Two particles, with charges of q_1 = 80.0 nC and q_2 = -80.0 nC, are placed at t

Question

Two particles, with charges of q_1 = 80.0 nC and q_2 = -80.0 nC, are placed at the points with coordinates (0, 8.00 cm) and (0, -8.00 cm) as shown in the figure below. A particle with charge q_3 = 40.0 nC is located at the origin. Find the electric potential energy of the configuration of the three fixed charges. A fourth particle, with a mass of 2.49 times 10^-13 kg and a charge of q_4 = 160.0 nC, is released from rest at the point (6.00 cm, 0). Find its speed after it has moved freely to a very large distance away.

Explanation / Answer

Potential at position 6cm=

k(q1/r1 +q2/r2 + q3/r3)

9*109(80nC/0.1 + -80nC/0.1 + 40nC/0.06) = 6000V

Now U = Vq = 6000*160nC = 9.6*10-4J

That will be its KE at infinity

So

KE = 0.5mv2 = 9.6*10-4

0.5*2.49*10-13v2 = 9.6*10-4

v = 87811.408 m/s

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