An infinite dielectric sheet having charge density `sigma` has a holeof radius R in it. An electrom is released from point P on the axis of the hole at a distance `sqrt(3)R` from the center. Find the speed with which it crosses the plane of the sheet.
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Explanation

The infinity charged sheet with a circular hole can be consider superposition of an infinite sheet of charge density `sigma` and charged disc of charge density `-sigma` . From the principle of superpoition, we have the following :
, ,
Net electric field at P is the vector sum of electric fields due to both infinite sheet of charge `(vecE_(1))` and disc `(vecE_(2))`

electric field due to infinite dielectric sheet is
`vecE_(1)=(sigma)/(2 epsilon_(0))(hat i)`
Electric field at the axis of a disc of radius R is
`vecE_(1)=(sigma)/(2 epsilon_(0))[1-(x)/(sqrt(x^(2)+R^(2)))](-hati)`
Resultant electric field is
`vecE_("net")=vecE_(1)+vecE_(2)=(sigma)/(2 epsilon_(0))(x)/(sqrt(x^(2)+R^(2)))(hat i)`
Force on the electron is
`vecF_(1)=(sigma ex)/(2 epsilon_(0)sqrt(x^(2)+R^(2)))(hat i)`
`mv(dv)/(dx)=-(sigma ex)/(2 epsilon_(0)sqrt(x^(2)+R^(2)))`
Let the speed of the electron when it crosses the sheet be v. Then
`m int_(0)^(v) vdv=-(sigma e)/(2epsilon_(0)) int_(sqrt(3)R)^(0) (x)/(sqrt(x^(2)+R^(2)))dx`
or, `m(v^(2))/(2)=(sigmae)/(2epsilon_(0))[sqrt(x^(2)+R^(2))]_(sqrt(3R))^(0)`
or, `v=sqrt((sigma e R)/(m epsilon_(0)))`