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Physics 2220 Physics for Scientists and Engineers II Physics for Scientists and Engineers II , Summer Semester 2009 Chapter 23: Electric Fields • Materials can be electrically charged. • Two types of charges exist: “Positive” and “Negative”. • Objects that are “charged” either have a net “positive” or a net “negative” charge residing on them. • Two objects with like charges (both positively or both negatively charged) repel each other. • Two objects with unlike charges (one positively and the other negatively charged) attract each other. • Electrical charge is quantized (occurs in integer multiples of a fundamental charge “e”). q = N e (where N is an integer) electrons have a charge q = - e protons have a charge q=+e neutrons have no charge Physics for Scientists and Engineers II , Summer Semester 2009 Material Classification According to Electrical Conductivity • Electrical conductors: Some electrons (the “free” electrons) can move easily through the material. • Electrical insulators: All electrons are bound to atoms and cannot move freely through the material. • Semiconductors: Electrical conductivity can be changed over several orders of magnitude by “doping” the material with small quantities of certain atoms, making them more or less like conductors/insulators. Physics for Scientists and Engineers II , Summer Semester 2009 Shifting Charges in a Conductor by “Induction” uncharged metal sphere Negatively charged rod + - - - - + + - - - - + Physics for Scientists and Engineers II , Summer Semester 2009 + -+ + Left side of metal sphere more positively charged - + + - + - Right side of metal sphere more negatively charged Coulomb’s Law (Charles Coulomb 1736-1806) Magnitude of force between two “point charges” q1 and q2 . Fe ke q1 q2 r 2 Coulomb constant Nm 2 1 ke 8.987 6 x 10 C2 4 o 9 where o 8.854 2 x 10 12 C2 N m2 Permittivity of free space Physics for Scientists and Engineers II , Summer Semester 2009 r = distance between point charges Charge Unit of charge = Coulomb Smallest unit of free charge: e = 1.602 18 x 10-19 C Charge of an electron: qelectron = - e = - 1.602 18 x 10-19 C Physics for Scientists and Engineers II , Summer Semester 2009 Vector Form of Coulomb’s Law Force is a vector quantity (has magnitude and direction). F 12 q1q2 ke 2 r̂12 r unit vector pointing from charge q1 to charge q2 Force exerted by charge q1 on charge q2 (force experienced by charge q2 ). Physics for Scientists and Engineers II , Summer Semester 2009 Vector Form of Coulomb’s Law Force is a vector quantity (has magnitude and direction). q1q2 q1q2 F 21 ke 2 rˆ21 ke 2 rˆ12 F 12 r r unit vector pointing from charge q2 to charge q1 Force exerted by charge q2 on charge q1 (force experienced by charge q1 ). Physics for Scientists and Engineers II , Summer Semester 2009 Directions of forces and unit vectors F 12 r̂21 r̂12 F 21 + q1 + q2 F 12 F 21 + q1 Physics for Scientists and Engineers II , Summer Semester 2009 - q2 Calculating the Resultant Forces on Charge q1 in a Configuration of 3 charges q1 + a = 1cm q3 - q2 + 0.5 cm q3 = - 2.0 mC q1 = q2 = +2.0 mC Physics for Scientists and Engineers II , Summer Semester 2009 0.5 cm Forces acting on q1 F 21 q1 + q2 + F 31 q3 Total force on q1: F 1 F 21 F 31 Physics for Scientists and Engineers II , Summer Semester 2009 Magnitude of the Various Forces on q1 2 12 2.0mC 2.0mC C2 9 Nm 4.0 10 2 F21 ke 8 . 99 10 3 . 596 10 N 2 2 4 2 (0.010m) C 1.0 10 m 2 12 2 2.0mC 2.0 mC Nm 4 . 0 10 C 9 2 F31 ke 8 . 99 10 7 . 