* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download PHYS113 Electricity
Survey
Document related concepts
Fundamental interaction wikipedia , lookup
Magnetic monopole wikipedia , lookup
Electrical resistivity and conductivity wikipedia , lookup
History of electromagnetic theory wikipedia , lookup
Potential energy wikipedia , lookup
Speed of gravity wikipedia , lookup
Electromagnetism wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Maxwell's equations wikipedia , lookup
Field (physics) wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Lorentz force wikipedia , lookup
Transcript
PHYS113 Electricity and Electromagnetism Semester 2; 2002 Professor B. J. Fraser 1 1. Electric Charge What is charge? 700 BC - Greeks write of effects of rubbing amber (Electrum) 1600’s - Gilbert shows electrification is a general phenomenon 1730 - C. Dufay concludes “there are 2 distinct Electricities” 1750 - Ben Franklin shows +ve & ve charges Electrostatics involves the forces between stationary charges. Charge is a basic atomic property forces between electrons & nuclei unlike charges attract like charges repel 2 Transfer of Charge Charge transfer touching charge sharing (conduction) only the electrons move Can appear as though positive charge has moved + ++ ++ + ++ + Unit of charge: Coulomb, C 1 Coulomb = 1 Ampere second Electronic charge, e = 1.602 x 10-19 C i.e. 1 C = 6.3 x 1018 electrons a small number!! 3 Conservation & Quantisation Charge is always conserved it cannot be created or destroyed Charge only comes in fixed packets the packet size is ± e It cannot wear off The light from distant quasars (billions of years old) shows evidence of exactly the same atomic charge. 4 Forces Between Charges Coulomb’s Law 1785: Coulomb experimentally determines force law between 2 charged point sources, q1 and q2. 1 F 2 F q1q2 r Thus: F k q1q2 r2 where k = 8.99 x 109 N m2 C-2 Electric force has direction (vector) Hence Coulomb’s Law is: qq F 12 k 1 2 2 r̂ 12 r F12 is the force on q1 due to q2 r12 is a unit vector from q2 to q1 along the line that joins them. 5 Hints for Problem Solving Draw a clear diagram Forces are vectors include coordinates i.e. Fx and Fy or i and j components add vectorially Shortcuts due to symmetry? Example: Electric Forces in a Plane Calculate the forces on q1 and q3 -3 nC q3 2.0 m q1 q2 +2 nC +2 nC 2.0 m 6 Solution: Forces in a plane Force on q1 This is due to q2 and q3 q3 F1 F13 F12 j i q1 q2 F 1 F 12 F 13 q q q q k 1 2 2 rˆ12 1 2 3 rˆ13 r13 r12 q2 q3 ˆ k q1 2 i 2 r13 r12 ˆj 9 9 2 x 10 3 x 10 ˆ 9 x 109 2 x 109 i 2 2 2 2 9.0 x 109 N iˆ 13.5 x 10 9 N ˆj F1 Fx 2 Fy 2 16.2 x 10 9 N 7 ˆj Solution: Forces in a plane Force on q3 This is due to q2 and q3 q3 F31 F32 F3 q1 j i q2 q q F 3 k q3 21 rˆ 31 22 rˆ 32 r32 r31 2 x 10 9 ˆ j 2 2 9 x 109 3 x 10 9 2 x 10 9 ˆ ˆ cos i sin j 2 2 2 2 13.5 x 10 9 N ˆj 4.8 x 10 9 N iˆ 4.8 x 10 9 N ˆj 4.8 x 10 9 N iˆ 18.3 x 10 9 N ˆj Find magnitude as before 8 2.The Concept of the Electric Field Why is there a force between charged particles? How does each particle know that the other one is there? What happens in space between charged particles? This is an example of an action-at-a-distance force. E.g. Gravitation, Magnetism These forces are described in terms of a field in space surrounding the particle or object. 