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Coulomb’s Law and Electric Field Chapter 24: all Chapter 25: all 1 Electric charge Able to attract other objects Two kinds Positive – glass rod rubbed with silk Negative – plastic rod rubbed with fur Like charges repel Opposite charge attract Charge is not created, it is merely transferred from one material to another 2 Elementary particles Proton – positively charged Electron – negatively charged Neutron – no charge Nucleus – in center of atom, contains protons and neutrons Quarks – fundamental particles – make up protons and neutrons, have fractional charge 3 ions Positive ions – have lost one or more electrons Negative ions – have gained one or more electrons Only electrons are lost or gained under normal conditions 4 Conservation of charge The algebraic sum of all the electric charges in any closed system is constant. 5 Electrical interactions Responsible for many things The forces that hold molecules and crystals together Surface tension Adhesives Friction 6 Conductors Permit the movement of charge through them Electrons can move freely Most metals are good conductors 7 Insulators Do not permit the movement of charge through them Most nonmetals are good insulators Electrons cannot move freely 8 Charging by induction See pictures on pages 539-540 9 Coulomb’s Law Point charge – has essentially no volume The electrical force between two objects gets smaller as they get farther apart. The electrical force between two objects gets larger as the amount of charge increases 10 Coulomb’s Law F k q1q2 r 2 r is the distance between the charges q1 and q2 are the magnitudes of the charges k is a constant 8.99 x 109 N∙m2/C2 11 Coulombs SI unit of charge, abbreviated C Defined in terms of current – we will talk about this later 12 Coulomb’s law constant k is defined in terms of the speed of light k = 10-7c k = 1/4pe0 e0 is another constant that will be more useful later e0 = 8.85 x 10-12 C2/N∙m2 13 The coulomb Very large amount of charge Charge on 6 x 1018 electrons Most charges we encounter are between 10-9 and 10-6 C 1 mC = 10-6 C 14 Examples See pages 543 - 546 15 Electric Field • A field is a region in space where a force can be experienced. • Or: a region in space where a quantity has a definite value at every point. 16 Electric Field • Produced by a charged particle. • The force felt by another charged particle is caused by the electric field. • We can check for an electric field with a test charge, qt. If it experiences a force, there is an electric field. 17 Electric field • The definite quantity is a ratio of the electric force experienced by a charge to the amount of the charge. • Vector quantity measured in N/C. F E qt F qt E 18 Electric field • To determine the field from a point charge, Q, we place a test charge, qt, at some position and determine the force acting on it. Q F qt 19 Direction of E • If the test charge is positive, E has the same direction as F. • If the test charge is negative, E has the opposite direction as F. 20 Electric Field - Point Charge F E qt Ek Fk qt Q r 2 Q r 2 E 1 Q 4peo r 2 21 Electric Field • The field is there, independent of a test charge or anything else! • The electric field vector points in the direction a positive charge would be forced. 22 Example 1 • Two charges, Q1 = +2 x 10-8 C and Q2 = +3 x 10-8 C are 50 mm apart as shown below. • What is the electric field halfway between them? Q1 E2 E1 50 mm Q2 23 Example 1 • At the halfway point, r1 = r2 = 25 mm. • Magnitudes of fields: kQ1 E1 2 r1 kQ2 E2 2 r2 (9 x 10 9 N•m C 2 2 8 )(2 x 10 C) 2 (2.5 x 10 m) (9 x 10 9 N•m C 2 2 2 8 )(3 x 10 C) 2 (2.5 x 10 m) 2 24 Example 1 • • • • • • E1 = 2.9 x 105 N/C E2 = 4.3 x 105 N/C E1 is to the right and E2 is to the left. E1 = 2.9 x 105 N/C E2 = - 4.3 x 105 N/C E = E1 + E2 = - 1.4 x 105 N/C 25 Example 2 • For the charges in Example 1, where is the electric field equal to zero? • Since the fields are in opposite directions between the charges, the point where the field is zero must be between them. Q1 E2 E1 Q2 26 Example 2 E1 E 2 kQ1 kQ 2 2 2 r1 r2 Q1 Q 2 2 2 r1 r2 r1 + r2 = s, so r2 = s – r1 27 Example 2 Q1 Q 2 2 2 r1 r2 Q1 Q2 2 2 r1 (s r1 ) 2 (s r1 ) Q2 2 r1 Q1 s r1 Q2 r1 Q1 r1 s Q2 1 Q1 r1 23 mm 28 Field Diagrams • To represent an electric field we use lines of force or field lines. • These represent the sum of the electric field vectors. 29 Field Diagrams 30 Field Diagrams 31 Field Diagrams • At any point on the field lines, the electric field vector is along a line tangent to the field line. 32 Field Diagrams 33 Field Diagrams • Lines leave positive charges and enter negative charges. • Lines are drawn in the direction of the force on a positive test charge. • Lines never cross each other. • The spacing of the lines represents the strength or magnitude of the electric field. 34 Point Charges • Lines leave or enter the charges in a symmetric pattern. • The number of lines around the charge is proportional to the magnitude of the charge. 35 Point Charges 36 Point Charges 37 Gauss’s Law • Electric flux through a closed surface is proportional to the total number of field lines crossing the surface in the outward direction minus the number crossing in the inward direction. EA Q e0 38 Example 25-9 (see page 563) Field of a charged sphere is the same as if it were a point charge 1 q E 4pe0 r 2 39 Example 25-10 (see page 564) Field of a infinite line of charge is E 1 2pe0 r 40 Other scenarios • See table on page 567 41 Example 3 • Two parallel metal plates are 2 cm apart. • An electric field of 500 N/C is placed between them. • An electron is projected at 107 m/s halfway between the plates and parallel to them. • How far will the electron travel before it strikes the positive plate? 42 Example 3 • Two charged parallel plates create a uniform electric field in the space between them. 43 Example 3 vo E This is just like a projectile problem except that the acceleration is not a given value. 44 Example 3 a= a= F F = qE = eE m eE m = (1. 6 x 10 –19 C )(500 N /C ) 9. 1 x 10 –31 kg = 8.8 x 1013 m/s2 45 Example 3 • 8.8 x 1013 m/s2 is the vertical acceleration of the electron. • Horizontally, the acceleration is zero. • x = vt • v = 1 x 107 m/s & t = ? 46 Example 3 • Back to vertical direction: • y = yo + vot + 1/2at2 • y = 1/2at2 2(0.01 m) 2y 13 2 t 8.8x10 m / s a = 1.5 x 10-8 s 47 Example 3 • Back to horizontal direction: • x = vt • x = (1 x 107 m/s)(1.5 x 10–8 s) • x = 0.15 m = 15 cm 48 Dipoles • A pair of charges with equal and opposite sign. • Induced dipoles, molecular dipoles, etc.… 49