Electricity And Magnetism (part2)

PHYSICS II 1

ELECTRICITY AND MAGNETISM

ELECTROSTATICS Electrostatics: the area of physics that deals with objects that have an electric charge Electric charge: a property of matter that is responsible for all electric and magnetic forces and interactions. Static electricity: electric

a build-up of stationary charge on a substance. 2

PRINCIPAL CONCEPTS IN THE BOHRRUTHERFORD MODEL OF THE ATOM Matter is composed of sub-microscopic particles called atoms. Electric charges are carried by particles within the atom that are called electrons and protons. Protons are found in a small central region of the atom called the nucleus. They are small, heavy particles, and each one carries a positive electric charge of a specific magnitude, called the elementary charge (e). Electrons move around the nucleus. They are small, very light particles (≈ 1/2000 the mass of 3 a proton) yet each carries a negative electric charge equal to the magnitude to that of the proton.

…BOHR-RUTHERFORD MODEL OF THE ATOM

5. Atoms are normally electrically neutral, because the number of (positive) protons is equal to the number of (negative) electrons. 6. Neutrons are small, heavy particles (slightly heavier than the protons) found in the nucleus. They carry no electric charge. 7. If an atom gains an extra electron, it is no longer neutral but has an excess of electrons and a net negative charge. Such an atom is called a negative ion. 8. If an atom loses an electron, it will have a deficit of electrons and a net positive charge. Such an e\atom is called a positive ion. 4

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Masses and Electric Charges of Sub-Atomic Particles

Particle

Mass

Electric Charge

Proton

1.673 x 10-27 kg

Positive

Electron

9.11 x 10-31 kg

Negative

Neutron

1.675 x 10-27 kg

No net charge

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The magnitude of the charge on the proton exactly equals the magnitude of the charge on the electron; the proton carries a charge +e, and the electron carries a charge of –e. In SI units the electric charge is:

e = 1.60 x 10-19 Coulomb The charge on an electron or proton is the smallest amount of free charge. Charges of larger magnitude are built up on an object by adding or removing electrons. Thus, any charge of magnitude q is an integer multiple of e, i.e.,

q = Ne, where N is an integer.

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Example. How many electrons are there in one coulomb of negative charge?

[514p]

REASONING: The negative charge is due to the presence of excess electrons, since they carry negative charge. Because an electron has a charge whose magnitude is e = 1.60 x 10-19 C the number of electrons is equal to the charge q divided by the charge e on each electron. Solution:

The number N of electrons is N = q/e = 1.00 C/1.60 x 10-19 C = 6.25 x 1018

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LAW OF CONSERVATION OF ELECTRIC CHARGE 

During any process, the net electric charge of an isolated system remains constant (is conserved).

When an ebonite rod is rubbed with animal fur, electrons from the fur are transferred to the rod. However, no electrons or protons are destroyed. Since the charges on the electrons and protons have identical magnitudes, but opposite signs, the algebraic sum of the two charges is zero. The transfer does not change the net charge of the fur/rod system. If each material contains an equal number of protons and electrons to begin with, the net charge of the system is zero initially and remains zero at all times during the rubbing process. 9

FUNDAMENTAL LAWS OF ELECTRIC CHARGES Opposite electric charges attract each other. Similar electric charges repel each other.

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Charged objects repel some neutral objects. This occurs because the charged object has the effect of "rearranging" the charges inside the originally neutral object. When two different insulators are rubbed together, electrons can be transferred from one insulator to the other. The body which has gained electrons has a negative charge and the one which lost electrons has a positive charge of equal magnitude. This process is called charging by friction.

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THE ELECTROSTATIC SERIES When charging by friction, the type of charge that develops can be determined using a chart called the electrostatic series. When rubbed together, substances at the top of the series will lose Acetate Weak hold on electrons to substances lower in electrons Glass the series, meaning that the Wool substance higher in the table Cat fur, human hair becomes positively charged while the substance from lower in the table become negatively Calcium, Mg, Pb charged. Silk increasing Aluminum, zinc tendency Cotton to Paraffin wax gain Ebonite Polyethylene (plastic) Carbon, Co., Ni. Rubber Platinum, gold

electrons

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A body can also be charged by simply placing it near a charged body. This process is called charging by induction. This is the process of "rearranging" the charges mentioned above. The diagram below represents the situation when a piece of metal is charged by induction.

