plates is d and applied voltage `V_((t)) =V_(0) 2pi vt`.
thus electric field
`E=(V_(0))/(d) " sin "(2pivt)`
`"Now using Ohm's law"" "J_(c) =(1)/(phi) (V_(0))/(d) "sin "(2pivt)`
`rArr " "=(V_(0))/(phid) "sin " (2pi vt) =J_(0)^(c) "sin"2pivt`
`"Here"" " J_(0)^(c) =(V_(0))/(rhod)`
Now the displacement current density is given as
`J_(d)= epsilon(deltaE)/(dt)=(epsilondelta)/(dt)" "[(V_(0))/(dt)"sin"(2pivt)]`
`=(epsilon2pivV_(0))/(d) "cos"(2pivt)`
`rArr" "=J_(0)^(d)"cos"(2pivt)`
`"Where"" " =J_(0)^(d) =(2pi VepsilonV_(0))/(d)`
`rArr" "(J_(0)^(d))/(J_(0)^(c))=(2pivepsilonV_(0))/(d) .(pd)/(V_(0))=2pivepsilonrho`
`=2pi xx 80epsilon_(0)v xx 0.25 =4pi epsilon_(0) v xx 10`
`=(10v)/(9xx10^(9))=(4)/(9)`