Chapter 18 Electric Force and Electric Fields Sections 18.1 18.6
Objectives: After finishing this unit, you should be able to: Explain and demonstrate the First law of electrostatics and discuss charging by contact and by induction. Write and apply Coulomb s Law and apply it to problems involving electric forces. Define the electron, the coulomb, and the microcoulomb as units of electric charge.
Electric Charge When a rubber rod is rubbed against fur, electrons are removed from the fur and deposited on the rod. Electrons move from fur to the rubber rod. positive negative The rod is said to be negatively charged because of an excess of electrons. The fur is said to be positively charged because of a deficiency of electrons.
Glass and Silk When a glass rod is rubbed against silk, electrons are removed from the glass and deposited on the silk. glass silk Electrons move from glass to the silk cloth. negative positive The glass is said to be positively charged because of a deficiency of electrons. The silk is said to be negatively charged because of a excess of electrons.
Two Negative Charges Repel 1. Charge the rubber rod by rubbing against fur. 2. Transfer electrons from rod to each pith ball. The two negative charges repel each other.
Two Positive Charges Repel 1. Charge the glass rod by rubbing against silk. 2. Touch balls with rod. Free electrons on the balls move to fill vacancies on the cloth, leaving each of the balls with a deficiency. (Positively charged.) The two positive charges repel each other.
The Two Types of Charge Rubber fur Attraction glas s silk Note that the negatively charged (green) ball is attracted to the positively charged (red) ball. Opposite Charges Attract!
The First Law of Electrostatics Like charges repel; unlike charges attract. Neg Pos Neg Pos Neg Pos
Charging by Contact 1. Take an uncharged electroscope as shown below. 2. Bring a negatively charged rod into contact with knob. 3. Electrons move down on leaf and shaft, causing them to separate. When the rod is removed, the scope remains negatively charged.
Charging Electroscope Positively by Contact with a Glass Rod: Repeat procedures by using a positively charged glass rod. Electrons move from the ball to fill deficiency on glass, leaving the scope with a net positive charge when glass is removed.
Charging Spheres by Induction Induction Electrons Repelled Uncharged Spheres Separation of Charge Isolation of Spheres Charged by Induction
Induction for a Single Sphere Induction Uncharged Sphere Separation of Charge Electrons move to ground. Charged by Induction
The Quantity of Charge The quantity of charge (q) can be defined in terms of the number of electrons, but the Coulomb (C) is a better unit for later work. A temporary definition might be as given below: The Coulomb: 1 C = 6.25 x 10 18 electrons Which means that the charge on a single electron is: 1 electron: e = 1.6 x 10 19 C
Units of Charge The coulomb (selected for use with electric currents) is actually a very large unit for static electricity. Thus, we often encounter a need to use the metric prefixes. 1 μc = 1 x 10 6 C 1 nc = 1 x 10 9 C 1 pc = 1 x 10 12 C
Example: If 16 million electrons are removed from a neutral sphere, what is the charge on the sphere in coulombs? 1 electron: e = 1.6 x 10 19 C q 19 6 1.6 x 10 C (16 x 10 e ) 1 e q = 2.56 x 10 12 C Since electrons are removed, the charge remaining on the sphere will be positive. Final charge on sphere: q = 2.56 x 10 12 C or 2.56 pc
Coulomb s Law The force of attraction or repulsion between two point charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them. F F q q r q q F F qq ' r 2
Calculating Electric Force The proportionality constant, k, for Coulomb s law depends on the choice of units for charge. F kqq 2 r ' where k 2 Fr qq ' When the charge, q, is in Coulombs, the distance, r, is in meters and the force, F, is in Newtons, we have: k Fr qq ' Nm C 2 2 9 9 x 10 2
Example Two protons in a molecule are separated by a distance of 3.8 x 10 10 m. Find the electrostatic force exerted by one proton on the other. Answer: 1.596 x 10 9 N
Example A 6.7 μc charge is located 5.0 m from a 8.4 μc charge. Find the electrostatic force exerted by one charge on the other. Answer: 0.0203 N
Example A 1.3 C charge is located on the x axis at x = 0.5 m, a 3.2 μc charge is located on the x axis at x = 1.5 m, and a 2.5 μc charge is located at the origin. Find the net force on the 2.5 μc charge. All charges are positive. Answer: 0.085 N
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? Answer: 5.08 m
Example Three point charges of 2 μc, 7 μc, and 4 μc are located at the corners of an equilateral triangle as in the Figure. Calculate the net electric force on the 7 μc charge. Answer: 0.873 N at 30.03 o south of east or 120.03 o from true north
Example Four identical point charges (q = 10 μc) are located on the corners of a rectangle as shown in the figure. The dimensions of the rectangle are L = 60 cm and W = 15 cm. Calculate the magnitude and direction of the net electrostatic force exerted on the charge at the lower left corner of the rectangle by the other three charges. Answer: 40.85 N at 186.724 o from true north or 6.724 o west of south
Objectives: After finishing this unit you should be able to: Define the electric field and explain what determines its magnitude and direction. Write and apply formulas for the electric field intensity at known distances from point charges. Discuss electric field lines and the meaning of permittivity of space.
