A parallel plate capacitor with plate area A and plate separation d, is charged by a constant current I. Consider a plane surface of area `A//2` parallel to the plates and situated symmetrically between the plates. Determine the displacement current through this area.

Options

A. I
B. `I/2`
C. `2I`
D. `I/3`

Answer

B

Explanation

Let q be the charge on the capacitor plates at any instant and `sigma` be the surface charge density.
Then the electric field between the plates,
`E = (sigma)/(epsilon_0) = (q)/(A epsilon_(0)) "where" sigma = q/A`
`:.` Flux through the area `A/2 = E xx Area`
`phi_(E) = q/(A epsilon_(0)) xx A/2 = q/(2 epsilon_0)`
`:.` The displacement current
`=I_(D) = epsilon_(0) (d phi_(E))/(dt) = epsilon_(0)1/(2 epsilon_(0)) xx (dq)/(dt)`
`:. I_(D) = I/2`
where `I = (dq)/(dt)` = the rate at which the charge flows through the conducting wire.