21-1 Static electricity; Electric Charge and its Conservation :

3 21-1 Static electricity; Electric Charge and its Conservation Electrical forces are what makes chemistry happen.
Static electricity
Electrons rub off
Net electric charge
Positive and negative charges
Unlike charges attract and like charges repel

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5 Net Electric Charge

Conservation of Charge :

6 Conservation of Charge Law of conservation of electric charge:
The net electric charge produced in any process is zero.

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21-3 Insulators and Conductors :

8 21-3 Insulators and Conductors Insulators
Conductors
Semiconductors

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9 21-2 Electric Charge on the Atom

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21-4 Induced Charge; the Electroscope :

17 21-4 Induced Charge; the Electroscope Charging by conduction or contact.
Charging by induction.

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21 Coulomb’s Experiment

21-5 Coulomb’s Law :

22 21-5 Coulomb’s Law F = k
In SI units k = 8.98 x 109 N m2/C2.
Thus 1 C is the amount of charge which, if placed on each of two point objects 1.0 m apart, will result is each object exerting a force of:
(9.0 x 109 N m2/C2)(1.0 C)/(1.0 m)2 = 9.0 x 109 N on the other. Q1Q2 r2 . .

Conservative Forces :

23 Conservative Forces A force is conservative if the work it does on a particle moving between any two points is independent of the path taken by the particle.
The work done by a conservative force exerted on a particle moving through any closed path is zero.
Gravitational (Newton’s law of gravity) and Electrical (Coulomb’s law of electrical force) are both conservative forces.

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Coulomb’s Law :

26 Coulomb’s Law The constant k is often written in terms of another constant, e0, called the permittivity of free space.
e0 = = 8.85 x 10-12 C2/N m2 1 4pk

Example 21-1 :

27 Example 21-1 Electric force on electron by proton.
Determine the magnitude of the electric force on the electron of a hydrogen atom exerted by the single proton (Q2 = + e) that is its nucleus. Assume the electron “orbits” the proton at its average distance of r = 0.53 x 10-10 m.

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Principle of Superposition :

29 Principle of Superposition Coulomb force is a vector force.
If several charges are present the net force on any one of them will be the vector sum of the forces due to each of the others.
Fnet = F1+ F2 + F3 + . . .

Conceptual Example 21-2 :

30 Conceptual Example 21-2 Which charge exerts the greater force?
Two positive point charges, Q1 = 50 mC and Q2 = 1 mC, are separated by a distance l. Which is the larger magnitude, the force that Q1 exerts on Q2, or the force that Q2 exerts on Q1?

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Example 21-3 :

32 Example 21-3 Three charges in a line.
Three charges particles are arranged in a line, as shown. Calculate the net electrostatic force on particle three due to the two other charges.

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Example 21-4 :

34 Example 21-4 Electric force using vector components.
Calculate the net electrostatic force on charge Q3 due to charges Q1 and Q2.

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*Vector Form of Coulomb’s Law :

36 *Vector Form of Coulomb’s Law F12 = k r21 r21 Q1 Q2 2

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21-6 The Electric Field :

38 21-6 The Electric Field British scientist Michael Faraday (1791 – 1867) conceived the concept of an electric field. ELECTRIC
E GRAVITATIONAL F q F m g

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Why Calculate Electric Fields? :

40 Why Calculate Electric Fields? The concept of electrical fields was introduced to simplify the calculation for the effect one charge or a collection of charges would have on another charge in its (their) vicinity.
The presence of an already existing charge changes the electrical nature of the space surrounding it.
The force another charge would experience when it is near the already existing charge can be calculated either by Coulomb’s law or from the electric field.
The electric field is a vector quantity.

Electric Fields :

41 Electric Fields F = r2 E F q kQq q kQ = r2 = Coulomb’s Law: kQq Force produced by any charge Q on the positive test charge q. Electric field produced by Q at a point a distance r from Q. r2

Electric Fields :

42 Electric Fields + Positive charge repels (pushes away) the positive test charge. Positive charge creating field at P P Q q (+)

Electric Fields :

43 Electric Fields Negative charge attracts (pulls toward it) the positive test charge. Negative charge creating field at P P Q q (+)

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44 Important

Example 21-6 :

45 Example 21-6 Electric field of a single point charge.
Calculate the magnitude and direction of the electric field a a point P which is 30 cm to the right of a point charge Q = -3.0 x 10-6 C.

