A circular parallel plate capacitor with plate radius `R` is charged by means of a cell, at time `t=0`. The initial conduction current is `i_(0)`. Consider a circular area of radius `R//4` coplanar with the capacitor plates and located symmetrically between them. Find the time rate of electric flux change through this area after one time constant.

Explanation

The conduction current at the end of one time constant can be obtained by substituting t = `tau` in the expression `i=i_(0)e^(-t//t)`

as `i=(i_(0))/(e )`i. (say)
(Where is the base of natural logarithm).
If Q be the charge at the mentioned instant then, the electric field between the plates is
`E = (sigma)/(epsilon_(0)) = (Q)/(epsilon_(0)(pi R)^(2))`
`therefore` The electric flux through the specified area is
`phi_(E ). = Epi(R//4)^(2) = (Q)/(epsilon_(0)(pi R)^(2)) ( (pi R^(2))/(16)) = (Q)/(16 epsilon_(0))`
Rate of electric flux change is
`(d phi_(E ))/(dt ) = (1)/(16 epsilon_(0)) (dQ)/(dt ) = (1(i.))/(16 epsilon_(0)) = (i_(0))/(16 epsilon_(0))`