Posted by **katie** on Thursday, July 15, 2010 at 12:02am.

An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.3×10^7 m/s. What was the electron's speed as it left the negative plate?

the voltage is 650V

and the charge of q1 is 1.7*10^-9 and the charge of q2 is -1.7*10^-9

- college physics -
**drwls**, Thursday, July 15, 2010 at 12:42am
The kinetic energy gained going from the negative plate to the postivie plate is

E = e*V,

where e is the electron charge.

(1/2)*m*(V2^2 - V1^2) = E

Solve for the final velocity, V2

V1 = 2.3*10^7 m/s

m is the electron mass and e is its charge.

You don't need to know the charges on the two plates.

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