Two particles, with charges of q1 = 40.0 nC and q2 = 40.0 nC, are placed at the

Question

Two particles, with charges of
q1 = 40.0
nC and
q2 = 40.0
nC, are placed at the points with coordinates (0, 4.00 cm) and
(0, 4.00 cm)
as shown in the figure below. A particle with charge
q3 =
20.0 nC is located at the origin.

(a) Find the electric potential energy of the configuration of the three fixed charges.

-18*10**-5

Correct: Your answer is correct.
J

(b) A fourth particle, with a mass of 1.83 10-13 kg and a charge of q4 = 80.0 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

Explanation / Answer

a.) Potential energy of a system of three charges q1, q2 and q3 is

U = (1/4o) (q1q2/r12 + q2q3/r23 + q1q3/r13 )

U = 9 x 109 ( 40 x10-9 x -40 x 10-9 / 0.08 + -40 x10-9 x 20x 10-9 / 0.04 + 40 x 10-9  x 20 x 10-9/ 0.04 )

U = - 18 x 10-5 J

b.) Potential at the rest point (0.03 m , 0) due to the first three charges

V3 = (1/4o) ( q1/r1 + q2 / r2 + q3 / r3 ) = 9 x 109  ( 40 x 10-9 /0.05 - 40x 10-9 / 0.05 + 20x 10-9 / 0.03 )

V3 = 6000 Volts

Potential at infinity = 0

So, the potential energy lost by the fourth charge when it freely moves from this rest position to infinity,

U' = q4 ( V3 - 0 ) = 80 x 10-9 x 6000 = 4.8 x 10-4 J

According to the law of conservation of energy, the total energy change here needs to be zero because of no external force (remember the fourth charge moves freely.)

Which implies that the Potential energy lost by q4 needs to equal the Kinetic energy gained by it.

So, if v is the final velocity of q4, then

U' =KE

4.8 x 10-4 = 0.5 x 1.83 x 10-13 x v2

v = 72428.59683 m/s

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