192 10 N 2 4 2 2 C 0.5 10 m (0.010 0.5 2 m) Note: I am temporarily carrying along extra significant digits in these intermediate results to avoid rounding errors in the final result. Physics for Scientists and Engineers II , Summer Semester 2009 Adding the Vectors Using a Coordinate System F 21 q1 + y q2 + F 31 q3 - Physics for Scientists and Engineers II , Summer Semester 2009 x Adding the Vectors Using a Coordinate System F 21 F21 iˆ 0 ˆj F 31 y 2 2 F31 iˆ F31 ˆj 2 2 F 21 F 1 F 21 F 31 2 2 ˆ F21 F31 i F31 ˆj 2 2 Physics for Scientists and Engineers II , Summer Semester 2009 x F 31 …doing the algebra… F 1 F 21 F 31 2 2 ˆ F21 F51 i F31 ˆj 2 2 3.596 10 2 N 5.086 10 2 N iˆ 5.086 10 2 N ˆj 1.5 10 2 N iˆ 5.110 2 N ˆj F1 has a magnitude of F1 5.3 102 N Physics for Scientists and Engineers II , Summer Semester 2009 Calculating the force on q2 … another example using an even more mathematical approach Charges Location of charges q1 = +3.0 mC x1=3.0cm ; y1=2.0cm ; z1=5.0cm q2 = - 4.0 mC x2=2.0cm ; y2=6.0cm ; z2=2.0cm In this example, the location of the charges and the distance between the charges are harder to visualize Use a more mathematical approach! Physics for Scientists and Engineers II , Summer Semester 2009 Calculating the force on q2 … another example using an even more mathematical approach q1q2 F2 F12 ke r̂ 2 12 d12 d12=distance between q1 and q2. rˆ12 unit vecto r pointing from q1to q 2 . Physics for Scientists and Engineers II , Summer Semester 2009 Calculating the force on q2 … mathematical approach We need the distance between the charges. d12 is distance between q1 and q2. y q1 + r2 r1 r1 - q2 r2 z Physics for Scientists and Engineers II , Summer Semester 2009 x d12 (r2 r1 ) (r2 r1 ) Calculating the force on q2 … mathematical approach Distance between charges q1 and q2 . 2.0 cm 3.0 cm 1.0 cm r2 r1 6.0 cm 2.0 cm 4.0 cm 2.0 cm 5.0 cm 3.0 cm 1.0 cm 1.0 cm d12 (r2 r1 ) (r2 r1 ) 4.0 cm 4.0 cm 3.0 cm 3.0 cm 1.0 cm 2 16.0 cm 2 9.0 cm 2 26 cm Physics for Scientists and Engineers II , Summer Semester 2009 Calculating the force on q2 … mathematical approach We need the unit vectors between charges. For example, the unit vector pointing from q1 to q2 is easily obtained by normalizing the vector pointing from from q1 to q2. y q1 + r1 r̂12 d12 (r2 r1 ) (r2 r1 ) r2 r1 - q2 r2 z Physics for Scientists and Engineers II , Summer Semester 2009 x r2 r1 rˆ12 d12 Calculating the force on q2 … mathematical approach The needed unit vector: 1.0 cm 1.0 r2 r1 1 1 rˆ12 4.0 cm 4.0 d12 26 cm 14 3 . 0 cm 3 . 0 Physics for Scientists and Engineers II , Summer Semester 2009 Calculating the force on q2 … mathematical approach You can easily verify that the length of the unit vector is “1”. 1 Length of rˆ12 4 3 1 26 4 26 3 26 Physics for Scientists and Engineers II , Summer Semester 2009 26 1 16 9 26 1 26 26 26 26 26 26 Calculating the force on q2 … another example using an even more mathematical approach F2 F12 ke q1q2 rˆ 2 12 d12 1.0 3.0 mC (4.0) mC 1 ke 4 . 0 26 cm 2 26 3 .0 2 1.0 2 Nm C 100cm 8.9876 109 2 0.462 10 12 2 4 . 0 C cm 1m 3.0 1.0 41.48 4.0 N 3.0 2 41.48 0.42 165.92 N 1.7 10 2 N 124.44 1.2 Physics for Scientists and Engineers II , Summer Semester 2009 Calculating the force on q2 … another example using an even more mathematical approach …and if you want to know just the magnitude of the force on q2 : F2 F2 F2 41.48 N 41.48 N 165.92 N 165.92 N 124.44 N 124.44 N 1721 N 2 27529 N 2 15485 N 2 2.1 10 2 N Physics for Scientists and Engineers II , Summer Semester 2009 23.4 The Electric Field Remember from Physics 2210 : The gravitatio nal field g Fg m at the location of the mass m. g is defined as the force Fg acting on a test particle of mass m divided by that mass. Similarly, Faraday defined : The electric field vector E Fe qo E is defined as the force Fe acting on a test charge q o divided by q o . It is convenient to use positive test charges. Then, the direction of the electric force on the test charge is the same as that of the field vector. Confusion is avoided. Physics for Scientists and Engineers II , Summer Semester 2009 23.4 The Electric Field N SI units of electric field E : C Note : The measured E is NOT produced by the test charge itself. The test charge only detects E produced by source charge Q. Of course, the test charge also produces an additionalelectric field. However, there is no force acting on the test charge due to the field produced by the test charge. Therefore, the force on the test charge is solely due to the electric field of the source charge Q alone.The