Question

Two electrons are held 3μm apart. When released from rest, what is the velocity of each electron when they are 8μm apart?​

Answer 1

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Let Ep1 be the potential electric energy at rest (distance r = 3μm) and Ep2 be the potential electric energy when they are 5μm apart and moving. The total (potential and kinetic energies) at each position are given by

Et1 = Ep1 + (1/2) m (0)2 = Ep1

Et2 = Ep2 + (1/2) m v2 + (1/2) m v2 = Ep2 + m v2

Formula for electric potential energy due to charges q1 and q2 distant by r is:

Ep = k q1 q2 /r

No external energy is used and no energy is lost, therefore there is conservation of energy such that potential energy is converted into kinetic energy.

Ep1 = Ep2 + m v2 , v is the velocity when 8μm apart.

charge of electron = - e = -1.60×10-19C , mass of electrom m = 9.109×10-31Kg

m v2 = Ep1 - Ep2 = k×e×e / (3×10-6) - k×e×e / (5×10-6) = 9×109(1.6×10-19)2 [ 1 / (3×10-6) - 1 / (8×10-6) ]

v ≈ 3.48×104 m/s

Answer 2

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Two electrons are held 3μm apart. When released from rest, what is the velocity of each electron when they are 8μm apart? ... E p1 = E p2 + m v 2 , v is the velocity when 8μm apart.