A particle of mass `9xx10^(-31)` kg having a negative charge of `1.6xx10^(-19)C` is projected horizontally with a velocity of `10^(6)ms^(-1)` into a region between two infinite horizontal parallel plates of metal. The distance between the plates is `d=0.3cm` and the particle enters 0.1cm below the top plate, THe top and bottom plates are connected, respectively to the positive and negative terminal of a 30V battery. Find the components of hte velocity of hte paricle just before it hits one of the plates.

Answer

`v_(x)=10^(6)m//s, v_(y)(4sqrt(2)xx10^(6))/(3) m//s`

Explanation

An electron is a negatively charged particle, so it will be attacted by the positive plate with force F=eE. Hence, acceleration of electron along y-axis will be
`a=(F)/(m)=(eE)/(m)=(eV)/(md)` `{as E=V/d}` (i)
So, from equation of motion , `v^(2)=u^(2)+2`as along the x-axis ,
Now as `y_(0)=1cm`(given ) and is given by eq.(i) the electrons will hit the top plate with velocity
`v_(y)=sqrt((2y_(0)eV)/(md))=sqrt((2xx1.6xx10^(-19)xx30xx1xx10^(-3))/(9xx10^(-31)xx3xx10^(-3)))`
`=(4sqrt(2)//3)xx10^(6)=1.885xx10^(6)ms^(-1)` ...(iii)
So, the electron will hit the upper plate with the velocity
`v_(x)=10^(6)(ms^(-1))`and `v_(y)=1.885xx10^(6)ms^(-1)`.