(a) Two particles which have the same magnitude charge but opposite sign are hel

Question

(a) 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 1.15 10-22 kg. Particle I's speed is 147 km/s when it is 4.9 nm away from Particle II. What is the magnitude of the charge on one of the particles?

(b) If the particles are still initially held 7.00 nm apart but both particles are released, when they are 4.9 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.)

im not sure if my calculations are wrong or if im using the wrong formulas but please help!

Explanation / Answer

(a)

Let the magnitude of charge on each particle be 'q'. Then intial potential energy of the system would be

U1 = -9*109*q2/7*10-9  = -(9/7)*q2   ; also the final potential energy would be:

U2  = -9*109*q2/4.9*10-9  = -(9/4.9)*q2

Now applying energy conservation between the two instances we have:

U1 = U2 + 0.5*m*v2   or -(9/7)*q2  = -(9/4.9)*q2 + 0.5*1.15*10-22*(1.47*105)2

So q2 = 1.24*10-12*49/27 = 2.25*10-12 C

(b)

Now the energy conservation would lead to the same kinetic enrgy of the system. This time however both the particles would have same same velocity say 'v0' (as they are completely identical except for the sign of charge).

So we would have 2*0.5*m*v02  = 0.5*m*v2   or v0  = v/20.5  

Moreover the velocities of the two particles are in opposite directions hence the relative velocity becomes:

V = 2*v0  = 20.5*v = 1.414*147 = 207.86 Km/s

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