Sea water at frequency $$V=4$$×$$108Hzhas$$ permittivity ∈≈$$80$$∈$$0$$ , permeability μ≈μ$$0$$ and resistivity ρ$$=0.25$$Ω−m. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating Voltage of $$400V$$ then the maximum conduction current density is $$_______Volt/ohm$$ $$m2($$ distance between the plates , $$d=1$$ $$m)$$

Question

Sea water at frequency $$V=4$$×$$108Hzhas$$ permittivity ∈≈$$80$$∈$$0$$ , permeability μ≈μ$$0$$ and resistivity  ρ$$=0.25$$Ω−m. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating Voltage of $$400V$$ then the maximum conduction current density is  $$_______Volt/ohm$$ $$m2($$ distance between the plates , $$d=1$$ $$m)$$

Infinity Learn · Accepted Answer

Applied electric field  $$E=V0dThe$$ conduction current density is  $$JC=E$$ρ$$=V0$$ρ$$d(maximum)JC=40025$$×$$10$$−$$2$$×$$1=16$$×$$102V$$Ω$$m2$$

Maxwell's displacement current

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Applied electric field $E = \frac{V_{0}}{d}$
The conduction current density is $J_{C} = \frac{E}{ρ} = \frac{V_{0}}{ρ d} (maximum)$
$J_{C} = \frac{400}{25 \times 10^{- 2} \times 1} = 16 \times 10^{2} \frac{V}{Ω m^{2}}$

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Maxwell's displacement current

Remember concepts with our Masterclasses.

Applied electric field E=V0dThe conduction current density is JC=Eρ=V0ρd(maximum)JC=40025×10−2×1=16×102VΩm2

Ready to Test Your Skills?

free study material

Applied electric field $E = \frac{V_{0}}{d}$
The conduction current density is $J_{C} = \frac{E}{ρ} = \frac{V_{0}}{ρ d} (maximum)$
$J_{C} = \frac{400}{25 \times 10^{- 2} \times 1} = 16 \times 10^{2} \frac{V}{Ω m^{2}}$