9 Electric Field Strength Test an invisible force field? Test for an electric field by measuring force experienced by a positive test charge. We know that E F and: E 1/q0 Q See if a test object experiences a force! where q0 = charge of test charge E = electric field, N C-1 Hence: E F q0 and since: F k qq0 r 2 ˆ r Then: q E k 2 rˆ r Electric field seen by q0 due to q. + + + + + ++ + + + + + Charged Object Electric Field Region q0 F + Test Charge E F q0 10 Electric Field Lines Electric field strength is a vector quantity. Much easier to represent using vectors pointing in field direction - electric field lines. Concept due to M. Faraday “lines of force” Electric Field lines point away from positive charges Field lines point in the direction of the force or electric field Density (spacing) of field lines depends upon magnitude of E. Field lines never intersect. 11 Electric Fields in Nature Field Description Interplanetary space Strength (NC-1=Vm-1) 10-3 – 10-2 At Earth’s surface in clear weather In a thunder storm 100 - 200 Electrical breakdown of dry air Van der Graaff generator 3 x 106 Fermilab accelerator Atom at electron orbit radius 103 106 1.2 x 107 109 All charges (fixed & moving) produce an electric field that carries energy through space at the speed of light. 12 Field Due to Point Charges Electric fields add vectorially: E = E1 + E2 + E3 + … Thus: qi E k 2 rˆi i ri Worked Example Find the electric field at point A for j the dipole shown + 6 cm 4 cm A q1 = +12 nC - i q2 = -12 nC Field at A due to q1. q1 E A1 k 2 rˆ r1 12 x 10 9 4 -1 ˆ ˆ 9 x 10 i 3 . 0 x 10 i NC 2 2 6 x 10 9 13 Field Due to Point Charges Field at A due to q2 E A2 q1 k 2 rˆ r2 9 12 x 10 9 4 -1 ˆ ˆ 9 x 10 i 6 . 8 x 10 i NC 2 2 4 x 10 Total electric field at A: E A 9.8 x 10 4 iˆ NC -1 EA1 A +q1 EA2 - q2 No component in the j direction Example of an electric dipole Often found in nature (e.g. molecules) For more: See Section 21.11 14 Field due to a line segment Charge, Q, distributed uniformly along length, L, with charge density: l = Q/L Worked Example What is the electric field at a distance R from a rod of length 2L carrying a uniform charge density, l? Consider an infinite collection of charge elements, dQ. j L ^ r dE y R O i dy P -L r dE dQ 15 Field due to line segment (contd) L dQ E k 2 rˆ r L But l = Q/L and thus dQ = l dy L dy kl 2 rˆ r L Centre rod at origin For every charge at + y, there is another corresponding charge at -y Thus, fields in j component add to 0. L L dy dy E kl 2 cos î - sin ĵ kl 2 cos î r r L L L dy R î 2 r r L kl L kl R dy î 3 r L R L kl L R 2 y cos = R/r r2 =(R2 + y2) 2 32 dy î Can you do this integral? 16 Solution to Field Due to a Line Segment The solution to the field due to a line segment is: 2lL Ek î 2 2 R L R So, what is the big deal? Well, what happens if L >> R? 2kl E î R Thus, the field from a line charge is proportional to 1/R and not 1/R2. 17 Field Due to a Surface Consider a charge Q uniformly distributed across surface of area A Surface charge density is: s = Q/A Worked Example Find the electric field at distance R from an infinite plane sheet with surface charge density s. Divide the sheet into an infinite P collection of line segments, L, long and, dx, wide R k^ ^j r L ^i x dx 18 Field Due to a Surface Charge on each strip: dQ = s dA = s L dx Charge per unit length: l = dQ/L = s dx From previous example, each strip sets up electric field: E = 2kl/r = 2 k s dx/r Summing for all the strips: dx E ks sin k̂ - cos î But i r components sin dx sum to 0 2ks k̂ r 2ks R 2ks R 2 x R 2 12 R R 2 x 2 dx dx 2 x 2 12 k̂ Can you do this integral? 