field measured is the actual field that exists once the test charge is removed again. Q + + + + + + + + qo E + test charge Source charge Physics for Scientists and Engineers II , Summer Semester 2009 23.4 The Electric Field Force on a charge q in an electric field : Fe q E For positive charge q : Fe is in the same direction as E . For negative charge q : Fe and E are in opposite directions. Once you know E at some point in space, you can calculate the force on any charged particle placed at that point. Physics for Scientists and Engineers II , Summer Semester 2009 23.4 The Electric Field of a “Point Charge” q Fe ke Force exerted on test charge q 0 by q : qq0 rˆ 2 r Fe q Electric field created by q at the place of q 0 : E ke 2 rˆ q0 r where r is the distance between q and q 0 and r̂ is the unit vecto r pointing from q to q 0 . q r̂ r Physics for Scientists and Engineers II , Summer Semester 2009 q0 23.4 The Electric Field of a Positive “Point Charge” q (Assuming positive test charge q0) Fe q0 + Force on test charge E + P Electric field where test charge used to be (at point P). The electric field of a positive point charge points away from it. Physics for Scientists and Engineers II , Summer Semester 2009 23.4 The Electric Field of a Negative “Point Charge” q (Assuming positive test charge q0) Fe - q0 Force on test charge E - P Electric field where test charge used to be (at point P). The electric field of a negative point charge points towards it. Physics for Scientists and Engineers II , Summer Semester 2009 23.4 The Electric Field of a Collection of Point Charges q E ke 2 rˆi ri i Physics for Scientists and Engineers II , Summer Semester 2009 23.4 The Electric Field of Two Point Charges at Point P y P E ? y q1 q2 a x b q q1 q2 E ke 2 rˆi ke 2 rˆ1 ke 2 rˆ2 ri r1 r2 i Physics for Scientists and Engineers II , Summer Semester 2009 23.4 The Electric Field of Two Point Charges at Point P y P r1 q1 r2 y a Pythagoras: r12 a 2 y 2 b q2 x q1 q2 ˆ E ke 2 r ke 2 rˆ 2 1 2 2 a y b y Physics for Scientists and Engineers II , Summer Semester 2009 r22 b 2 y 2 23.4 The Electric Field of Two Point Charges at Point P y rˆ1 r1 p P rˆ2 r1 p r1 p r1 r2 p r2 1 1 0 a 1 a 1 R C 1 r1 r1 y 0 r1 y a2 y2 R q1 0 R y r̂2 r̂1 C1 C2 q1 E ke 2 a y2 1 1 0 b 1 b 1 R C 2 r2 r1 y 0 r1 y b2 y 2 q2 a C1 0 a y b y b C 2 0 x a q2 ke 2 2 2 2 y b y a y 1 Physics for Scientists and Engineers II , Summer Semester 2009 b 2 2 y b y 1 23.4 The Electric Field of Two Point Charges at Point P E ke a q1 2 E x ke E y ke y a a 2 3 a q2 ke y b2 y 2 2 q1a 2 y 2 q1 y 2 y 2 3 3 ke 2 2 ke Physics for Scientists and Engineers II , Summer Semester 2009 b b q2b 2 y 2 q2 y 2 y 2 3 3 3 2 2 2 b y 23.4 The Electric Field of Two Point Charges at Point P Special case: q1= q and q2 = -q AND b = a E from + charge total E (has no y - component) E from - charge + - q -q E x ke E y ke a a q1a 2 y2 q1 y 2 y 2 3 3 2 2 ke ke b q2b 2 b y2 3 q2 y 2 y 2 2 3 ke 2 ke a qa 2 a Physics for Scientists and Engineers II , Summer Semester 2009 y2 3 qy 2 y 2 2 3 ke 2 a ke qa 2 y2 a 3 2 qy 2 y ke 2 3 a 0 2 2qa 2 y2 3 2 23.4 The Electric Field of Two Point Charges at Point P Special case: q1= q and q2 = q AND b = a total E E from other + charge E from + charge + + q q E x k e E y ke (has no x - component) a a q1a 2 y 2 q1 y 2 y2 3 3 2 2 ke ke b q2b 2 b y 2 q2 y 2 y2 3 2 3 ke 2 ke Physics for Scientists and Engineers II , Summer Semester 2009 a qa 2 a y 2 qy 2 y2 3 2 3 ke 2 a ke qa 2 y a 2 3 0 2 qy 2 y2 3 2 ke a 2qy 2 y2 3 2 This is called an electric DIPOLE Special case: q1= q and q2 = -q AND b = a E from + charge total E (has no y - component) E from - charge + - q -q E x ke a 2qa 2 y 2 3 Ey 0 2 For large distances y (far away from the dipole), y >> a: E x k e 2qa y 2 3 2 k e 2qa y3 E falls off proportional to 1/y3 Fall of faster than field of single charge (only prop. to 1/r2). From a distance the two opposite charges look like they are almost at the same place and neutralize each other. Physics for Scientists and Engineers II , Summer Semester 2009