Question

Two particles which have the same magnitude charge but opposite
sign are held *r* = 7 nm apart. Particle I is then released
while Particle II is held steady; the released particle has a mass
of 8.72 × 10^{-23} kg. Particle I's speed is 110 km/s when
it is 0.74*r* away from Particle II.

1. What is the magnitude of the charge on **one**
of the particles?

2. If the particles are still initially held 7 nm apart but
**both** particles are released, when they are 0.74 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.)

Answer #1

0.748 A Inm r= 7nm Potential Energy of the system of changes - untor I qe 7V 2 finally pifcis uño orar change in PE = change in KE Loto +3 = { move 9x109 x (82) (0.357 = 4 x 8:72x16 23 (110000) 7x109 L TQ = 1.083x10-15 cl

Iloemle is the speed when one of the the particle is kept steady But when Both the particles are free the rise in ke will be distributed for both the particle so, velocity of particle -I will lesser than lloem's because some of kincha Fenergy will be then one by second particles 022

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