The Concept of a Field A field is defined as a property of space in which a material object experiences a force. m P. F Above earth, we say there is a gravitational field at P. Because a mass m experiences a downward force at that point. No force, no field; No field, no force! The direction of the field is determined by the force.
The Gravitational Field A F F B Note that the force, F, is real, but the field is just a convenient way of describing space. Consider points A and B above the surface of the earth just points in space. If g is known at every point above the earth then the force, F, on a given mass can be found. The field at points A or B might be found from: g F m The magnitude and direction of the field, g, depends on the weight, which is the force, F.
The Electric Field 1. Now, consider point P a distance r from Q. 2. An electric field E exists at P if a test charge, q, has a force, F, at that point. 3. The direction of the E is the same as the direction of a force on (pos) charge. 4. The magnitude of E is given by the formula: E F P. q Q Electric Field F q r E N ; Units C
Field is Property of Space q F. E Force on q is with field direction. q. E r Q Electric Field Force on q is against field direction. F r Q Electric Field The field E at a point exists whether there is a charge at that point or not. The direction of the field is away from the Q charge.
Field Near a Negative Charge E q F r. Q Electric Field Force on q is with field direction. Force on q is against field direction. F q. E r Q Electric Field Note that the field E in the vicinity of a negative charge Q is toward the charge the direction that a q test charge would move.
The Magnitude of EField The magnitude of the electric field intensity at a point in space is defined as the force per unit charge (N/C) that would be experienced by any test charge placed at that point. Electric Field Intensity, E F N E ; Units q C The direction of E at a point is the same as the direction that a positive charge would move IF placed at that point.
Example What are the magnitude and direction of the electric field that will balance the weight of (a) an electron and (b) a proton? Answer: a. 5.572 x 10 11 N/C, points down, b. 1.024 x 10 7 N/C, points up
The EField at a distance r from a single charge Q Consider a test charge, q, placed at P a distance, r, from Q. The outward force on q is: F kqq 2 r The electric field, E, is therefore: Q 2 F kqq r kq E E 2 q q r q r E F E q. P kq 2 r
The Resultant Electric Field. The resultant field, E, in the vicinity of a number of point charges is equal to the vector sum of the fields due to each charge taken individually. Consider E for each charge. E 2 Vector Sum: E = E 1 E 2 E 3 q 1 E 1 E R A q 3 E 3 q 2 Magnitudes are from: E kq 2 r Directions are based on positive test charge.
Example A point charge of 5.2 μc is located at the origin. Find the electric field (a) on the x axis at x = 3 m, (b) on the y axis at y = 4 m, (c) at the point with coordinates x = 2 m, y =2 m. Answer: a. 5200 N/C, points toward the left b. 2925 N/C, points up y axis c. 5843 N/C at 225 o, points toward origin
Example Two charges are placed on the x axis. One charge (q = 8.5 μc) is a x = 3.0 cm and the other (q = 21 μc) is at x = 9.0 cm. Find the new electric field (magnitude and direction) at a. x = 0 cm, b. x = 6.0 cm. Answers: a. 6.17 x 10 7 N/C, to the left; b. 2.95 x 10 8 N/C, to the right.
Example: Find the resultant field at point A due to the 3 nc charge and the 6 nc charge arranged as shown. q 1 3 cm E 2 3 nc A E 1 5 cm 6 nc 4 cm q 2 Answer: 4.52 x 10 4 N/C, at 48.37 o West of North or 311.63 o from North
Electric Field Lines Electric Field Lines are imaginary lines drawn in such a way that their direction at any point is the same as the direction of the field at that point. Q Q Field lines go away from positive charges and toward negative charges.
Examples of EField Lines Two equal but opposite charges. Two identical charges (both ). Notice that lines leave charges and enter charges. Also, E is strongest where field lines are most dense.
Assignment Pages 564 568, #1, 4, 5, 8, 11, 18, 27, 29, 70, 73, 75