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The situation is different when an insulator is charged by induction, as shown by the following diagram.

Electric charges which remain still are called static charges. The only charges which can move freely through metals are negative charges carried by electrons. A flow of electric charges is called an electric current.

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Charging by Contact When a negatively charged ebonite rod is rubbed on a metal object (see figure), some of the excess electrons from the rod are transferred to the object. Once the electrons are on the metal sphere, where they can move freely, they repel one another and spread out over the sphere’s surface. When the rod is removed, the sphere is left with a negative charge distributed over its surface. Likewise, the sphere would be left with a positive charge after being rubbed with a positively charged rod. In this case, electrons from the sphere would be transferred to the rod. Charging by contact:

the process of giving one object a net electric charge by placing it in contact with another object that is already charged.

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Think: Two metal spheres are identical. They are electrically neutral and are touching. An electrically charged ebonite rod is then brought near the spheres without touching them, as the drawing shows. After a while, with the rod held in place, the spheres are separated, and the rod is then removed. Which one or more of the statements is true? (a) mA = mB , (b) mA > mB , if the rod is positive, (c) mA < mB , if the rod is positive, (d) mA > mB , if the rod is negative, (e) mA < mB if the rod is negative. BE 16

CONCEPT REVIEW (1)______________ is that area of Physics that deals with objects that have (2)____________ __________. Electric charge is a property of matter that is responsible for all (3) ________________ and (4) ________________ forces and interactions. Matter is composed of sub-microscopic particles called (5) _________. Electric charges are carried by particles within the atom that are called (6) ____________ and (7) ______________. Protons are found in a small central region of the atom called (8) _______________. They are small, heavy particles, and each carries a positive electric charge of a specific magnitude, called the (9) _____________ ______________. Electrons move around the (10) __________. They are small, very light particles, yet each carries a (11) ____________ electric charge equal in magnitude to that of the proton. Atoms are normally electrically (12) ____________, because the number of protons is equal to the number of electrons. (13) ____________ are small, heavy particles found in the nucleus. They carry no (14) ___________ ____________ If an atom gains an extra electron, it is no longer (15) ____________ but has now a net (16) ____________ ____________. Such an atom is called a (17) ____________ ____________. If an atom loses an electron, it shall have a net (18) __________ ____________. 17 Such an atom is called a (19) ____________ ____________.

Coulomb’s Law The figure shows two charged bodies. These objects are so small, compared to the distance r between them, that they can be regarded as mathematical points. The “point charges” have magnitudes I q1 I and I q2 I. If the charges have unlike signs, each object is attracted to the other by a force that is directed along the line between them; +F is the electric force exerted on object 1 by object 2 and –F is the electric force exerted on object 2 on object 1. If the charges have the same sign, each object is repelled from each other. Whether attractive or repulsive, the two forces are equal in magnitude but opposite in direction. These forces always exist as a pair, each one acting on a different object, in accord with Newton’s action-reaction law. 18

…Coulomb’s Law

F = k _Iq1I Iq2I_ r2 where k is a proportionality constant whose value in SI is k = 8.99 x 109 N-m2 /C2 It is common practice to express k in terms of another constant є0 , by writing k = 1/4‫תּ‬є0

Є0 is called the permittivity of free space and has a value of Є0 = 1/4‫תּ‬k = 8.85 x 10-12 C2 /N-m2

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Example: Two objects, whose charges are +1.0 C and -1.0 C, are separated by 1.0 km. Compared to 1.0 km, the sizes of the objects are small. Find the magnitude of the attractive force that the other charge exerts on the other. REASONING: Considering that the sizes of the objects are small compared to the separation distance, we can treat the charges as point charges. Coulomb’s Law may then be used to find the magnitude of the attractive force, provided that only the magnitudes of the charges are used for the symbols Iq1I and Iq2I that appear in the law. Solution:

The magnitude of the force is

F = k (q1 q2) /r2 = (8.99 x 109 N-m2/C2)( 1.0 C)(1.0C)/ (1000m)2 = 9.0 x 103 N 20

