Question

A parallel plate capacitor with plate area A and separation between the plates d, is charged by a constant current I. Consider a plane surface of area A/2 parallel to the plates and drawn between the plates. The displacement current through the area is:

• A. I
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{8}$

Answer 1

Answer: (B)

$\frac{1}{2}$

Answer 2

: Charge on capacitor plates at time t is, q = It. Electric field between the plates at this instant is

$E = \frac{q}{A\varepsilon_{0}} = \frac{It}{A\varepsilon_{0}}$ …(i)

Electric flux through the given area $\frac{A}{2}$is

$\varphi_{E} = \left( \frac{A}{2} \right)E = \frac{It}{2\varepsilon_{0}}$ Using (i) …(ii)

Therefore, displacement current,

$I_{D} = \varepsilon_{0}\frac{d\varphi_{E}}{dt}$

$= \varepsilon_{0}\frac{d}{dt}\left( \frac{It}{2\varepsilon_{0}} \right) = \frac{I}{2}$ [Using (ii)]