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Sinai University -Faculty of Dentistry

Sinai University -Faculty of Dentistry. Course name: General Physics II, Dentistry. Text references 1- Physics for biology and premedical students,Burns 2- College Physics, Serway 3- Lecture notes 4- Internet sites. ellawindy@su.edu. eg Department site: www.engineering.su.edu.eg

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Sinai University -Faculty of Dentistry

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  1. Sinai University -Faculty of Dentistry

  2. Course name: General Physics II, Dentistry Text references 1- Physics for biology and premedical students,Burns 2- College Physics, Serway 3- Lecture notes 4- Internet sites ellawindy@su.edu.eg Department site: www.engineering.su.edu.eg www.dendistry.su.edu.eg

  3. Assessment %15 15 marks class work 5% 5 marks Pre. term exam 1 10% 10 marks Mid term exam 2 20% 20 marks Lab. Works 50% 50 50 % 50 marks Final yea exam 100% 100 marks total

  4. Electricity Course objectives At the end of this course, students will be able to: 1- Define and apply coulomb force law. 2- Illustrate electric field, electric potential, and Ohm law. 3- Analyze simple electric circuits. 3- Explain the conduction mechanisms. 4- Use the gained information to understand the operational functions of electrical instruments. 5- To link the course to future applications

  5. Sinai University Prof Ahmed El-Lawindy Faculty of dentistry Physics II Code: SGS 122Lectures time table Subject Weeks Date Ch 1 Electrostatics W1+W2 5/3/2011-12/3/2011 Ch 2 Potential, current and their sourcesW3 19/3/2011 Quiz 1 (15 min) W4 26/3/2011 Ch 3 Ionic conduction W4-W5 26/3/2011-2/4/2011 Ch 4 Magnetic fields and magnetic forces W6-W7 9/4/2011- 16/4/2011 Midterm Exam (30 min) W8 23/4/2011 Ch 5 Magnetic induction W8 23/4/2011 Quiz 2 (15 min) W9 Ch 6 X-rays W9-W10 30/4/2011 -7/5/2011 Quiz 3 (15 min) W11+ +W13 14/5/2011 -28/5/2011 Ch 7 Radioactivity W11-W12-W13 14/5/2011 -28/5/2011

  6. Chapter 1 Electrostatics

  7. Introduction • Up to now, we have for the most part dealt with topics which are macroscopic, • Force, pressure, work etc… • For the next few chapters, we will be concerned with the microscopic world, which is largely hidden from our senses and our common sense. • Gravity, electric fields, magnetic fields, etc..

  8. Introduction The basic problem which electricity presents to our intuition is: how can the electric force act at a distance, without direct contact? When you get closer to a charged sphere, your hair will rise, defying gravity. • What actually happens is • a separation of “electric charges" from the boy's hair such that the charges on hair nearer to the charged sphere are of opposite type of the charge on it. This is an attractive force between what is called positive and negative charges.

  9. Coulomb force

  10. Coulomb force

  11. Q Q Q/2 Q/2 Q/2 Q/4 Q/4 Q/4 Q/8 Q/8 Q/8 Q/16 Q/16 23.3 Coulomb’s Law t=kq =FxR F aq

  12. Coulomb force Law • At the heart of this simple phenomenon is one of the most universally applicable ideas in physics: action at a distance is caused by charges, which are the sources of forces. • or, in scalar form, F = q1 q2 /( 4 pe0 r 2). it is a vector quantity =K q1 q2 / r 2 K=1/4peo= 8.99 x l09 N.m2/C2, eo= 8.85419x10-12 C V-1 m-1 is the electric permittivity of space eo

  13. ++++++++++++ ++++++++++++ + Electric charges detection Neutral bodies are charged by different ways. The most common ways are charging by 1- conduction, 2- induction 3- rubbing 1-Charging by conduction

  14. 2-Charging by induction

  15. 3-Charging by rubbing • # Glass rubbed with silk +ve • # Hard rubber rubbed with fur –ve

  16. Charge is Quantized • Electric charge was thought to be a continuous fluid q = ne n=0,+1,+2,+ 3, ….. Millikan’s Experiment

  17. Milliken's apparatus V drift velocity, constant velocity, D, is the distance traveled by an electron in a time t

  18. Derivation • Step 1 Free Fall Mg= bv • Step 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)

  19. Sample Problem 1 • A penny, being electrically neutral, contains equal amounts of positive and negative charge. What is the magnitude of these equal charges? • Solution • The charge q is given by NZe,in which N is the number of atoms in a penny and Ze is the magnitude of the positive and the negative charges carried by each atom. • The number N of atoms in a penny, assumed for simplicity to be made of copper, is NAm/M, in which NAis the Avogadro constant. The mass m of the coin is 3.11 g, and the mass M of 1 mol of copper (called its molar mass) is 63.5 g. We find • N= NAm = (6.02 x 1023 atoms/mol x (3.11 g) • M 63.5g/mol • = 2.95 x 1022atoms. • Q= NZe=2.95 x 1022 atomsx26x2 atoms-1x1.6x10-19 C • = 245440 C • It is a huge quantity of charge, but remember that these charges neutralize each other so that we don’t feel it.

  20. The force of gravity and the Coulomb force Fx = - G m1 m2 Rx / r3 Or F = - G m1 m2 / r2. ("Newton's Constant", equal to 6.673 x 10 - 11 N m2 / kg2) Fx = q1 q2 R x / 4 p e r 3,or, in scalar form, F = q1 q2 / 4 pe r 2.

  21. The force of gravity and the 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 / rE2, or approximately 9.8 m / s2. m1g= G m1mE / rE2 g= G mE / rE2 =9.8 m / s2.

  22. Sample Problem 2 • The average distance r between the electron and the proton in the hydrogen atom is 5.3x10-11 m. (a) What is the magnitude of the average electrostatic force that acts between these two particles? (b) What is the magnitude of the average gravitational force that acts between these particles? • Solution(a) From Eq. (2-1) we have, for the electrostatic force, • Fe= 1 q1 q2 • 4peo r2 • = (8.99 X 109N.m2/C2)(1.60X l0-19C) • (5.3 X 10-11 m)2 • = 8.2 X 10-8 N. • 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. • (b) For the gravitational force, Eq. (3-1), we have • Fg= G m1m2/ r2 • = (6.67 x 10-11 N.m/kg2)(9.ll x l0-31 kg) ( 1.67 X 10-27kg) • (5.3 X 10-11 m)2 • = 3.6 X l0-47 N. • Fe/ Fg= 8.2 X 10-8 /3.6 X l0-47=~1039

  23. Sample Problem 3 • The nucleus of an iron atom has a radius of about 4x10-15 m and contains 26 protons. What repulsive electrostatic force acts between two protons in such a nucleus if a distance of one radius separates them? • Solution From Eq. (2-1), we have • F = 1 qp qp • 4peo r2 • = (8.99 x l09N.m2/C2)(l.60x l0-19 C)2 • (4 X l0-15)2 • =14 N. • This enormous force, more than 1 kg and acting on a single proton, 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 cannot be felt very far outside the nucleus, is known as the "strong nuclear force" and is very well named.

  24. Problems Solve the following problems 1-3-5 Hand it tomorrow

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