19 Solution to Field Due to a Surface The solution to the field due to a surface is: E 2 ks k̂ So, what’s the big deal this time? How does the field vary with R? Thus, the field from a surface in independent of the distance R! 20 Particles in an Electric Field A particle of charge, q, in an electric field, E, experiences a force: E = F/q F = qE = ma The particle accelerates at a +ve particle moves in direction of E I.e. from +ve to -ve charge regions. Thus an electron will be deflected toward a +ve charged plate as its moving past it. Examples: operation of CRT’s, TV tubes, etc. + + + + + + + + + screen e- - - - - - - - 21 Particles in an Electric Field Worked Example An electron in near-Earth space is accelerated Earthward by an electric field of 0.01 NC-1. Find its speed when it strikes air molecules in the atmosphere after travelling 3 Earth radii (19 000 km). The electron experiences a force: F = ma = qE a = qE/m For motion at constant acceleration: v2 = u2 + 2as = 2as 1 19 2 7 2qEs 21.6 x 10 10 1.9 x 10 2 v 31 m 9 . 1 x 10 v = 2.6 x 108 ms-1 i.e. 0.8 x speed of light 22 Fishnets and Flux: The Gaussian Surface Consider a fishnet with water flowing through it. The rate of flow through net is the flux. fw = vA v = velocity of flow A = area of net If the net is angled at to the flow: fw = vA cos A In vector form: fw = v A where the direction of A is normal to net 23 Defining Electric Flux For an irregular shape, area A is sum of infinitesimal elements dA. Thus, summing over 2-D surface S: fw v d A S Now, replace water with electric field, i.e. there is no physical motion. The electric flux through a surface of area A is: fE E d A S The electric flux through a surface is proportional to the number of field lines passing through a surface. If the fishnet is formed into a closed shape (e.g. lobster pot) its called a Gaussian surface. 24 The Gaussian Surface dA E The total electric flux (number of field lines) passing through this surface is: f E E d A where A points perpendicularly away from each element dA. If the flux in one side is the same as that out then the total flux is zero. If there is no net charge inside a Gaussian surface the electric flux through it adds to zero. Gaussian surfaces are imaginary constructions! 25 3. Gauss’ Law Consider a point charge surrounded by a Gaussian sphere. The electric field is: 1 q q E k 2 rˆ rˆ 2 4e 0 r r where e0 = permittivity of free space = 8.85 x 10-12 C2 N-1 m-2 The electric flux through the surface is then: f E E d A Radial field lines are always normal to sphere S Flux lines E dA S q 4e 0 r 2 dA sphere q 2 4 r 2 4e 0 r Gaussian surface 26 Gauss’ Law in General f E E d A S Qencl qi i Qencl e0 Gauss’ Law states that the electric flux through any closed surface enclosing a point charge Q is proportional to Q. The surface need not be centred on Q and can be any shape. Example: Coulomb’s Law from Gauss’ Law What is the electric field due to a point charge? Consider a Gaussian sphere of radius r centred on a charge q. Only interested in radial field direction. All fields in other directions cancel. 27 Coulomb from Gauss Consider surface elements dA If E is along dA then: E.dA = E dA cos(0º) = E dA Hence: E E d A E dA S S E dA +q S E 4 r 2 dA From Gauss’ Law: E 4 r 2 q r e0 Gaussian surface Rearranging: q E 4 e 0 r 2 Which, since F = qE, gives Coulomb’s Law, where we put E radially outward from the charge q. 28 Applications of Gauss’ Law Use Gauss’ law to find electric flux or field in a symmetrical situation. Shape of the Gaussian surface is dictated by the symmetry of the problem. Worked Example Find the electric field due to an infintely long rod, positively charged, of constant charge density, l. + + + + + + + + + + + + P 29 Electric Field of Long Rod Consider motion of a test charge. Only field lines radially away from the rod are important. Consider a Gaussian cylinder around part of the rod, radius r, height, h. Total