Two particles, with charges of 40.0 nC and -40.0 nC, are placed at the points wi

Question

Two particles, with charges of 40.0 nC and -40.0 nC, are placed at the points with coordinates (0, 4.00) cm and (0, -4.00) cm as shown in Figure P20.18. A particle with charge 20.0 nC is located at the origin.

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

(b) A fourth particle, with a mass of 2.28 10-13 kg and a charge of 80.0 nC, is released from rest at the point (3.00, 0) cm. Find its speed after it has moved freely to a very large distance away.

______ m/s

Explanation / Answer

(a)
U1 = 0
U2 = -40nC* [k* 40nC/0.08] = - 1.8*10-4
U3 = 20nC*[k*20nC/0.04] + 20nC*[k*(-40nC)/0.04] = 0.9*10-4 - 1.8*10-4
UT = U1 + U2 + U3
UT = - 2.7*10-4 J

(b) The fourth particle has a net force on it due to the Coulomb force of each of the other three (fixed) particles. Initially the 80nC particle is at rest, so vo = 0 m/s. Thus the problem is to find the final speed (the speed when the particle is at ?).

Since, the electricfield is conservative.

Thus, ?E = ?U + ?K = 0, or

?K = -?U.

Since vo = 0, then ?K = 1/2 * m *vf2

Equating this to -?U, which is just the initial electricpotential energy of the fourth particle (the final electricpotential energy of the foruth particle at ? is 0) will allow you to solve for the final speed of the fourth particle.

Now,

1/2 * m * vf2 = -Uf +Ui

= 0 + Ui

= - 2.7*10-4 + U4

where U4 is the electric potential on q4 due to all the other three.

vf = ? [ (2/m)* (- 2.7*10-4 + U4) ]

vf = ? [ (2/2.28*10-13)* (- 2.7*10-4 + U4) ]

vf = ? [ 8.77*1012* (- 2.7*10-4 + U4) ]

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