Sinai University -Faculty of Dentistry Sinai University -Faculty of Dentistry

Course name: General Physics Course name: General Physics II, DentistryII, Dentistry

Text referencesText references

1- Physics for biology and premedical students,Burns 1- Physics for biology and premedical students,Burns 2- College Physics, Serway2- College Physics, Serway3- Lecture notes3- Lecture notes4- Internet sites4- Internet sites

ellawindy@su.edu.egegDepartment site: Department site:

www.engineering.su.edu.egwww.dendistry.su.edu.eg

AssessmentAssessment

%%15 15 marks class work15 15 marks class work 5% 5 marks Pre. term exam 15% 5 marks Pre. term exam 1 10% 10 marks Mid term exam 2 10% 10 marks Mid term exam 2 20% 20 marks Lab. Works20% 20 marks Lab. Works 50% 5050% 5050 % 50 marks Final yea exam50 % 50 marks Final yea exam

100% 100 marks total100% 100 marks total

Electricity Course objectivesElectricity Course objectivesAt the end of this course, students will be able to:At the end of this course, students will be able to:

1- Define and apply coulomb force law. 1- Define and apply coulomb force law.

2- Illustrate electric field, electric potential, and Ohm 2- Illustrate electric field, electric potential, and Ohm law.law.

3- Analyze simple electric circuits. 3- Analyze simple electric circuits.

3- Explain the conduction mechanisms.3- Explain the conduction mechanisms.

4- Use the gained information to understand the 4- Use the gained information to understand the operational functions of electrical instruments.operational functions of electrical instruments.

5- To link the course to future applications5- To link the course to future applications

Sinai University Prof Ahmed El-Lawindy Sinai University Prof Ahmed El-Lawindy

Faculty of dentistry Physics II Code: SGS 122Faculty of dentistry Physics II Code: SGS 122 Lectures time tableLectures time table

Ch 1 Electrostatics W1+W2 5/3/2011-12/3/2011 Ch 2 Ch 1 Electrostatics W1+W2 5/3/2011-12/3/2011 Ch 2

Potential, current and their sourcesPotential, current and their sources W3 19/3/2011W3 19/3/2011

Quiz 1 (15 min)Quiz 1 (15 min) W4 26/3/2011 W4 26/3/2011

Ch 3 Ionic conductionCh 3 Ionic conduction W4- W4- W5 W5 26/3/2011-2/4/201126/3/2011-2/4/2011

Ch 4 Magnetic fields and magnetic forces W6-W7 9/4/2011- 16/4/2011Ch 4 Magnetic fields and magnetic forces W6-W7 9/4/2011- 16/4/2011

Midterm Exam (30 min) W8 23/4/2011Midterm Exam (30 min) W8 23/4/2011

Ch 5 Magnetic induction W8 23/4/2011Ch 5 Magnetic induction W8 23/4/2011

Quiz 2 (15 min) W9Quiz 2 (15 min) W9

Ch 6 X-rays W9-W10 30/4/2011 -7/5/2011Ch 6 X-rays W9-W10 30/4/2011 -7/5/2011

Quiz 3 (15 min) W11+ +W13 14/5/2011 -28/5/2011Quiz 3 (15 min) W11+ +W13 14/5/2011 -28/5/2011

Ch 7 Radioactivity W11-W12-W13 14/5/2011 -28/5/2011Ch 7 Radioactivity W11-W12-W13 14/5/2011 -28/5/2011

Subject Weeks DateSubject Weeks Date

Chapter 1Chapter 1

ElectrostatiElectrostaticscs

IntroductionIntroduction

Up to now, we have for the most part dealt Up to now, we have for the most part dealt with topics which are with topics which are macroscopic,macroscopic,

Force, pressure, work etc…Force, pressure, work etc…

For the next few chapters, we will be For the next few chapters, we will be concerned with the concerned with the microscopicmicroscopic world, world, which is largely hidden from our senses which is largely hidden from our senses and our common sense.and our common sense.

Gravity, electric fields, magnetic fields, etc..Gravity, electric fields, magnetic fields, etc..

IntroductionIntroduction

What actually happens isWhat actually happens isa separation of “electric a separation of “electric charges" from the boy's charges" from the boy's hair such that the hair such that the charges on hair nearer to charges on hair nearer to the charged sphere are the charged sphere are of opposite type of the of opposite type of the charge on it. This is an charge on it. This is an attractive force between attractive force between what is called positive what is called positive and negative chargesand negative charges..

