Two particles, with charges of 20.0 nC and –20.0 nC, are placed at the points wi

Question

Two particles, with charges of 20.0 nC and –20.0 nC, are placed at the points with coordinates (0, 4.00 cm) and (0, -4.00 cm), as shown in the figure to the right. A particle with charge 10.0 nC is located at the origin. a) Find the electric potential energy of this configuration of the three fixed charges. b) A fourth particle, with a mass of 2.00x10-13 kg and a charge of 40.0 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.

Explanation / Answer

q1 = 20.0 nC @ (0,4.0)
q2 = -20.0 nC @ (0,-4.0)
q3 = 10.0 nC @ (0,0)

r1 = r2 = 4.0 cm = 4.0*10^-2 m
r3 = 0.4*2 = 8.0 cm = 8.0*10^-2 m

(a)
Electric potential energy of this system of charges is gievn by

P.E =  K*q1q3/r1 + k*q2q3/r2 + k*q1q2/r3
P.E = 8.9*10^9 * 10^-9 * 10^-9 * 10^2 * [(20*10)/4.0 - (20*10)/4.0 - (20*20)/8.0]
P.E = - 4.45 * 10^- J

(b)
m = 2.0 * 10^-13 Kg
q4 = 40.0 nC @ (3.0, 0)

Using Energy Conservation,
Initial Potential Energy = Final Kinetic Energy
1/2 * mv^2 = P.Ein

Now,
Potential of the System at x = 3.0 cm
V = k*q1/r1 + k*q2/r2 + k*q3/r3
V = 8.9 * 10^9 * 10^-9 * 10^3[ 20/5 - 20/5 + 10/3]
V = 8.9 * 10^3 * 10/3 Volt
V = 2966.7 Volt

P.E of the fourth Charge at that Point, P.E = q4*V
P.Ein = 40.0 * 10^-9 * 2966.7 J
P.Ein =  1.18 * 10^-4 J

1/2 * mv^2 = 1.18 * 10^-4
1/2 * 2.0 * 10^-13 * v^2 = 1.18 * 10^-4
v = 3.44 * 10^4 m/s
Speed of the particle, v = 3.44 * 10^4 m/s

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