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The Superposition Principle :

47 The Superposition Principle The electric field due to a positive charge points away from the charge.
The electric field from a negative charge point toward the charge.
If the field is due to more than one charge, the individual fields due to each charge are added vectorially to get the total field at any point.
E = E1 + E2 +
The validity of this superposition principle for electric fields is fully confirmed by experiment. . . .

Example 21-7 :

48 Example 21-7 E in between two point charges.
Two point charges are separated by a distance of 10.0 cm. One has a charge of -24mC and the other +50 mC. (a) What is the direction and magnitude of the electric field at a point P in between them that is 2.0 cm from the negative charge. (b) If an electron is placed at rest at P, what will be its acceleration?

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Example 21-8 :

51 Example 21-8 E above two point charges.
Calculate the total electric field (a) at point A and (b) at point B due to both charges Q1 and Q2.

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Problem Solving :

53 Problem Solving Electrostatics
Draw a free-body diagram for each object, showing all the forces acting on that object, or the electric field at a point due to all forces.
Apply Coulomb’s law to get the magnitude of each force on a charged object, or the electric field at a point. Deal only with magnitudes of charges (leaving out minus signs), and obtain the magnitude of each force or electric field.

Problem Solving :

54 Problem Solving Show and label each vector force or field on your diagram. Then add vectorially all forces on an object, or the contributing fields at that point, to get the resultant.
Use symmetry (say in the geometry) whenever possible.

21-7 Electric Field Calculations for Continuous Charge Distributions :

55 21-7 Electric Field Calculations for Continuous Charge Distributions In many cases we can treat charge as being continuously distributed.
Divide up a charge distribution into infinitesimal charges dQ, each of which acts as a tiny point charge.
The contribution to the electric field at a distance r from each dQ is
dE = 4pe0 r2 1 dQ E = dE

Example 21-9 :

56 Example 21-9 A ring of charge.
A thin ring-shaped object of radius a holds a total charge Q distributed uniformly around it. Determine the electric field at a point P on the axis, a distance x from the center. Let l be the charge per unit length (C/m).

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Example 21-10 :

58 Example 21-10 Long line of charge.
Determine the magnitude of the electric field at any point P a distance x from a very long line of uniformly distributed charge. Assume x is much smaller than the length of the wire, and let l be the charge per unit length (C/m).

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Example 21-11 :

60 Example 21-11 Uniformly charged disk.
Charge is distributed uniformly over a thin circular disk of radius R. The charge per unit area (C/m2) is s. Calculate the electric field at a point P on the axis of the disk, a distance z above its center.

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Example 21-12 :

62 Example 21-12 Two parallel plates.
Determine the electric field between two large parallel plates, which are very thin and are separated by a distance d which is very small compared to their height and width. One plate carries a uniform surface charge density s and the other carries a uniform surface charge density -s.

Infinite Plane of Charge :

63 Infinite Plane of Charge E = s 2e0

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21-8 Field Lines :

67 21-8 Field Lines Since the electric field is a vector, it is sometimes referred to as a vector field.
Electric field lines.

Electric Field Lines :

68 Electric Field Lines The field lines indicate the direction of the electric field.
The lines are drawn so that the magnitude of the electric field, E, is proportional to the number of lines crossing a unit area perpendicular to the lines. The closer the lines, the greater the field.
Electric field lines start on positive charges and end on negative charges; and the number stating or ending is proportional to the magnitude of the charge.

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21-9 Electric Fields and Conductors :

74 21-9 Electric Fields and Conductors The electric field inside a good conductor is zero in a static situation—that is, when the charges are a rest.
If the electric field was not zero there would be a force on the free electrons. They would redistribute themselves until the net force was zero.
As a result, any net charge on a good conductor distributes itself on the surface.
The electric field is always perpendicular to the surface outside of a conductor. VERY IMPORTANT

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76 An electric field parallel to the surface would accelerate electrons. Therefore, in the static case it must be zero

Conceptual Example 21-13 :

77 Conceptual Example 21-13 Shielding and safety in a storm.
A hollow metal box is placed between two parallel charged plates. What is the field inside the box?