The basic problem which electricity presents to our The basic problem which electricity presents to our intuition is: how can the electric force act at a distance, intuition is: how can the electric force act at a distance, without direct contact?without direct contact?

When you get closer to a charged sphere, your hair will When you get closer to a charged sphere, your hair will rise, defying gravityrise, defying gravity. .

Coulomb forceCoulomb force

Coulomb forceCoulomb force

23.3 Coulomb’s Law23.3 Coulomb’s Law

Q

Q Q/2 Q/2

Q/2 Q/4 Q/4

Q/4 Q/8 Q/8

Q/8 Q/16 Q/16

=k

FxR

F

Coulomb force LawCoulomb force LawAt the heart of this simple phenomenon is one of the most At the heart of this simple phenomenon is one of the most universally applicable ideas in physics: universally applicable ideas in physics: action at a distance action at a distance is caused by charges, is caused by charges, which are the sourceswhich are the sources of of forcesforces..

or, in scalar form, or, in scalar form,

F = qF = q11 q q22 /( 4 /( 4 r r 22). ). it is a it is a vector quantity quantity =K q=K q11 q q22 / r / r 22

K=1/4K=1/4oo= = 8.99 x l08.99 x l099 N.m N.m22/C/C22, , oo= 8.85419x10= 8.85419x10-12-12 C V C V-1-1 m m-1-1

is the electric permittivity of spaceis the electric permittivity of spaceoo

Electric charges detectionElectric charges detection

++++++++++++++++++++++++

+

Neutral bodies are charged by different ways. The most common ways are charging by 1- conduction, 2- induction

3- rubbing

1-1-Charging by conductionCharging by conduction

2-Charging by induction2-Charging by induction

# Glass rubbed with silk +ve

# Hard rubber rubbed with fur –ve

3-Charging by rubbing3-Charging by rubbing

Charge is QuantizedCharge is Quantized

Electric charge was thought to be a Electric charge was thought to be a continuous fluid continuous fluid

q = ne n=0,+1,+2,+ 3, …..

Millikan’s ExperimentMillikan’s Experiment

Milliken's apparatusMilliken's apparatus

V drift velocity, constant velocity, D, is the distance traveled by an electron in a time t

DerivationDerivation

Step 1 Free Fall Mg= bvStep 2 Motion in an Electric Field, E QE-Mg=- bu

Then: QE-bv=bu

QE=bu-bv=b(u-v) QV/d=b(D/t1-D/t2)

Q=ne= bdD(1/t1-1/t2)

Sample Problem 1Sample Problem 1A penny, being electrically neutral, contains equal A penny, being electrically neutral, contains equal amounts of positive and negative charge. What is the amounts of positive and negative charge. What is the magnitude of these equal charges?magnitude of these equal charges?

Solution Solution The charge The charge q q is given by is given by NZe,NZe, in which in which N N is the number of atoms in a penny and is the number of atoms in a penny and ZeZe is is the magnitude of the positive and the negative charges carried by each atom.the magnitude of the positive and the negative charges carried by each atom.The number The number NN of atoms in a penny, assumed for simplicity to be made of copper, is of atoms in a penny, assumed for simplicity to be made of copper, is NNAAm/M, m/M, in which in which NNAA is the is the AvogadroAvogadro constant. The mass constant. The mass m m of the coin is 3.11 g, and the of the coin is 3.11 g, and the mass mass MM of 1 mol of copper (called its of 1 mol of copper (called its molar mass) molar mass) is 63.5 g. We findis 63.5 g. We find

N= N= N NAAmm = = (6.02 x 10(6.02 x 102323 atoms/mol x (3.11 g atoms/mol x (3.11 g)) M M 63.5g/mol63.5g/mol = 2.95 x 10= 2.95 x 102222 atoms.atoms.Q=Q= NZe= NZe=2.95 x 102.95 x 102222 atomsx26x2 atoms atomsx26x2 atoms-1-1x1.6x10x1.6x10-19-19 C C = 245440 C= 245440 CIt is a huge quantity of charge, but remember that these charges neutralize each other It is a huge quantity of charge, but remember that these charges neutralize each other so that we don’t feel it.so that we don’t feel it.

