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## Homework Statement

An electron is emitted by an electron gun which is midway between two parallel charged plates which are 20 cm apart. The electrical field strength between the plates is20 N C^-1. The electron is attracted towards the positive plate and strikes the plate as shown. If the electron is emitted at a speed of 3 x 10^6 ms^-1, find the speed of the electron just before striking the positive plate.

photo(not necessary) : http://s613.photobucket.com/user/Yan_Wa_Chung/media/03_zpskzvvveir.png.html

## Homework Equations

v^2 = 2as(not sure) , e=1.60 x 10^(-19) C , mass of electron = 9.11 x 10^(-31) kg

## The Attempt at a Solution

Why is the answer 8.9 x 10^6 ms-1? I got a value about 3x10^6 because the width of the plates is small and it cant gain much vertical speed.

vertical acceleration = 20 x 1.60 x 10^(-19) / 9.11 x 10^(-31) = 3.51 x 10^(12) m s^-2

by v^2 = 2as

v^2 = 2 (3.51 x 10^(12)) (10/100)

= 7.02 x 10^(11)

speed = [[7.02 x 10^(11)] + (3x 10^6)^2] 1/2 =3.11 x 10^6 ms-1

or by energy conservation,

(3x10^6)^2/2 + 20 x 1.60 x 10^(-19) / 9.11 x 10^(-31) (work done) = speed^2/2

speed=3.11 x 10^6 ms-1

what's wrong with my calculation?