A partical 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 suffiecienly 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 velosity of the particle parallel and perpendicular to the plate, we get
`u_(||)=u cos 45^(@)=(u)/(sqrt(2)) and u_(_|_)=u sin 45^(@)=(u)/(sqrt(2))`
force on the charged particle in the downward direction and normal to the plate is eE. Therefore, acceleration `a=eE//m,` where m is the mass of the charged paricle. the particle will not hit the upper plate if the velocity component normal to hte plate becomes zero before reaching it . i.e. `0=-u_(_|_)^(2)-2ay` with `y led`, where d is the distnce between the plates. therefore, the maximum velocity for the particle not to hit the upper plate (for this `y=d=2 cm`)is
`u_(_|_)=sqrt(2ay)=sqrt((2xx1.6xx10^(-19)xx10^(3)xx2xx10^(-2))/(1.6xx10^(-30)))`
`=2xx10^(6)ms^(-1)`
or, `u_(max)=u_(_|_)//cos45^(@)=2sqrt(2)xx10^(6)ms^(-1)`.