A particle having charge that of an electron and mass `1.6xx10^(-30)`kg is projected with an initial speed u at the angle `45^(@)` to the horizontal from the lower plate of a parallel-plate capacitor as shown in fig. The plates are sufficiently long and have a separation of 2 cm, find the maximum value of the velocity of he particle so that it does not hit the upper plate. Take the electric field between the plates as `10^(3)Vm^(-1)` directed upward.

Explanation

Resolving the velocity of particle parallel and perpendicular to the plate,
`mu_(II)=ucos45^(@)=(u)/(sqrt(2))` and `u_(bot)=usin45^(@)=(u)/(sqrt(2))`
Force on the charged particle in the downward direction normal is to the plate `eE` Acceleration is `a=eE//m`, where m is the mass of charged particle. The particle will nothit the upper plate, if the valocity component normal to plate becomes zero before reaching it `i.e., 0=u_(bot)^(2)-2ay` with `yled` where d distance between the plates. therefore the maximum velocity for the particle not to hit the upper plate (for thi `y=d=2cm`) is
`u_(bot)=sqrt(2ay)=sqrt((2xx1.6xx10^(-19)xx10^(3)xx2xx1)^(-2))/(1.6xx10^(-30))`
`=2xx10^(6)ms^(-1)`
`u_(max)=u_(bot)//cos45^(@)=2sqrt(2)xx10^(6)ms^(-1)`