Two parallel large metal plates, each of area `A` are piaced at a distance `d` apart. Light of intensity `I` and wavelength `lambda` falls on the plate `1` which is photosensitive. The efficiency of photoelectron emission is `100%`. The photoelectrons are immediately swept away from plate `1` in such a way that the plate `2` remains neutral. Find the potential difference between the two plates at any time `t`, if plate `1` is neutral initially.

Explanation

`N=(lambdaIAt)/(hc)`, where `N` is number of photons falling on the plate.
`Q_(1)=(lambdaIAt)/(hc)e`, `v_(1)-v_(2)=(Q//2)/(C )`, where `C=(epsilon_(0)A)/(d)`
`V_(1)-V_(2)=(lambdaIAte)/(2hcepsilon_(0)A)d=(lambdaIdte)/(2hcepsilon_(0))`