Maxwell's displacement current

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Question

Sea water at frequency $V = 2 \times 10^{8} H z$ has permittivity $\in \approx 60 \in_{0}$ , permeability, $μ \approx μ_{0}$ and resistivity $ρ = 0.50 Ω - m$ .Imagine a parallel plate capacitor with plate gap of 1m, immersed in sea water and driven by an alternating Voltage of $200 V$ then the maximum conduction current density is $_______\frac{V}{Ω m^{2}}$

Moderate

Solution

Applied electric field, $E = \frac{V_{0}}{d}$ .
The max conduction current density is $J_{C} = \frac{E}{ρ} = \frac{V_{0}}{ρ d}$
$J_{C} = \frac{200 \times 2}{1 \times 1} = 400 \frac{V}{Ω m^{2}}$

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Maxwell's displacement current

Remember concepts with our Masterclasses.

Applied electric field, E=V0d. The max conduction current density is JC=Eρ=V0ρd JC=200×21×1=400VΩm2

Ready to Test Your Skills?

free study material

Applied electric field, $E = \frac{V_{0}}{d}$ .
The max conduction current density is $J_{C} = \frac{E}{ρ} = \frac{V_{0}}{ρ d}$
$J_{C} = \frac{200 \times 2}{1 \times 1} = 400 \frac{V}{Ω m^{2}}$