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An electron is released from the bottom ...

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

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`

Text Solution

Verified by Experts

The correct Answer is:
A
`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`
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