A parallel- plate capacitor having plate-area A and plate separation d is joined to a battery of emf epsilon and internal resistance R at t=0. Consider a plane surface of area A/2, parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time.

Explanation

Given, `q = CV (1-e^(-t^//tau))`
`:.` Surface charge dencity, `sigma = q/A= (CV)/A (1-e^(-t//tau))`
Electric field between the plates of capacitor,
`E=sigma/(epsilon_0)=(CV)/(epsilon_0A) (1-e^(-t//tau))`
Electric flux from the given area
`phi_E =(EA)/2=(CV)/(2epsilon_0)(1-e^(-t//tau))`
Displacement current, `i_d=epsilon_0 (dphi_E)/dt`
or `i_d=epsilon_0 d/dt[(CV)/(2epsilon_0) (1-e^(-t//tau))]=(CV)/(2 tau)e^(e^(-t//tau))`
Substituting, `tau = CR`
We have,
`i_(d) = (V)/(2R) e^(-t//CR)`
Again substituting, `C = (epsilon_(0)A)/(d)`
`i_(d)=(V)/(2R)e^(-(td)/(epsilon_(0)AR))`.