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Question
Sea water at frequency ν = 4 × 108 Hz has permittivity ε ≈ 80 εo, permeability µ ≈ µo and resistivity ρ = 0.25 Ω–m. Imagine a parallel plate capacitor immersed in seawater and driven by an alternating voltage source V(t) = Vo sin (2πνt). What fraction of the conduction current density is the displacement current density?
Solution
Let the separation between the plates of capacitor immersed in seawater plates is d and applied voltage across the plates is V(t) = V0 sin (2πνt).
Thus, electric field, E = `(V(t))/d`
⇒ E = `V_0/d sin (2πνt)`
Now using Ohm's law, the conduction current density Jc = `E/ρ = 1/ρ V_0/d sin (2πνt)`
⇒ Jc = `V_0/(ρd) sin (2πνt) = J_0^c sin 2πνt`
Here, `J_0^c = V_0/(ρd)`
The displacement current density is given as
Jd = `ε (dE)/(dt) = ε d/(dt) [V_0/d sin (2πνt)]`
= `(ε2πν V_0)/d cos (2πνt)`
⇒ Jd = `J_0^d cos (2πνt)`
Where, `J_0^d = (2πνεV_0)/d`
⇒ `J_0^d/J_0^c = ((2pivεV_0)/d)/((V_0)/(ρd)) = 2πνερ`
= `2pi xx 80ε_0v xx 0.25 = 4piε_0v xx 10`
⇒ `J_0^d/J_0^c = (10 xx 4 xx 10^8)/(9 xx 10^9) = 4/9`
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