Norah Ali Al-moneef king Saud unversity Chapter 23 electric field 23.1 Properties of Electric Charges 23.2 Charging Objects By Induction 23.3 Coulomb’s Law 23.4 The Electric Field 23.6 Electric Field Lines 23.7 Motion of Charged Particles in a Uniform Electric Field 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity Electric Charge Types: Positive Glass rubbed with silk Missing electrons Negative Rubber/Plastic rubbed with fur Extra electrons Arbitrary choice convention attributed to ? Units: amount of charge is measured in [Coulombs] Empirical Observations: Like charges repel Unlike charges attract 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity Charge in the Atom Protons (+) Electrons (-) Ions Polar Molecules 1/10/2006 Norah Ali Al-moneef king Saud unversity

23.1 Properties of Electric Charges Conservation electricity is the implication that electric charge is always conserved. • That is, when one object is rubbed against another, charge is not created in the process. The electrified state is due to a transfer of charge from one object to the other. • One object gains some amount of negative charge while the other gains an equal amount of positive charge. Quantization The smallest unit of charge is that on an electron or proton. (e = 1.6 x 10-19 C) It is impossible to have less charge than this It is possible to have integer multiples of this charge 1/10/2006 Norah Ali Al-moneef king Saud unversity

Conductors and Insulators 23.2 Charging Objects By Induction Conductors and Insulators Conductor transfers charge on contact Insulator does not transfer charge on contact Semiconductor might transfer charge on contact 1/10/2006 Norah Ali Al-moneef king Saud unversity

Charge Transfer Processes Conduction Polarization Induction 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity 23-3 Coulomb’s Law Empirical Observations Formal Statement Direction of the force is along the line joining the two charges 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity Consider two electric charges: q1 and q2 The electric force F between these two charges separated by a distance r is given by Coulomb’s Law The constant k is called Coulomb’s constant and is given by The coulomb constant is also written as 0 is the “electric permittivity of vacuum” A fundamental constant of nature 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity Double one of the charges force doubles Change sign of one of the charges force changes direction Change sign of both charges force stays the same Double the distance between charges force four times weaker Double both charges force four times stronger 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity Example: What is the force between two charges of 1 C separated by 1 meter? Answer: 8.99 x 109 N, Blackboard calculation 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity Coulomb’s Law Example What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e) and its innermost electron if the distance between them is 1.5 x 10-12 m The magnitude of the Coulomb force is F = kQ1Q2/r2 = (9.0 x 109 N · m2/C2)(26)(1.60 x 10–19 C)(1.60 x 10–19 C)/(1.5x10–12 m)2 = 2.7 x 10–3 N. 1/10/2006 Norah Ali Al-moneef king Saud unversity

Example - The Helium Nucleus Part 1: The nucleus of a helium atom has two protons and two neutrons. What is the magnitude of the electric force between the two protons in the helium nucleus? Answer: 58 N Part 2: What if the distance is doubled; how will the force change? Answer: 14.5 N Blackboard calculation Inverse square law: If the distance is doubled then the force is reduced by a factor of 4. 1/10/2006 Norah Ali Al-moneef king Saud unversity

Example - Equilibrium Position Consider two charges located on the x axis The charges are described by q1 = 0.15 C x = 0.0 m q2 = 0.35 C x = 0.40 m Where do we need to put a third charge for that charge to be at an equilibrium point? At the equilibrium point, the forces from the two charges will cancel. Here the forces from q1 and q2 can balance. q3 1/10/2006 Norah Ali Al-moneef king Saud unversity

Zero Resultant Force, Example Two fixed charges, 1mC and -3mC are separated by 10cm as shown in the figure (a) where may a third charge be located so that no force acts on it?  The magnitudes of the individual forces will be equal Directions will be opposite Will result in a quadratic Choose the root that gives the forces in opposite directions 1/10/2006 Norah Ali Al-moneef king Saud unversity

Example - Charged Pendulums Consider two identical charged balls hanging from the ceiling by strings of equal length 1.5 m (in equilibrium). Each ball has a charge of 25 C. The balls hang at an angle  = 25 with respect to the vertical. What is the mass of the balls? Step 1: Three forces act on each ball: Coulomb force, gravity and the tension of the string. x y 1/10/2006 Norah Ali Al-moneef king Saud unversity

Example - Charged Pendulums (2) Step 2: The balls are in equilibrium positions. That means the sum of all forces acting on the ball is zero! d=2 l sin θq Blackboard calculation Answer: m = 0.76 kg A similar analysis applies to the ball on the right. 1/10/2006 Norah Ali Al-moneef king Saud unversity

