Two charges q 1 = 3.20 nC and q 2 = +8.36 nC are at a distance of 1.35 µm from e

Question

Two charges q1 = 3.20 nC and q2 = +8.36 nC are at a distance of 1.35 µm from each other. q1 is fixed at its location while q2 is released from rest.

(a) What is the kinetic energy of the charge q2 when it is 0.340 µm from q1?

What is the electric potential at the initial and final locations of the moving charge due to the fixed charge? What is the change in electric potential energy of the moving charge? J

(b) The charge q2 has a mass m2 = 7.85 µg. What is its speed when it is 0.340 µm from q1?

How is kinetic energy related to the speed of the particle? m/s

Explanation / Answer

initial potential energy Ui = k*q1*q2/r1 = -9*10^9*3.2*10^-9*8.36*10^-9/(1.35*10^-6) = -0.178 J

final potential energy Uf = k*q1*q2/r2 = -9*10^9*3.2*10^-9*8.36*10^-9/(0.34*10^-6) = -0.708 J

initial kinetic energy Ki = 0

final kinetic energy = Kf

Ui + Ki = Uf + Kf

Kf = Ui - Uf = -0.178 + 0.708 = 0.53 J

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part b

Kf = (1/2)*m2*v^2

(1/2)*7.85*10^-6*v^2 = 0.53

v = 367.5 m/s

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