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 symmetrically between the plates. Find the displacement current through this area.

Options

A. `I`
B. `(I)/(2)`
C. `(I)/(4)`
D. `(I)/(8)`

Answer

B

Explanation

Charge on capacitor plates at time t is q, = It.
Electric field between the plates at this instant is
`E=(q)/(Aepsilon_(0))=(It)/(Aepsilon_(0))" ....(i)"`
Electric flux through the given area `(A)/(2)` is
`phi_(E)=((A)/(2))E=(It)/(2epsilon_(0))" Using (i) ....(ii)"`
Therefore, displacement currrent
`I_(D)=epsilon_(0)(dphi_(E))/(dt)`
`=epsilon_(0)(d)/(dt)((It)/(2epsilon_(0)))=(I)/(2)" [Using (ii)]"`