Two particles which have the same magnitude charge but opposite sign are held 7.

Question

Two particles which have the same magnitude charge but opposite sign are held 7.00 nm apart. Particle I is then released while Particle II is held steady; the released particle has a mass of 1.35 x 10-22 kg. Particle I's speed is 82.0 km/s when it is 3.71 nm away from Particle II. What is the magnitude of the charge on one of the particles? If the particles are still initially held 7.00 nm apart but both particles are released, when they are 3.71 nm away from each other, how would Particle I's speed compare to the speed used in part (a) above? (Assume that Particle II's mass is not the same as Particle I's; you should be able to answer this without performing a detailed calculation.) Particle I would be moving at the same speed that it had in (a) Particle I would be faster than it was in (a) To answer this at all, Particle II's mass would be needed Particle I would be slower than it was in (a) Particle I's speed is unknown but it will definitely be the same as Particle II's

Explanation / Answer

a) r1 = 7 nm

r2 = 3.71 nm

let q1 = q, q2 = -q.


U1 = -k*q^2/r1

U2 = -k*q^2/r2

workdone = U1-U2

w = -k*q^2/r1 + k*q^2/r2

w = q^2*(k/r2 - k/r1)

w = q^2**1.14*10^18 J

work done = change in kinetic enrgy

0.5*m*v^2 = q^2*1.14*10^18

0.5*1.35*10^-22*(82*10^3)^2 = q^2*1.14*10^18

q^2 = 3981.32*10^-36

q = 6.3097*10^-17 C

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b)
2*0.5*m*v^2 = q^2**1.14*10^18 J

2*0.5*m*v^2 = (6.3097*10^-17)^2*1.14*10^18


m*v^2 = 4.539*10-15

v^2 = 3.362*10^7

v = 5.799 km/s

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