Question

A specified volume of space contains an electric field for which the magnitude is given by E=E0cos(ωt). Suppose that E0 = 20 V/m and ω = 1.0 × 107 s−1. What is the maximum displacement current through a 0.40 m2 cross-sectional area of this volume?

Answer 1

Answer: $0.708\ mA$

Explanation:

Given

$E_o=20\ V/m$

$\omega =10^7\ s^{-1}$

Cross-sectional area $A=0.40\ m^2$

Current density is given by

$J=\epsilon_o \dfrac{dE}{dt}$

Displacement current

$\Rightarrow I=JA\\\Rightarrow I=8.854\times 10^{-12}\times 20\times 10^7\times 0.4\\\Rightarrow I=0.708\times 10^{-3}\ A$

Answer 2

The required value of the maximum displacement current of the given space is  $7.08 \times 10^{-4} \;\rm A$.

Given data:

The intensity of electric field is, $E_{0}=20 \;\rm V/m$.

The angular frequency of electric field is, $\omega=1.0 \times 10^{7} \;\rm s^{-1}$.

The cross-sectional area of space is, $A =0.40 \;\rm m^{2}$.

In the given problem, the instantaneous electric field is given by $E = E_{0} \times cos(\omega t)$

So, the expression for the current density is,

$J= \epsilon_{0} \times \dfrac{dE}{dt}$

Here, $\epsilon_{0}$  is the permittivity of free space. Solving as,

$J= \epsilon_{0} \times \dfrac{d(E_{0} \times cos(\omega t))}{dt}\\\\J= -\epsilon_{0} \times E_{0} \times \omega \times sin(\omega t)$

And the expression for the maximum displacement current is,

$I = J \times A$

And the maximum displacement current is possible only when, J is positive and J will be positive for  $sin(\omega t)=-1$.

Then solving as,

$I = (-\epsilon_{0} \times E_{0} \times \omega \times sin(\omega t)) \times A\\\\I = (-8.85 \times 10^{-12} \times 20 \times (1.0 \times 10^{7}) \times (-1)) \times 0.40\\\\I = 7.08 \times 10^{-4} \;\rm A$

Thus, we can conclude that the required value of the maximum displacement current of the given space is  $7.08 \times 10^{-4} \;\rm A$.

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