An electron is released from the bottom plate A as shown in the figure `(E = 10^(4) N//C)`. The velocity of the electron when it reaches plate B will be nearly equal to

Options

A. `0.85 xx10^7 m//s`
B. `1.0 xx10^7 m//s`
C. `1.25 xx10^7 m//s`
D. `1.65 xx10^7 m//s`

Answer

A

Explanation

`F=eE implies ma = eE implies a = (eE)/(m) , S = 2cm " & " u =0 `
`v^2=u^2+2aS implies v=sqrt(2aS) = sqrt(2xx(eE)/m xx2 xx10^(-2))`
`=sqrt((2 xx1.6 xx10^(-19) xx10^4xx2 xx10^(-2))/(9xx10^(-31)))`
`=0.843 xx10^7 m//s ~~ 0.85 xx10^7 m//s`