Two particles, with charges of q1 10.0 nC and 10.0 nc, are placed at the points

Question

Two particles, with charges of q1 10.0 nC and 10.0 nc, are placed at the points with coordinates 0, 4.00 cm) and 0 4.00 cm as shown in the figure belov. A particle with charge q, 5.0 nC s located at the origin. 9, (a) Find the electric potential energy of the configuration of the three fixed charges (b) A fourth particle, with a mass of 2.10 x 10-13 kg and a charge of 20.0 nC, is released from rest at the point (3.00 cm, 0). Find its speed after it has moved freely to a very large distance away m/s Need Help?

Explanation / Answer

x13 = distance between charge q1 and q3 = 4 cm = 0.04 m

x23 = distance between charge q2 and q3 = 4 cm = 0.04 m

x12 = distance between charge q1 and q2 = 8 cm = 0.08 m

electric potential energy of system of fixed charges is given as

U = k q1 q2/x12 + k q1 q3/x13 + k q2 q3/x23

U = (9 x 109) (10 x 10-9) (- 10 x 10-9)/(0.08) + (9 x 109) (10 x 10-9) (5 x 10-9)/(0.04) + (9 x 109) (5 x 10-9) (- 10 x 10-9)/(0.04)

U = - 11.25 x 10-6 J

b)

r14 = distance of charge q1 from q4 = sqrt(42 + 32) = 5 cm = 0.05 m

r24 = distance of charge q2 from q4 = sqrt(42 + 32) = 5 cm = 0.05 m

r34 = distance of charge q3 from q4 = 3 cm = 0.03 m

Ui = electric potential energy of charge q4 at given location

Uf = electric potential energy far away = 0 J

Ui = k ((q1/r14) + (q2/r24) + (q3/r34) ) q4

Ui = (9 x 109) (((10 x 10-9)/(0.05)) + ((- 10 x 10-9)/(0.05)) + ((5 x 10-9)/(0.03)) ) (20 x 10-9)

Ui = 30 x 10-6 J

using conservation of energy

kinetic energy gained = electric potential energy lost

(0.5) m v2 = Ui

(0.5) (2.1 x 10-13) v2 = (30 x 10-6)

v = 1.7 x 104 m/s

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