(a) Two particles which have the same magnitude charge but opposite sign are hel
ID: 1422582 • Letter: #
Question
(a) Two particles which have the same magnitude charge but opposite sign are held 6.00 nm apart. Particle I is then released while Particle II is held steady; the released particle has a mass of 1.13 10-22 kg. Particle I's speed is 101 km/s when it is 3.96 nm away from Particle II. What is the magnitude of the charge on one of the particles?
-ok for this, I think I'm setting it up right but I keep getting the wrong answer.
I'm setting it up: 9e9*q^2 (1/3.96-1/6) = 0.5*1.13e-22*(101e3)^2
I'm just not sure what I'm doing wrong
(b) If the particles are still initially held 6.00 nm apart but both particles are released, when they are 3.96 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.)
A) Particle I would be moving at the same speed that it had in (a)
B)To answer this at all, Particle II's mass would be needed
C)Particle I would be slower than it was in (a)
D) Particle I's speed is unknown but it will definitely be the same as Particle II's
E)Particle I would be faster than it was in (a)
Explanation / Answer
a)
v^2 = 2a*s
where v = 101000 m/s, s = distance travelled = 6 - 3.96 nm = 2.04*10^-9 m
Particle I acceleration is a = 101000^2 / 2.04*10^-9 m/s^2 = 5*10^18 m/s^2
so accelerating force is f = m*a = 1.13*10^-22 * 5*10^18 N = 5.65*10^-4N
f is also the electrostatic force f = Eo*q^2 / r^2.
At rest r = 6nm, but the acceleration will actually increase as r decreases due to the increasing electrostatic force, so we should take r = 3.96 nm to equate the forces.
So charge q = r * sqrt(f*/Eo) = 3.96*10^-9 * sqrt(5.65*10^-4/8.85*10^-12) Coulomb
q = 3.1640 * 10^-5 Coulomb
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