Sea water at frequency `v=4xx10^(8)` Hz has permittivity `in ~~ 80 in_(0)`, permeability `mu = mu_(0)` and resistivity `rho = 0.25 Omega`, Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source `V(t)=V_(0)sin 2pi vt`. What fraction of the conduction current density is the displacement current current density ?

Explanation

Let distance between two plates of capacitor be .d. and voltage applied is `V(t)=V_(0)sin 2pi vt`
Hence electric field,
`E=(V_(0))/(d)sin 2pi v t " "` ….(1)
By Ohm.s law,
`I_(c )=sigma E=(E )/(rho)`
`therefore I_(c ) = (V_(0))/(rho d) sin 2pi v t`
Let `(V_(0))/(rho d) = J_(0)^(c ) " "` ....(2)
`therefore J_(c ) = J_(0)^(c )sin 2pi vt " "` .....(3)
Now displacement current density,
`J_(d)=in(delta E)/(dt)=(in delta)/(dt)[(V_(0))/(d)sin 2pi vt] " "` [`because` From equation (1)]
`=(in 2piv V_(0))/(d)cos 2pi vt`
Let `(2pi v V_(0))/(d)=J_(0)^(d) " "` ....(4)
`therefore J_(d) = J_(0)^(d)cos pi vt " "` ...(5)
By taking ratio of (4) and (2),
`(J_(0)^(d))/(J_(0)^(2))=(2pi v in V_(0))/(d)xx(rho d)/(V_(0))`
`= 2 pi v in rho`
`= 2pi v xx 80 in_(0) xx 0.25 [because in = 80 in_(0), rho = 0.25 Omega m]`
`= 4pi in_(0) v xx 10`
`=(10 v)/(9xx 10^(9)) " " [because 4pi in_(0)=(1)/(9xx10^(9))]`
`= (10xx4xx10^(8))/(9xx10^(9))[because v=4xx10^(8)Hz]`
`=(4)/(9)`