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.