Suppose that the particle is an electron projected with velocity `V_(x)=2.0xx10^(6) ms^(-1)`. If E between the plates separated by 0.5 cm is `9.1xx10^(2)N//C`, where will the electron strike the upper plate? `(|e|=1.6xx10^(-19) C, m_(e)=9.1xx10^(-31) kg)`.

Explanation

Given `V_(x)=2xx10^(6) m//s, E=9.1xx10^(2) N/C`
`q=e=1.6xx10^(-19) C, m_(e) =9.1xx10^(-31) kg`
`d=0.5 cm =0.5xx10^(-2) m=5xx10^(-3) m`
The electron will strike the upper plate at its other end at X = L as it get deflected.
`Y=d/2=(5xx10^(-2))/2=2.5xx10^(-3) m`
Using `Y=(qEL^(2))/(2m V_(x)^(2))`
`L=sqrt((2m V_(x)^(2).Y)/(qE))=sqrt((2mY)/(qE)).V_(x)`
`L=sqrt((2xx9.1xx10^(-31) xx2.5xx10^(-3))/(1.6xx10^(-19) xx9.1xx10^(2)))xx2xx10^(6)`
`L=1.12xx10^(-2) m`
`=1.12 cm`.