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Two charges q1 = −3.20 nC and q2 = +8.23 nC are at a distance of...

Two charges q1 = −3.20 nC and q2 = +8.23 nC are at a distance of 1.72 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?
J

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

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Answer #1

(a) at 1.72 µm, the potential energy is

EPE = kQq / d = 8.99e9N·m²/C² * 3.20e-9C * 8.23e-9C / 1.72e-6m

EPE = 2.37e-7N·m² / 1.72e-6m = 0.138 J

at 0.340 µm,

EPE = 2.37e-7N·m² / 0.340e-6m = 0.697 J

KE = ΔEPE = 0.559 J

(b) v = √(2*KE / m) = √(2 * 0.559J / 7.85e-9kg) = 1.19e4 m/s

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