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Part 1: Electric Force
1.1: Review of Vectors
Review your vectors! You should know how to
· convert from polar form to component form and vice versa
· add and subtract vectors
· multiply vectors by scalars
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Example:
Find the resultant vector R = r2 – 3r1 where r1 = 10 m, 45° and r2 = 20 m, 270°.
Ans. 46.3 m, 243°
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1.2 Charge
Symbol q, Q
mks units [Coulomb = C]
Fundamental charge e = 1.6x10-19 C
Charge of proton qp = +e
Charge of electron qe = -e
An object is neutral if it has the same number of electrons and protons.
An object has a net positive charge if it has more protons than electrons.
An object has a net negative charge if it has more electrons than protons.
Objects usually are charged up through the transfer of electrons from one object to the other.
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Example:
An object has a net charge of -8 nC. Does it have more protons or electrons? How many more?
Ans. The object has 50 billion more electrons than protons.
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1.3 Electric Force
Coulomb’s Law describes the electric force between charged objects.
Size:
Coulomb’s constant k = 9x109 N·m2/C2
Direction:
Opposite charges attract along line r.
Like charges repel along line r.
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Example:
Consider the ground state of a H-1 hydrogen atom that has one electron orbiting a single proton in the lowest orbit allowed for the electron. Compare the size of the electric force between the electron and proton to that of the gravitational force. You will need the following information.
mass of electron 9.11x10-31 kg mass of proton 1.67x10-27 kg
distance between electron and proton in ground state 0.05 nm
Are we justified in saying that it is the electric force that holds atoms together and that the gravitational force is negligible in comparison?
Ans. electric force ~ 9x10-8 N, gravitational force ~4x10-47 N. The electric force is larger by 39 orders of magnitude. We can definitely ignore gravity.
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Example:
The following three charges are held fixed on an x-y grid.
q1 = -2 mC at (0,0) q2 = +3 mC at (2 m,0) q3 = -4 mC at (1 m, -2 m)
Find the net force exerted by charges 2 and 3 on charge 1.
Ans. 0.0147 N, 61.3°
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1.4 Electric Field
Symbol E
mks units [N/C]
A charged object with charge qo produces an electric field vector at every point in space except at its position. This object exerts an electrical force on another charged object with charge q. This force is given by
where is the electric field of qo at the position of q.
If q is +, the force points in same direction as the field vector.
If q is -, the force points in the opposite direction as the field vector.
Point Charges
A charged object that can be approximated as a point charge produces an electric field at point P.
Size of field:
Direction of field:
If q is +, field vector points away from charge along line r.
If q is -, field vector points towards charge along line r.
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Example:
The following three charges are held fixed on an x-y grid.
q1 = -2 mC at (0,0) q2 = +3 mC at (2 m,0) q3 = -4 mC at (1 m, -2 m)
Note that this is the same charge distribution used in the previous example in the force section.
(a) Find the net electric field at the origin due to these point charges.
(b) Find force exerted by charges 2 and 3 on charge 1 by using and show that this answer is identical to the one you found in the previous example.
Ans. (a) 7344 N/C, 241.3° (b) 0.0147 N, 61.3
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Continuous Charge Distributions
If point P is near a large charged object, then to find the electric field at P you chop up the object into many point charges. You find the electric field due to each point charge and sum up all these field vectors.
Example 1: Field on the central axis of a uniformly charged rod of length L and charge Q
If Q is +, field vector points away from rod along axis.
If Q is -, field vector points towards rod along axis.
Example 2: Field on the central axis of a uniformly charged loop of radius a and charge Q
If Q is +, field vector points away from loop along axis.
If Q is -, field vector points towards loop along axis.
Gauss’s Law
Gauss’s Law is another way to find the electric field near a large charged object.
Results of the application of Gauss’s Law follow.
INSULATORS
CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM
1. Electric field inside the conductor is zero.
2. Net charge inside the conductor is zero.
3. Any net charge resides on the exterior surface(s) of the conductor.
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Example:
A metal spherical shell with a net charge of +90 nC and inner and outer radii of 4 cm and 5 cm is centered around a solid metal ball with a net charge of -50 nC and a radius of 2 cm.
(a) Find the size and direction of the electric field at the following distances from the center of the solid ball.
(i) 1.5 cm (ii) 3 cm (iii) 4.2 cm (iv) 6 cm
(b) Find the amount of charge on the following surfaces.
(i) outside surface of ball (ii) inside surface of shell (iii) outside surface of shell
Ans. (a) (i) 0 (ii) 500,000 N/C, radially inward (iii) 0 (iv) 100,000 N/C, radially outward (b) (i) -50 nC (ii) +50 nC (iii) +40 nC
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Motion of Charged Particles in E-Fields
A charged particle of mass m moving in a region of electric field experiences an acceleration with a size of
If the electric field is uniform in this region then the acceleration is constant and the equations of motion for constant acceleration can be used to describe the particle’s motion.
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Example:
An electron, initially at rest, is accelerated horizontally from left to right by a uniform electric field of 2500 N/C between two metal plates that are separated by 2 cm. After leaving these plates, the electron then enters a uniform field of 1000 N/C oriented vertically with the electric field pointing upwards.
(a) Find the speed of the electron as it enters the second field region.
(b) Find the position of the electron 2 ns after it enters the second field region.
Ans. (a) 4.19x106 m/s (b) 8.4 mm to the right and 0.35 mm down
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