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Two stationary positive point charges, charge 1 of magnitude 3.15 nC and charge 2 of magnitude...

Two stationary positive point charges, charge 1 of magnitude 3.15 nC and charge 2 of magnitude 1.65 nC , are separated by a distance of 53.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

What is the speed vfinal of the electron when it is 10.0 cm from charge 1?

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Answer #1

Potential, V = k (q1/r + q2/r)

V1 = (9 x 10^9)[(3.15 x 10^-9/(0.53/2)) + (1.65 x 10^-9/(0.53/2))] = 163.02 V

V2 = (9 x 10^9)[(3.15 x 10^-9/0.10) + (1.65 x 10^-9/0.43)] = 318.03 V

(V2 - V1) q = 0.5 mv^2

v = sqrt[(2 x (318.03 - 163.02) x 1.6 x 10^-19)/(9.11 x 10^-31)] = 7.38 x 10^6 m/s

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