A particle of mass ma and charge q is released from rest in a uniform filed of magnitude E. the uniform field is created between two parallel plates of charge densities `+sigma` and `-sigma`, respectively. The particle accelerates toward the other plate a distance d apart. Determine the speed at which it stirkes the opposite plate.

Explanation

The applied electric field is `vec(E )=-E_(0)hat j`. The force experienced by the charge q is
`vec(F)=qvec(E )=-qE_(0)hat j`
The force is constant, and so the acceleration is constant as well, i.e.
`vec(a)=(vec(F))/(m) =-(qE_(0))/(m)hat j`
Because of constant acceleration the particle moves in the negative y-direction , the problem is analogous to the motion of a mass released from rest in a gravitational field. from equations of motion we get
`v_(y)=v_(y0)+a_(y)t=0--(qE_(0))/(m) t`...(i)
and `y=y_(0)+v_(0)t+(1)/(2)a_(y)t^(2)` or `0=d+0-(1)/(2)(qE_(0))/(m) t^(2)`...(ii)
The particle starts at `y_(0)=d` and impact occure at y=0 from eq. (ii) we get,
`t=((2dm)/(qE_(0)))^(1//2)`
From eq. (i) we get
`v_(y)=-(qE_(0))/(m)((2dm)/(qE_(0)))^(1//2)=-sqrt((2qE_(0)d)/(m))`.