Two particles, with charges of q_1 = 30.0 nC and q_2 = -30.0 nC, are placed at t

Question

Two particles, with charges of q_1 = 30.0 nC and q_2 = -30.0 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 = 15.0 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 1.50 times 10^-13 kg and a charge of q_4 = 60.0 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

The potential at the (6,0) point is
V = k*q1/r1 + q2/r2 + q3/r3) = 9.0c10^9*(30x10^-9/0.02 - 30x10^-9/0.14 + 15x10^-9/0.06) =
13821V

So U = V*q = 13821*40x10^-9 = 5.52x10^-4J

This will be the kinetic energy at infinity

so 1/2*m*v^2 = 5.52x10^-4J

v = sqrt(2*5.52x10^-4/1.50x10^-13) = 8.5 x10^4m/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.