flux through cylinder is: ftotal E d A top h + + + + + + + + + + dA dA r dA E d A E d A bottom side But, @ top & bottom E dA E.dA=0 For the side E is parallel to dA E.dA=E dA 30 Field due to a Long Rod fE E d A E dA E dA total side side dA side area of a cylinder of height, h side 2π rh lh fE 2 rh e0 e0 q Gauss’ Law Rod Charge Density l E 2 e 0 r Compare this with our previous result. E varies as 1/R 31 Charged Spherical Shell Worked Example E-field inside & outside a charged spherical shell (e.g. plane, car) Outside the shell Use a Gaussian sphere of radius r centred on the shell. Then: E.dA = E dA (since E ||dA) Q e0 E d A E dA r E 4 r 2 R E E Q 4 e 0 r 2 32 Inside a Charged Spherical Shell Inside the shell r < R so consider a Gaussian sphere inside the shell. no net charge enclosed by sphere Qencl = 0, so f E Q e0 0 Inside the shell the field is zero: a physically important result. No field inside the shell Faraday Cage!! E 33 Solid Polarisable Sphere Worked Example What is the electric field outside & inside a solid nonconducting sphere of radius R containing uniformly distributed charge Q. Outside the sphere: r > R consider spherical Gaussian surface fE E d A Qencl E 4 e 0 r 2 As before Qencl e0 0 + 34 Inside the Solid Sphere Inside the sphere r < R Charge enclosed by a Gaussian sphere of radius r<R is: Q charge density volume Q 4 r 3 3 4 R 3 3 r3 Q 3 R r R 35 Field Inside Solid Charged Sphere Hence, from Gauss’ Law: Q r3 1 E Q 3 2 4 e 0 r R 4 e 0 r 2 E Q E r 3 4 e 0 R Q 4 e 0 R 2 ~r 1 ~ 2 r r R The same behaviour is found for other forces, e.g. gravity. 36 Behaviour of Charges & Fields Near Conductors The electric field is zero everywhere inside a conductor. Electrons move to create an E field which opposes any external field. Free charges move to the outside surfaces of conductors A result of Gauss’ law. The electric field near a conductor is perpendicular to its surface. A parallel component would move charges and establish an electric field inside. 37 Why Doesn’t My Radio Work The electric field outside a charged conductor is: where: Q s s E e0 area Proof Consider a Gaussian cylinder straddling the conductor’s surface. fC En dA ++++++ + ++ +E = 0 + + + + + + + +++ Q e0 En A En dA Q s En e0 A e0 Closed hollow conductors admit no electric field EM shielding “Faraday Cages” Car Radios and biomagnetics 38 Importance and Tests of Gauss’ Law Coulomb’s law experimental evidence of Gauss’s law 1/r2 law is the key prediction Gauss’ law is so basic that its essential to test its validity Tests of F 1/r2±d Robinson 1769 d = 06 Cavendish 1773 0.02 Coulomb 1785 0.10 Maxwell 1873 5 x 10-5 Plimpton & Lawton 1936 2 x 10-9 Williams, Faller & Hill 1971 3 x 10-16 39 4. Electric Potential: Technology Can’t Live Without It! Technology relies on using energy associated with electrical interactions Work is done when Coulomb forces move a charged particle in an electric field. This work is expressed in terms of electric potential (energy) Electric potential is measured in Volts. Basic to the operation of all electric machines and circuits. 