# Problemas de tippens resueltos ley de columb

Solo disponible en BuenasTareas
• Páginas : 12 (2848 palabras )
• Descarga(s) : 0
• Publicado : 1 de marzo de 2012

Vista previa del texto
Chapter 23. The Electric Force

Coulomb's Law
23-1. Two balls each having a charge of 3 C are separated by 20 mm. What is the force of repulsion between them?
; F = 202 N

23-2. Two point charges of -3 and +4 C are 12 mm apart in a vacuum. What is the electrostatic force between them?
; F = 750 N, attraction

23-3. An alpha particle consists of two protons (qe = 1.6 x 10-19 C)and two neutrons (no charge). What is the repulsive force between two alpha particles separated by 2 nm?
q = 2(1.6 x 10-19 C) = 3.2 x 10-19 C
; F = 2.30 x 10-10 N
23-4. Assume that the radius of the electron's orbit around the proton in a hydrogen atom is approximately 5.2 x 10-11 m. What is the electrostatic force of attraction?
; F = 8.52 x 10-8 N
23-5. What is the separationof two -4 C charges if the force of repulsion between them is 200 N?
; r = 26.8 mm
23-6. Two identical charges separated by 30 mm experience a repulsive force of 980 N. What is the magnitude of each charge?
; q = 9.90 C

*23-7. A 10 C charge and a -6 C charge are separated by 40 mm. What is the force between them. The spheres are placed in contact for a few moments andthen separated again by 40 mm. What is the new force? Is it attractive or repulsive?
2 C
2 C
-6 C
10 C
0.08 m
2 C
2 C
0.08 m
2 C
2 C
-6 C
10 C
0.08 m
2 C
2 C
0.08 m
; F = 338 N, attraction
When spheres touch, 6 C of charge are neutralized,
leaving 4 C to be shared by two spheres, or
+2 C on each sphere. Now they are again separated.
;F = 5.62 N, repulsion

*23-8. Two point charges initially attract each other with a force of 600 N. If their separation is reduced to one-third of its original distance, what is the new force of attraction?
; r1 = 3 r2
F2 = 5400 N

The Resultant Electrostatic Force
F1
F2
q3
q2
q1
-35 C
+20 C
+60 C
60 mm
F1
F2
q3
q2
q1
-35 C
+20 C
+60 C
60 mm23-9. A +60 C charge is placed 60 mm to the left of a +20 C charge. What is the resultant force on a -35 C charge placed midway between the two charges?

F13 = 2.10 x 104 N, directed to the left
; F13 = 2.10 x 104 N, directed to right.
FR = F13 + F23 = (-2.10 x 104 N) + (0.700 x 104 N); FR = -1.40 x 104 N, left.

F1
F2
q3
q2
q1
10 C
-22 C
+36 C
80 mm
F1
F2
q3q2
q1
10 C
-22 C
+36 C
80 mm
23-10. A point charge of +36 C is placed 80 mm to the left of a second point charge of -22 C. What force is exerted on third charge of +10 C located at the midpoint?

F13 = 2025 N, directed to the right
; F13 = 1238 N, directed to right.
FR = F13 + F23 = 2025 N + 1238 N; FR = 3260 N, left.
q3 = 12 C
-22 C
60 mm
F1
F2
q2
q1
+36 C80 mm
q3 = 12 C
-22 C
60 mm
F1
F2
q2
q1
+36 C
80 mm
23-11. For Problem 23-10, what is the resultant force on a third charge of +12 C placed between the other charges and located 60 mm from the +36 C charge?

;
Both to right, so FR = F13 + F23 = 1080 N + 5940 N; F = 7020 N, rightward.
q3 = -2 C
6 C
20 mm
F1
F2
q2
q1
-8 C
44 mm
24 mm
q3 = -2 C
6 C20 mm
F1
F2
q2
q1
-8 C
44 mm
24 mm
23-12. A +6 C charge is 44 mm to the right of a -8 C charge. What is the resultant force on a -2 C charge that is 20 mm to the right of the -8 C charge?

;
Both to right, so FR = F13 + F23 = 360 N + 187.5 N; F = 548 N, rightward

s
50 mm
q3 = -12 C
16 C
30 mm
F1
F2
q2
q1
+64 C

s
50 mmq3 = -12 C
16 C
30 mm
F1
F2
q2
q1
+64 C
*23-13. A 64-C charge is locate 30 cm to the left of a 16-C charge. What is the resultant force on a -12 C charge positioned exactly 50 mm below the 16 C charge?

F13 = 2033 N, 59.00 N of W
= 691 N, upward.
Fx = 0 – F13 cos 59.00 = -(2033 N) cos 590 ; Fx = -1047 N
Fy = F23 +...