Example In the Bohr model of the atom the electron (-e) is in orbit around the nuclear proton (+e) at a radius of r = 5.29 x 10-11 m, as shown in the drawing. Determine the speed of the electron, assuming the orbit to be circular. Insert: Any object moving with speed v on a circular path of radius r has a centripetal acceleration of ac = v2/r. This acceleration is directed toward the center of the circle. Newton’s second law specifies that the net force ΣF needed to create this acceleration is ΣF = mac = mv2/r, where m is the mass of the object. Solving for speed, v = (ΣFxr/m)1/2 . ΣF Reasoning: Since the mass of the electron is m = 9.11 x 10-31 kg and the radius is given, we can calculate the speed, once ΣF is known. For hydrogen atom, the net force is provided exclusively by the electrostatic force, as given by Coulomb’s law. This force is toward the center of the circle, since the electron and the proton have opposite signs. The electron is also pulled toward the proton by gravitational force. However, the gravitational force is negligible in comparison 21 to the electrostatic force.

Solution: The electron experiences an electrostatic force of attraction because of the proton, and the magnitude of this force is F = k x (q1q2 )/r2 = _(8.99 x 109 N-m2 /C2)(1.60 x 10-19 C)(1.60 x 10-19 C)_ (5.29 x 10-11 m)2 = 8.22 x 10-8 N

1/2

v = (rΣF/m)1/2 = _(8.22 x 10-8 N) (5.29 x 10-11 m)2 _ 9.11 x 10-31 kg

= 2.18 x 106 m/s This orbital speed is almost five million miles per hour!

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THE FORCE ON A POINT CHARGE DUE TO TWO OR MORE OTHER POINT CHARGES. Suppose that a third point charge, q3 , is also present, what would be the net force on q1 due to both q2 and q3 ? First, find the magnitude and direction of the force exerted on q1 by q2 (ignoring q3). Then, determine the force exerted by q3 on q1 (ignoring q2). The net force on q1 is the vector sum of these forces. Example The drawing shows three point charges that lie along the x axis in a vacuum. Determine the magnitude and direction of the net electrostatic force on q1. Reasoning: Part b of the drawing shows a free body diagram of the forces that act on q1. Since q1 and q2 have opposite signs, they attract each other. Thus, the force exerted by q2 on q1 is F12, and points to the left. Also, the force exerted by q3 on q1 is F13 and is also an attractive force The net force is the vector sum of F12 and F13. .

Solution:

The magnitudes of the forces are

F = k (q1 q2)/r12 2 = __(8.99 x 109 N-m2 /C2)(3.0 x 10-6 C)(4.0 x 10-6 C)__ = 2.7 N (0.20 m)2 F = k (q1 q3)/r13 2 = __(8.99 x 109 N-m2 /C2)(3.0 x 10-6 C)(7.0 x 10-6 C)__ = 8.4 N (0.15 m)2 Since F12 points to the negative x direction, and F13 points in the positive x direction, the net force F is

F = F12 + F13 = (-2.7 N) + (8.4 N) = +5.7 N

Answer

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Example: The drawing shows three point charges that lie in the x, y plain in a vacuum. Find the magnitude and direction of the net electrostatic force on q1 ? Reasoning: The force exerted by q2 on q1 is F12 and is an attractive force because the two charges have opposite signs. It points along the line between q1 and q2. The force exerted by q3 on q1 is F13 and is also an attractive force. It points along the line between the two charges. Since the forces point in different directions, we use vector components to find the net force.

Solution…

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…solution F12 = k (q1 q2)/r12 2 = __(8.99 x 109 N-m2 /C2)(4.0 x 10-6 C)(6.0 x 10-6 C)__ = 9.6N (0.15 m)2 F13 = k (q1 q3)/r13 2 = __(8.99 x 109 N-m2 /C2)(4.0 x 10-6 C)(5.0 x 10-6 C)__ = 8.4 N (0.10 m)2 The net force F is the vector sum of F13 and F12 . The components of F that lie in the x and y directions are Fx and Fy , respectively. The forces F12 and F13 are resolved into x and y components. Force

x component

y component

F12

+(9.6) sin 730 = +9.2 N

F13

+(9.6 N) cos 730 = +2.8 N +18 N

F

Fx = 20.8 N

Fy = +9.2 N

0.00 N

F = (20.82 + 9.22 )1/2 = 23 N θ = tan-1 (Fy /Fx) = tan-1 (9.2 N/20.8 N) = 240 Answer

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