Electric Force and Gravitational Force Coulomb’s Law that describes the electric force and Newton’s gravitational law have a similar functional form Both forces vary as the inverse square of the distance between the objects. Gravitation is always attractive. k and G give the strength of the force. 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity Example: An electron is released above the surface of the Earth. A second electron directly below it exerts an electrostatic force on the first electron just great enough to cancel out the gravitational force on it. How far below the first electron is the second? Fe 5.1 m e mg r = ? e 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity The electron and proton of a hydrogen atom are separated (on the average) by a distance of approximately 5.3 x10-11 m. Find the magnitudes of the electric force and the gravitational force between the two particles. Compare the electrostatic and gravitational the forces Fe/Fg = 2 x 1039  The force of gravity is much weaker than the electrostatic force 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity 1/10/2006 Norah Ali Al-moneef king Saud unversity

Electric Forces and Vectors Electric Fields and Forces are ALL vectors, thus all rules applying to vectors must be followed. Consider three point charges, q1 = 6.00 x10-9 C (located at the origin),q3 = 5.00x10-9 C, and q2 = -2.00x10-9 C, located at the corners of a RIGHT triangle. q2 is located at y= 3 m while q3 is located 4m to the right of q2. Find the resultant force on q3. Which way does q2 push q3? Which way does q1 push q3? 4m q2 q3 3m Fon 3 due to 1 5m q q1 Fon 3 due to 2 q = 37 q3 q= tan-1(3/4) 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity q2 q3 3m Fon 3 due to 1 5m q q1 F3,1sin37 Fon 3 due to 2 q = 37 q= tan-1(3/4) q3 F3,1cos37 5.6 x10-9 N 7.34x10-9 N 1.1x10-8 N 64.3 0 above the +x 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity Coulomb’s Law Example Q = 6.0 mC L = 0.10 m What is the magnitude and direction of the net force on one of the charges? We find the magnitudes of the individual forces on the charge at the upper right corner: F1= F2 = kQQ/L2 = kQ2/L2 = (9 x109 N · m2/C2)(6 x10–3 C)2/(0.100 m)2 = 3.24 x107 N. F3= kQQ/(L√2)2 = kQ2/2L2 = (9 x109 N · m2/C2)(6 x10–3 C)2 /2(0.100 m)2 = 1.62 x107 N. We find the magnitudes of the individual forces on the charge at the upper right corner: F1= F2 = kQQ/L2 = kQ2/L2 = (9.0 ´ 109 N · m2/C2)(6.00 ´ 10–3 C)2/(0.100 m)2 = 3.24 ´ 107 N. F3= kQQ/(L√2)2 = kQ2/2L2 = (9.0 ´ 109 N · m2/C2)(6.00 ´ 10–3 C)2/2(0.100 m)2 = 1.62 ´ 107 N. The directions of the forces are determined from the signs of the charges and are indicated on the diagram. For the forces on the upper-right charge, we see that the net force will be along the diagonal. For the net force, we have F = F1 cos 45° + F2 cos 45° + F3 = 2(3.24 ´ 107 N) cos 45° + 1.62 ´ 107 N = 6.20 ´ 107 N along the diagonal, or away from the center of the square. From the symmetry, each of the other forces will have the same magnitude and a direction away from the center: The net force on each charge is 6.20 ´ 107 N away from the center of the square. Note that the sum for the three charges is zero. 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity along the diagonal, or away from the center of the square. From the symmetry, each of the other forces will have the same magnitude and a direction away from the center: The net force on each charge is= 6.20 ء 107 N away from the center of the square. . 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity Example - Four Charges Consider four charges placed at the corners of a square with sides of length 1.25 m as shown on the right. What is the magnitude of the electric force on q4 resulting from the electric force from the remaining three charges? Answer: F (on q4) = 0.0916 N … and the direction? Blackboard calculation 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity HOMEWORK : 23-7; Three point charges are located at the corners of an equilateral triangle. Calculate the net electric force on the 7.00 uC charge. 23-8: Two small beads having positive charges 3q and q are fixed at the opposite ends of a horizontal insulating rod extending from the origin to the point x =d. a third small charged bead is free to slide on the rod. At what position is the third bead in equilibrium? Can it be in stable equilibrium? 1/10/2006 Norah Ali Al-moneef king Saud unversity

Norah Ali Al-moneef king Saud unversity 23-12; An object having a net charge of 24.0 C is placed in a uniform electric field of 610 N/C that is directed vertically. What is the mass of this object if it “floats” in the field? 3-18; Two 2.00uC point charges are located on the x axis. One is at x = 1.00 m, and the other is at x =- 1.00 m. (a) Determine the electric field on the y axis at y =0.500 m. (b) Calculate the electric force on a - 3.00uC charge placed on the y axis at y = 0.500 m. 23-41; An electron and a proton are each placed at rest in an electric field of 520 N/C. Calculate the speed of each particle 48.0 ns after being released. 23-44; The electrons in a particle beam each have a kinetic energy of 1.60 x 10-17 J. What are the magnitude and direction of the electric field that stops these electrons in a distance of 10.0 cm? 1/10/2006 Norah Ali Al-moneef king Saud unversity