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1. B
11. C
21. B
2.D
12.A
22.C
27. F = 480 N
28.
3.C 4.D
13.C 14.D
23.D 24.D
5.B 6.C
15.D 16.E
25.A
7.D 8.B 9.A 10.B
17.D 18.A 19.E 20.A
26.
F = -45 N (attractive)
F
1
= 9x10 9 N F
2
= 9x10 9 N
F = 12.7 x 10 9 N θ = 45 0
The resultant force is 1.3 x 10 10 N in a direction of 45 0 S of W.
29. F
1
= 9x10 9 N F
2
= 9x10 9 N
F = 12.7 x 10 9 N θ = 45 0
The resultant force is 1.3 x 10 10 N vertical downward
30. C 31.A 32.A 33.B
34. E = 4.0 N/C (in the direction of the force).
35. E= 4.0 x 10 3 N/C (radially outwards).
36. E = 10.4 x 10

= 44 0
2 N/C
E = 10.4 x 10 2 N/C at 1 0 below a horizontal line drawn through X.
37. Its mass decreases. This is because a positive charge is generally due to a loss of electrons, each of which has a small mass.
38. Its mass increases. This is because a negative charge is generally due to a gain of electrons, each of which has a small mass.
39. The charged comb causes the paper to become polarized, with the side nearest to the comb acquiring a charge opposite to the charge of the comb. The result is an attractive interaction between the comb and the paper.
40. the paper gets the same charge so there is repulsive force now between balloon and paper bits.
41. Loosely held electrons are transferred from the carpet to the socks when the socks are rubbed against the carpet.
At that point, the body and socks have an excess of electrons and are negatively charged. Touching the doorknob allows the electrons to escape. The shock felt is the sudden movement of charges as the body and socks return to a neutral state.
42. Yes. If the suspended object were neutral it would be attracted to the charged rod by polarization effects.
The fact that the suspended object is repelled indicates that it has a charge of the same sign of that of the rod.

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43. No. It can be neutral and then polarized and consequently attracted to the rod.
44. By Newton’s third law, the forces exerted on the electron and the nucleus are equal in magnitude and opposite in direction.
45. the force is the same; electron is faster – it has greater acceleration (smaller mass) so (c)
46. They’ll repel each other, and eventually start moving in opposite direction.
47. Because of very strong repulsion between outer electrons in our body’s molecules and floors’ molecules.
48. The charge in the middle experiences the greatest force. Next in magnitude is the net force exerted on the charge at the left end, and the smallest net force is exerted on the charge to the right.
49. (a) Their shoes should be conduction so any charge they transfer during walking can travel into the ground.
50. Any material through which charge can move easily is called conductor.
In conductors many of electrons are bound very loosely to the nuclei and can move about freely within the material (although they cannot leave the material easily). They are often called free electrons or conduction electrons.
Metals are generally good conductors.
51. Any material that impedes the flow of charge is called insulator.
Electrons in insulators are tightly bound to atomic nuclei and so cannot easily be made to drift from one atom to the next.
Most materials other then metals are insulators.
52. A good electrical conductor is any material composed of atoms with "loose" outer electrons that are free to move about in the material, such as copper. A good electrical insulator is a material composed of atoms that tightly
bind electrons. Wood and distilled water are examples of good electrical insulators.
53. 10 N/C
54. 400 N
55. 30 N/C
56. 1.6 x 10 -15 C
57. 4.8 x 10 -17 N
58. D 59.A 60.B 61.C,D,A 62.D 63.C 64.E
65. E 66.E 67.A 68.C

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1. C Highly important example A uniform electric field with a magnitude of 500 N/C is directed parallel to the positive x-axis. If the potential at x = 5 m is 2500 V, what is the potential at x = 2 m?
Steps how to solve this problem are: a. Drawing a sketch will help a lot: b. Imagine a positive charge q at position x = 5m. c. Force on q due to electric field E is F = qE to the right. d. To move charge from x = 5m position to x = 2m position, one has
to exert applied force in the direction against electric field. e. Work done by applied force against electric field will increase
electric potential energy of the charge q in electric field E:
W = F app
d cos θ = (cos θ = 1) = qEd = ∆ U f. Use definition for change in potential
∆ 𝑉 =
∆𝑈 𝑞
=
𝑊 𝑞
= E d = 1500 V g. Potential energy increases to the left (work by applied force)
so does potential:
V
2m
= V
5m
+ ∆V = 2500 V + 1500 V = 4000 V h. If there was a positive charge of 2C at position x = 5m, the work W = qEd done on that charge by applied force qE is converted into increase of potential energy: ∆U = W = qEd = 3000 J i. ◊ relationship between uniform electric field E and potential difference between two points distance d from each other along electric field line:
ΔV = Ed → E = ΔV/d and N/C = V/m
2.A 3.B 4.C
14. E 15. A 16. A 17. A 18. C 19. B 20. B
21.A22.D 23.D 24.C
5.B 6.C 7.B 8.C 9.C 10.B 11.A 12.A 13.B
25.D 26.A 27. B 28. D 29. E

1. C 2. A 5. A 8. A 12. A 15. A 16. B
17. A
two forces are acting on proton. Gravitational downward mg. In order for the proton to be in equilibrium electrical force must act upward. So electric field is upward. Magnitudes must be the same:
grav. force = electric force mg=eE
E=mg/e = (1.67x10
-27 )x(9.81)/(1.6x10
-19 )=10.2x10
-8 V/m = 102 nanovolts/m upward
18. E potential energy difference = kinetic energy
qV = KE
alpha particle: KE = (2e)(2500 V)= 5000 eV
electron: KE = (1e)(2500 V)= 2500 eV
19. A Rememberthat electric field in capacitor with parallel plates is constant. So if we know electric force acting on electron we can find electric field:
F = eE E = F/e = 6 × 10 −13 /1.6x10
-19 = 3.75x10
6 V/m
You see from the units that formula for potential difference is:
∆ 𝑉 = 𝐸𝑑 ∆ 𝑉 = (3.75x10
6 V/m)(1.2x10
-3 m)= 4.5x10
3 V
20. C potential energy difference = kinetic energy
qV = KE
alpha particle: KE = q ∆ V = 2e ∆ V
proton: KE = q ∆ V = e ∆ V
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