Question 455: Electrostatics & Capacitance

Question

A parallel plate capacitor with plate area A and plate separation $d$ is charged by constant current $I$. A plane surface of area $\frac{\mathrm{A}}{2}$, parallel to the plates is drawn simultaneously between the plates. The displacement current through this area is

Accepted Answer

Displacement current between the plates is equal to the conduction (charging) current I for the full area. The electric field and flux are uniform between the plates, so the displacement current through any surface is proportional to the area of that surface .
Displacement current $I_d=\frac{A / 2}{A} 	imes I=\frac{1}{2} I$
$$ 	herefore \quad I_D=\frac{I}{2} $$

Answer

I

Answer

$\frac{\mathrm{I}}{2}$

Answer

$\frac{\mathrm{I}}{4}$

Answer

$\frac{1}{8}$

Electrostatics & Capacitance

A parallel plate capacitor with plate area A and plate separation d is charged by constant current I. A plane surface of area \frac{\mathrm{A}}{2}, parallel to the plates is drawn simultaneously between the plates. The displacement current through this area is

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