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 Department site:

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

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

Sinai University Prof Ahmed El-Lawindy Faculty of dentistry Physics II Code: SGS 122 Lectures time table Ch 1 Electrostatics W1+W2 5/3/ /3/2011 Ch 2 Potential, current and their sources W4 26/3/2011 Quiz 1 (15 min) W5 2/4/2011 Ch 3 Ionic conduction W5- W6 2/4/2011-9/4/2011 Ch 4 Magnetic fields and magnetic forces W7-W8 16/4/ /4/2011 Midterm Exam (30 min) W9 30/4/2011 Ch 5 Magnetic induction W9 30/4/2011 Quiz 2 (15 min) W10 Ch 6 X-rays W10-W11 7/5/ /5/2011 Quiz 3 (15 min) W12+ +W14 21/5/ /6/2011 Ch 7 Radioactivity W12-W13-W14 21/5/ /6/2011 Subject Weeks Date

Chapter 1 Electrostatics

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. 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..

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

Coulomb force

23.3 Coulomb’s Law QQ Q/2 Q/4 Q/8 Q/16  =k    FxR F  

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 = q 1 q 2 /( 4    r 2 ). it is a vector quantity F = q 1 q 2 /( 4    r 2 ). it is a vector quantityvector =K q 1 q 2 / r 2 =K q 1 q 2 / r 2 K=1/4  o = 8.99 x l0 9 N.m 2 /C 2,  o = x C V -1 m -1 is the electric permittivity of space oooo

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

2-Charging by induction

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

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

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

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/t 1 -D/t 2 ) Q=ne= bdD(1/t 1 -1/t 2 )

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 N A m/M, in which N A is 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= N A m = (6.02 x atoms/mol x (3.11 g) M 63.5g/mol M 63.5g/mol = 2.95 x atoms. = 2.95 x atoms. Q= NZe=2.95 x atomsx26x2 atoms -1 x1.6x C = C = C It is a huge quantity of charge, but remember that these charges neutralize each other so that we don’t feel it.

The force of gravity and the Coulomb force Fx = - G m 1 m 2 Rx / r 3 Or F = - G m 1 m 2 / r 2. Fx = - G m 1 m 2 Rx / r 3 Or F = - G m 1 m 2 / r 2. ("Newton's Constant", equal to x N m 2 / kg 2 ) Fx = q1 q2 R x / 4 p e r 3,or, in scalar form, F = q 1 q 2 / 4   r 2.

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 m 2 equal to the earth's mass (m E ) and r to its average radius (r E ). The acceleration g is then G m E / r E 2, or approximately 9.8 m / s 2. m 1 g= G m 1 m E / r E 2 g= G m E / r E 2 =9.8 m / s 2.

Sample Problem 2 The average distance r between the electron and the proton in the hydrogen atom is 5.3x 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, F e = 1 q 1 q 2 F e = 1 q 1 q 2 4  o r 2 4  o r 2 = (8.99 X 109N.m 2 /C 2 )(1.60X l0 -19 C) = (8.99 X 109N.m 2 /C 2 )(1.60X l0 -19 C) (5.3 X m) 2 (5.3 X m) 2 = 8.2 X N. = 8.2 X 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 F g = G m 1 m 2 / r 2 F g = G m 1 m 2 / r 2 = (6.67 x N.m/kg 2 )(9.ll x l0 -31 kg) ( 1.67 X kg) = (6.67 x N.m/kg 2 )(9.ll x l0 -31 kg) ( 1.67 X kg) (5.3 X m) 2 (5.3 X m) 2 = 3.6 X l0 -47 N. = 3.6 X l0 -47 N. F e / F g = 8.2 X /3.6 X l0 -47 =~10 39 F e / F g = 8.2 X /3.6 X l0 -47 =~10 39

Sample Problem 3 The nucleus of an iron atom has a radius of about 4x 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 q p q p F = 1 q p q p 4  o r 2 4  o r 2 = (8.99 x l0 9 N.m 2 /C 2 )(l.60x l0 -19 C) 2 = (8.99 x l0 9 N.m 2 /C 2 )(l.60x l0 -19 C) 2 (4 X l0 -15 ) 2 (4 X l0 -15 ) 2 =14 N. =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.

Problems Solve the following problems Hand it tomorrow