Two charge particles of equal mass of 2.6 x 10^-27 kg and equal charge are accel

Question

Two charge particles of equal mass of 2.6 x 10^-27 kg and equal charge are accelerated toward one another with initial speeds of 1.2 x 10^6 m/s relative to the earth. The charges are initially very far apart. As they get closer together, the two charges stow until momentarily stop and than begin to return back. The distance of the closest approach between the two charges is 309 nm. a. What is the charge on each particle? b. What is the acceleration on one of the two charge particles? c. How does the acceleration vary as the two charges move closer to each other? Explain

Explanation / Answer

m = mass of each particle = 2.6 x 10-27 kg

q = charge

Vi = initial speed = 1.2 x 106 m/s

r = distance when they are closest = 309 nm = 309 x 10-9 m

a)

Using the conservation of energy

Total Kinetic energy of system = Total electric potential energy

(0.5) m V2 + (0.5) m V2 = kq2 /r

inserting the values

( 2.6 x 10-27 ) (1.2 x 106 )2 = (9 x 109) q2 /(309 x 10-9 )

q = 3.6 x 10-16 C

b)

Force between the charge is given as ::

F = k q2 /r2

inserting the values

F = (9 x 109) (3.6 x 10-16)2 /(309 x 10-9 )2

F = 1.22 x 10-8 N

acceleration is given as ::

a = F/m = 1.22 x 10-8 / (2.6 x 10-27)

a = 4.7 x 1018 m/s2

c) as the charges move closer the force between them increases, and hence the acceleration increases

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