The force of gravity and The force of gravity and the Coulomb forcethe Coulomb force

Fx = - G mFx = - G m11 m m22 Rx / r Rx / r33 Or F = - G m Or F = - G m11 m m22 / r / r22..

("Newton's Constant", equal to 6.673 x 10 ("Newton's Constant", equal to 6.673 x 10 - 11- 11 N m N m22 / kg / kg22))

Fx = q1 q2 R x / 4 p e r 3,or, in scalar form, F = qFx = q1 q2 R x / 4 p e r 3,or, in scalar form, F = q11 q q22 / 4 / 4 r r 22..

The force of gravity and The force of gravity and the Coulomb forcethe Coulomb force

We can now see how to compute the value of g, the "constant" acceleration due to gravity.

The force due to a spherical source is equivalent to a point source at its center. Approximating the earth by a sphere, we set m2 equal to the earth's mass (mE) and r to its average radius (rE). The acceleration g is then G mE / rE

2, or approximately 9.8 m / s2.

m1g= G m1mE / rE2

g= G mE / rE2

=9.8 m / s2.

Sample Problem 2Sample Problem 2The average distance The average distance r r between the electron and the proton in the between the electron and the proton in the hydrogen atom is 5.3x10hydrogen atom is 5.3x10-11-11 m. m. (a) (a) What is the magnitude of the average What is the magnitude of the average electrostatic force that acts between these two particles? electrostatic force that acts between these two particles? (b) (b) What is the What is the magnitude of the average gravitational force that acts between these magnitude of the average gravitational force that acts between these particles?particles?

SolutionSolution (a) (a) From Eq. (2-1) we have, for the electrostatic force, From Eq. (2-1) we have, for the electrostatic force,

FFee = = 1 1 qq11 q q22 44o o rr2 2 = (= (8.99 X 109N.m8.99 X 109N.m22/C/C22)(1.60X l0)(1.60X l0-19-19C)C) (5.3 X 10(5.3 X 10-11 -11 m)m)22

= 8.2 X 10= 8.2 X 10-8-8 N. N.While this force may seem small (it is about equal to the weight of a speck of dust), it produces an immense effect, While this force may seem small (it is about equal to the weight of a speck of dust), it produces an immense effect, namely, the acceleration of the electron within the atom.namely, the acceleration of the electron within the atom.(b) (b) For the gravitational force, Eq. (3-1), we haveFor the gravitational force, Eq. (3-1), we have

FFgg= = G mG m11mm22/ r/ r22

= (= (6.67 x 106.67 x 10-11-11 N.m/kg N.m/kg22)(9.ll x l0)(9.ll x l0-31-31 kg) ( 1.67 X 10 kg) ( 1.67 X 10-27-27kgkg) ) (5.3 X 10(5.3 X 10-11-11 m) m)22

= 3.6 X l0= 3.6 X l0-47-47 N. N. FFee/ F/ Fgg= = 8.2 X 108.2 X 10-8-8 /3.6 X l0 /3.6 X l0-47-47=~10=~103939

Sample Problem 3Sample Problem 3The nucleus of an iron atom has a radius of about 4x10The nucleus of an iron atom has a radius of about 4x10-15 -15 m and contains 26 m and contains 26 protons. What repulsive electrostatic force acts between two protons in such a protons. What repulsive electrostatic force acts between two protons in such a nucleus if a distance of one radius separates them?nucleus if a distance of one radius separates them?SolutionSolution From Eq. (2-1), we have From Eq. (2-1), we have

F = F = 1 1 q qpp q qpp 44oo r r22

= (= (8.99 x l08.99 x l099N.mN.m22/C/C22)(l.60x l0)(l.60x l0-19-19 C) C)22 (4 X l0(4 X l0-15-15))22

=14 N.=14 N.This enormous force, more than This enormous force, more than 1 kg1 kg and acting on a single proton, and acting on a single proton, must be more than balanced by the attractive nuclear force that binds must be more than balanced by the attractive nuclear force that binds the nucleus together. This force, whose range is so short that its effects the nucleus together. This force, whose range is so short that its effects cannot be felt very far outside the nucleus, is known as the "cannot be felt very far outside the nucleus, is known as the "strong strong nuclear forcenuclear force"" and is very well named. and is very well named.

ProblemsProblems

Solve the following problemsSolve the following problems

1-3-51-3-5

Hand it tomorrowHand it tomorrow