f = k Q1 Q2 / d^2
change in Ke = 1/2 m v^2
= change in U = k Q1 Q2 (1/r2 -1/r1)
r1 = 2*10^-14 + 8.22*10^-13
r2 = 2*10^-14
An alpha particle(charge +2e) is sent at a speed towards a gold nucleus( +79e ).
1) what is the electric force acting on the alpha particle when it is 2.0 x 10^-14 m from the gold nucleus.
2) if the alpha particle came to a stop at this position after traveling 8.22 x 10^-13, what was the initial velocity of the particle. (mass= 6.64 x 10^-27 kg)
1 answer
1 answer