Ncert Exercises & Solutions

$Seawater at frequency ν = 4 \times 10^{8} Hz has permittivity ε \approx 80 ε_{0}, permeability μ \approx μ_{0}$ and resistivity $ρ$ =0.25 $Ω$ -m. Imagine a parallel plate capacitor immersed in seawater and driven by an alternating voltage source $V (t) = V_{0} \sin (2 πνt) .$ What fraction of the conduction current density is the displacement current density?

Hint: The displacement current density depends on the variation of the electric field between the plates of the capacitor.

Step 1 : Find the conduction current density .

Suppose the distance between the parallel plates is d and applied voltage V_{(t)} = V_{0} \sin (2 πνt) .

Thus, electric field,

E = \frac{V_{0}}{d} \sin (2 πνt)

Now using Ohm' s law, J_{0} = \frac{1}{ρ} \frac{V_{o}}{d} \sin (2 πνt)

\Rightarrow = \frac{V_{0}}{ρd} \sin (2 πνt) = {(J_{0})}_{c} \sin (2 πνt)

Here, {(J_{0})}_{c} = \frac{V_{0}}{ρd}

Step 2 : Find the displacement current density .

Now the displacement current density is given as;

J_{0} = ε \frac{dE}{dt} = ε \frac{d}{dt} [\frac{V_{0}}{d} \sin (2 πνt)]

= \frac{2 {πνεV}_{0}}{d} \cos (2 πνt)

\Rightarrow = {(J_{0})}_{d} \cos (2 πνt)

Where, {(J_{0})}_{d} = \frac{2 {πνεV}_{o}}{d}

\Rightarrow \frac{{(J_{0})}_{d}}{{(J_{0})}_{c}} = \frac{2 {πvεV}_{0}}{d} . \frac{ρd}{V_{0}} = 2 πνερ

= 2 π \times 80 ε_{0} ν \times 0.25 = 4 {πε}_{0} ν \times 10

= \frac{10 ν}{9 \times 10^{9}} = \frac{4}{9}

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