40 Mechanical Analogue In mechanics Work done in moving from point ab b Wab F d S a results in a change in potential energy: W a b= Ua - Ub When W >0 Ua > Ub e.g. a mass falling under gravity ab 41 What is Electric Potential? In electricity Consider a test charge q0 moving with respect to a charge, q, fixed at the origin. The work done is: b Wab q0 E d S a When integrated along the path and thus: b U q0 E d S a This is the change in electric potential energy, for a charge q0 moving from a b. 42 Electric Potential Energy Since: b U F d S a qq0 d S qq0 1 1 U 2 4 e 0 a rab 4 e 0 ra rb b By definition, a charge infinitely far away has zero potential energy. The electric potential energy between 2 charges is then: 1 qq0 qq0 U r k 4 e 0 r r Since this is a scalar the total potential energy for a system of charges is: q1q2 q1q3 U k r13 r12 43 Uranium Nucleus Example Worked Example Calculate the electrostatic potential energy between 2 protons in a Uranium nucleus separated by 2 x 10-15 m. q1q2 U r k r 1.6 x 10 9.0 x 10 -19 2 9 2 x 10 -15 13 ~ 10 J 44 Electric Potential Definition: Electric potential is potential energy per unit charge: V r U r q0 where U(r) is the potential energy of test charge q0 due to a charge distribution. V(r) is a property of the charges producing it, not q0. Volt = unit of electric potential 1 V = 1 volt = 1 J/C Note also that 1 V/m = 1 N/C 45 Potential & Charge Distribution For a single point charge; q, a distance r away, the electric potential is: q q V r k r 4 e 0 r Potential is zero if r = For a collection of charges: n qi V k i 1 ri For a charge distribution: dq V k r 46 Electric Potential Difference Difference in electric potential for a charge q between points a and b. 1 1 V Vb Va kq rb ra U Ua 1 V b F ds q0 q0 a b b V E d s Ed a For a uniform field, d || E i.e potential difference can be expressed as a path-independent integral over an electric field. All charge distributions have an electric potential The potential difference Va - Vb is the work/unit charge needed to move a test charge from a b without changing its kinetic energy. 47 The electron volt For the definition of volt, 1J of work is needed to move 1 C of charge through a potential difference of 1V A more convenient unit at atomic scales is the electron-volt: The energy gained by an electron (or proton) moving through a potential difference of 1 volt: 1 eV = (1.6 x 10-19 C)(1 V) = 1.6 x 10-19 J Not an SI unit but a very useful one! Worked Example In a hydrogen atom the e- revolves around the p+ at a distance of 5.3 x 10-11 m. Find the electric potential at the e- due to the p+, and the electrostatic potential energy between them. 48 Worked Examples e- A very simplistic picture p+ r Electric potential due to proton: q 9 x 109 1.6 x 10-19 V r k 27 V -11 r 5.3 x 10 Electrostatic p.e. is given by: q1q2 U12 k r12 e V p 1.6 x 10-19 27 4.3 x 10-18 J 49 Forces on Charged Particles Worked Example In a CRT an electron moves 0.2 m in a straight line (from rest) driven by an electric field of 8 x 103 V/m. Find: (a) The force on the electron. (b) The work done on it by the E-field. (c) Its potential difference from start to finish. (d) Its change in potential energy. (e) Its final speed. 50 Worked Examples (a) Force is in opposite direction to the E-field, magnitude: F qE 1.6 x 10-19 8 x 103 1.3 x 10-15 N (b) Work done by force: Work Fs 1.3 x 10-15 0.2 2.6 x 10-16 J (c) Potential difference is defined as work/unit charge: W 2.6 x 10-16 3 V 1.6 x 10 V -19 q 1.6 x 10 Alternatively (e- opposite to p+): b d a 0 V E d s E dx Ed 8 x 103 0.2 1.6 x 103 V 51 Worked Examples (d) Change in potential energy: b U q0 E d s a q0 V - 1.6 x 10-19 1.6 x 103 - 2.6 x 10-16 J work done (e) Loss of PE = gain in KE = ½mv2 2KE v m 22.6 x 10-16 9.1 x 10-31 2.4 x 107 ms -1 52 Worked Examples Worked Example A proton is accelerated across a potential difference of 600 V. Find its change in K.E. and its final velocity. By definition, 1 eV = 1.6 x 10-19 J. Acceleration across 600 V Proton gains 600 eV. K.E. = 600(1.6 x 10-19) = 9.6 x 10-17 J Final velocity is: 29.6 x 10-17 v 1.7 x 10-27 3.4 x 105 ms -1 If it started from rest 53 Equipotentials Regions of equal electric potential may be joined by contour lines. These are equipotentials. In 3-D these can form equipotential surfaces where the potential is the same at each point on the surface. Field lines and equipotentials are always perpendicular. No work is done in moving a charge along an equipotential surface because there is no change in potential. The surface of a conductor is an equipotential since charge is uniformly distributed across the surface of conductors. 