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21-10 Motion of a Charged particle in an Electric Field :

79 21-10 Motion of a Charged particle in an Electric Field If an object having an electric charge q is at a point in space where the electrical force is E, the force on the object is given by
F = qE

Force and Electric Fields :

80 Force and Electric Fields F = qE E = F q

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Example 21-14 :

82 Example 21-14 Electron accelerated by an electric field.
An electron (mass m = 9.1 x 10-31 kg) is accelerated in a uniform electric field E (E = 2.0 X 104 N/C) between two parallel charged plates. The separation of the plates is 1.5 cm. The electron is accelerate from rest near the negative plate and passes through a tiny hole in the positive plate. (a) With what speed does it leave the hole? (b) Show that the gravitational force can be ignored. Assume the hole is so small that it does not affect the uniform field between the plates.

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Example 21-15 :

84 Example 21-15 Electron moving perpendicular to E.
Suppose an electron (say from example 21-14) traveling with a speed v0 = 1.0 x 107 m/s enters a uniform electric field E at right angles to v0 as shown. Describe its motion by giving the equation of its path while in the electric field. Ignore gravity.

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21-11 Electric Dipoles :

88 21-11 Electric Dipoles A combination of two equal charges of opposite sign, + Q and – Q, separated by a distance l, is referred to as an electric dipole.
The quantity Ql is called the dipole moment and is represented by the symbol p.
The dipole moment can be considered to be a vector p, of magnitude Ql, that point from the negative to the positive charge.

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Dipole in an External Field :

90 Dipole in an External Field If a dipole is placed in a uniform electric field there is not net force, but there is a torque
t = p x E.
Work done by the field decreases the potential energy, U, of the dipole in this field.
U = - W = - p E.

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Electric Field Produced by a Dipole :

92 Electric Field Produced by a Dipole An external electric field produces a torque on a dipole.
A dipole itself produces an electric field.

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Example 21-16 :

94 Example 21-16 Dipole in a Field.
The dipole moment of a water molecule is 6.1 x 10-30 C•m. A water molecule is placed in a uniform electric field with a magnitude 2.0 x 105 N/C. (a) What is the magnitude of the maximum torque that the field can exert on the molecule? (b) What is the potential energy when the torque is at its maximum? (c) In what position will the potential energy take on its greatest value? Why is this different than the position where the torque is maximized?

Homework Problem 19 :

95 Homework Problem 19 Two charges, - Q0 and - 3Q0 are a distance l apart. These two charges are free to move but do not because there is a third charge nearby. What must be the charge and the placement of the third charge for the firs two to be in equilibrium?

Homework Problem 35 :

96 Homework Problem 35 An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 145 m/s2. What is the magnitude and direction of the electric field?

Homework Problem 42 :

97 Homework Problem 42 Two parallel circular rings of radius R have their centers on the x axis separated by a distance l. If each ring carries a uniformly distributed charge Q, find the electric field E(x), at points along the x axis.

Homework Problem 56 :

98 Homework Problem 56 A water droplet of radius 0.020 mm remains stationary, in the air. If the electric field of the Earth is 150 C/N downward, how many excess electron charges must the water droplet have?

Homework Problem 57 :

99 Homework Problem 57 At what angle will the electrons in Example 21-15 leave the uniform electric field at the end of the parallel plates? Assume the plates are 6.0 cm long and E = 5.0 x 103 N/C.

Homework Problem 58 :

100 Homework Problem 58 Suppose electrons enter a uniform electric field mid-way between two plates, moving at an upward angle as shown. What maximum speed can the electrons have if they are to avoid striking the upper plate? Ignore fringing of the field?

Homework Problem 60 :

101 Homework Problem 60 A dipole consists of charges + e and - e separated by 0.68 nm. It is in an electric field E = 2.7 x 104 N/C. (a) What is the value of the dipole moment? (b) What is the torque on the dipole when it is perpendicular to the field? (c) What is the torque on the dipole when it is at an angle of 45o to the field? (d) What is the work required to rotate the dipole from being oriented parallel to the field to being antiparallel to the field?

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By: adnankhantomani (23 month(s) ago)

Kindly allow me to download it to add it to the introduction of my course. Thanks aboelaf@hotmail.com