• Author / Uploaded
• Diego fernando Leòn ariza

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1. A particle (charge = 50 μC) moves in a region where the only force on it is an electric force. As the particle moves 25 cm from point A to point B, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference, ABVV.

2. A particle (mass = 6.7 10–27kg, charge = 3.2 10–19C) moves along the positive x axis with a speed of 4.8 105m/s. It enters a region of uniform electric field parallel to its motion and comes to rest after moving 2.0 m into the field. What is the magnitude of the electric field?

Convertir 1206 N/C = 1.206 kN/C 3. An electron (m= 9.1 10–31kg, q= –1.6 10–19C) starts from rest at point A and has a speed of 5.0 106m/s at point B. Only electric forces act on it during this motion. Determine the electric potential difference Va-Vb.

4. If a= 30 cm, b= 20 cm, q= +2.0 nC, and Q= –3.0 nC in the figure, what is the potential difference BAVV?

5.

Several charges in the neighborhood ofpoint P produce an electric potential of 6.0 kV (relative to zero at infinity) and an electric field of 36iN/C at point P. Determine the work required of an external agent to move a 3.0-μC charge along the xaxis from infinity to point P without any net change in the kinetic energy of the particle.

W=(q)deltaV W=(3.0x10^-6C)(6000V)=.018J = 18mJ.

6. Three charged particles are positioned in the xyplane: a 50-nC charge at y= 6 m on the yaxis, a – 80-nC charge at x= –4 m on the xaxis, and a 70-nc charge at y=–6 m on the yaxis. What is the electric potential (relative to a zero at infinity) at the point x= 8 m on the xaxis?

7. Point charges of equal magnitudes (25 nC) and oppositesigns are placed on (diagonally) opposite corners of a 60-cm 80-cm rectangle. If point A is the corner of this rectangle nearest the

positive charge and point B is located at the intersection of the diagonals of the rectangle, determine the potential difference, VB–VA.

8. A +4.0-μC charge is placed on the xaxis at x= +3.0 m, and a –2.0-μC charge is located on the yaxis at y= –1.0 m. Point A is on the yaxis at y= +4.0 m. Determine the electric potential at point A (relative to zero at the origin).

9. A particle (charge = Q) is kept in a fixed position at point P, and a second particle (charge = q) is released from rest when it is a distance Rfrom P. If Q= +2.0 mC, q=–1.5mC, and R= 30 cm, what is the kinetic energy of the moving particle after it has moved a distance of 10 cm?

10. Particle A (mass = m, charge = Q) and B (mass = m, charge = 5 Q) are released from rest with the distance between them equal to 1.0 m. If Q= 12 μC, what is the kinetic energy of particle B at the instant when the particles are 3.0 m apart? Método para resolver K = 9 * 10 ^ 9 d2 = 3.0 m d1 = 1.0 m Ambos tendrán la misma velocidad porque la conservación del impulso da que m1v1 = m2v2 y m1 = m2 entonces v1 = v2 por lo tanto -K * 5Q ^ 2 / d2 + K5Q ^ 2 / d1 = mv ^ 2 Energía cinética de la partícula B = .5mv ^ 2 = .5 * (- K * 5Q ^ 2 / d2 + K5Q ^ 2 / d1) al resolver obtenemos Energía cinética = 7.591 J

Falta resolver con los valores dados 11. A particle (charge 7.5 μC) is released from rest at a point on the x axis, x= 10 cm. It begins to move due to the presence of a 2.0-μC charge which remains fixed at the origin. What is the kinetic energy of the particle at the instant it passes the point x= 1.0 m?

12. Two identical particles, each with a mass of 4.5 mg and a charge of 30 nC, are moving directly toward each other with equal speeds of 4.0 m/s at an instant when the distance separating the two is equal to 25 cm. How far apart will they be when closest to one another?

13. A particle (q= +5.0 μC) is released from rest when it is 2.0 m from a charged particle which is held at rest. After the positively charged particle has moved 1.0m toward the fixed particle, it has a kinetic energy of 50 mJ. What is the charge on the fixed particle? Carga dada q = 5 * 10 ^ -6 C Sea Q la carga de la partícula fija. Energía potencial inicial U1 = KqQ / r1 donde K = 9 * 10 ^ 9 Nm ^ 2 / C ^ 2                                            = [(9 * 10 ^ 9 Nm ^ 2 / C ^ 2) * (5 * 10 ^ -6 C) Q] / (2 m)                                            = 22500Q J Energía potencial final U2 = KqQ / r2                                          = [(9 * 10 ^ 9 Nm ^ 2 / C ^ 2) * (5 * 10 ^ -6 C) Q] / (1 m)                                          = 45000Q J Cambio en la energía potencial ΔU = U2 - U1                                                   = 45000Q J - 22500Q J                                                   = 22500Q J Cambio de energía cinética ΔK = 50 mJ                                                = 50 * 10 ^ -3 J De Conservatrion de energía                                        ΔU = - ΔK                                22500Q = - 50 * 10 ^ -3                                           Q = - 2,2 * 10 ^ -6 C                                               = - 2,2 μC La opción a es correcta.