54 Obtaining E from the Electric Potential Recall: b b a a Vb Va dV E d s b b a a dV E d s If the direction s is parallel to E for infinitesimal elements ds from a to b dV E ds dV E ds Electric field is the rate of change of potential V in the direction ds. In 3-D space we use x, y & z components to express in terms of partial derivatives. dV Ex , dx dV Ey , dy dV Ez dz 55 Vector Notation In vector notation: V ˆ V ˆ V ˆ E i j k x y z ˆ ˆ ˆ i j k V y z x V Where is the gradient operator. 56 Vector Notation Example Worked Example A potential distribution in space is described by: V = Axy2 - Byz where A and B are constants. Find the electric field. E V dV Ay 2 dx dV 2Axy - Bz dy dV -By dz E Ay 2 iˆ 2 Axy Bz ˆj By kˆ 57 Potential Due to Charge Distributions If E is known, use: b V Vb Va E d s a If E is not known, use: n qi V k i 1 ri for continuous charge distributions: dq V k r 58 Parallel Plates Worked Example Two parallel metal plates have an area A = 225 cm2 and are l=0.5 cm apart, with a p.d. of 0.25 V between them. Calculate the electric field. ds V Vleft Vright 0V 0.25V b E d s a l E dx 0 l E dx 0.1V 0.2V 0 El x =0 x =0.5m V 0.25 50 Vm -1 l 0.5 This is obvious from the definition of units of electric field = V/m. E 59 Uniformly Charged Disc Worked Example Find the electric potential and electric field along the axis of a uniformly charged disc of radius R and total charge Q. R y (y2 + x2)½ x P dy Consider the disc divided into rings of radius,r, width, dr. dq k V k dq 2 2 r y x 60 Uniformly Charged Disc (contd) For the ring shown: dV For the total potential we integrate over all rings: R V k 0 k dq y2 x2 dq y2 x2 By definition of charge density: Q Q s for disc 2 area R For the ring: dq = s 2 y dy s 2 y dy V 4e 0 0 y 2 x 2 R s 2 2 y x 0 2e 0 s R2 x2 x 2e 0 R 61 Uniformly Charged Disc The field is only in the x direction. E Ex dV dx 1 Q 2 2 2 E R x x 2 x 2e 0 R x 1 1 2 2e 0 R R 2 x 2 2 Q 62 Why Sparks Occur at Pointed Tips Recall: Conducting objects contain zero electric field. Charge resides on outer surface This surface is an equipotential. Equipotential surfaces outside the conductor are parallel to its surface. For curved conductors, surface 1 charge density: s r 1 Hence: E (radius of curvature) r Small radius implies s and E are large E.g. at points and tips 63 St. Elmo’s Fire Regions of strong E-field Corona discharge Ionisation of air greenish glow (St. Elmo’s Fire) E > 3 x 106 V/m Ionisation Current flow Carry away excess charge Lightning conductors Do not attract lightning Introduce a lower potential difference region close to clouds. + ++ +++ + ++ + + + + + ++ ++ ++ ++ + 64 Uses in Technology Accelerators (1929) (Giancoli Section 44.2, p1115) Van der Graaf HV Accelerator Works because E-field inside Gaussian sphere is zero 1m sphere 3 x 106 V Up to 20 MV produced Precipitators (See Figure shown) Remove dust and particles from coal combustion -ve wire @ 40 - 100 kV E-field particles to wall > 99% effective. Photocopiers (1940) (Giancoli Example 21.5, p555) Image on +ve photoconductive drum Charge pattern -ve toner pattern Heat fixing +ve paper. 65