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PHYS113 Electricity and Electromagnetism Semester 2; 2002

PHYS113 Electricity and Electromagnetism Semester 2; 2002. Professor B. J. Fraser. 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

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PHYS113 Electricity and Electromagnetism Semester 2; 2002

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  1. PHYS113 Electricityand ElectromagnetismSemester 2; 2002 Professor B. J. Fraser

  2. 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

  3. 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!! + + + + + + + + +

  4. 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.

  5. Forces Between Charges • Coulomb’s Law • 1785: Coulomb experimentally determines force law between 2 charged point sources, q1 and q2. • Thus: • where k = 8.99 x 109 N m2 C-2 • Electric force has direction (vector) • Hence Coulomb’s Law is: • F12 is the force on q1 due to q2 • r12 is a unit vector from q2 to q1 along the line that joins them.

  6. q3 -3 nC 2.0 m q1 q2 +2 nC +2 nC 2.0 m 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

  7. j i Solution: Forces in a plane Force on q1 • This is due to q2 and q3 q3 F1 F13 F12 q1 q2

  8. j i Solution: Forces in a plane Force on q3 • This is due to q2 and q3 • Find magnitude as before q3 F31 F32 F3 q q1 q2

  9. 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.

  10. q0 Electric Field Region + + Q + F + + + + + + + + + + Test Charge Charged Object Electric Field Strength • Test an invisible force field? • See if a test object experiences a force! • Test for an electric field by measuring force experienced by a positive test charge. • We know that E F and: E 1/q0 • where q0 = charge of test charge • E = electric field, N C-1 • Hence: and since: • Then: • Electric field seen by q0 due to q.

  11. 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.

  12. Electric Fields in Nature • All charges (fixed & moving) produce an electric field that carries energy through space at the speed of light.

  13. j i 6 cm 4 cm + - A q1 = +12 nC q2 = -12 nC Field Due to Point Charges • Electric fields add vectorially: E = E1 + E2 + E3 + … • Thus: Worked Example • Find the electric field at point A for the dipole shown Field at A due to q1.

  14. EA1 A +q1 - q2 EA2 Field Due to Point Charges Field at A due to q2 • Total electric field at A: • No component in the j direction • Example of an electric dipole • Often found in nature (e.g. molecules) • For more: See Section 21.11

  15. j i ^ r dy dE L y R q O P -L r dE dQ 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.

  16. Field due to line segment (contd) But l = Q/L and thus dQ = l dy • Centre rod at origin • For every charge at + y, there is another corresponding charge at -y • Thus, fields in j component add to 0. cosq = R/r r2 =(R2 + y2) Can you do this integral?

  17. Solution to Field Due to a Line Segment • The solution to the field due to a line segment is: • So, what is the big deal? • Well, what happens if L >> R? Thus, the field from a line charge is proportional to 1/R and not 1/R2.

  18. P ^ ^ k j r R ^ i q L x dx 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 collection of line segments, L, long and, dx, wide

  19. 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: But i components sum to 0 Can you do this integral?

  20. Solution to Field Due to a Surface • The solution to the field due to a surface is: • 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!

  21. screen + + + + + + + + + e- - - - - - - - - - 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.

  22. 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 v = 2.6 x 108 ms-1 • i.e. 0.8 x speed of light

  23. A q 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 q to the flow: fw = vA cosq • In vector form: fw = vA • where the direction of A is normal to net

  24. Defining Electric Flux • For an irregular shape, area A is sum of infinitesimal elements dA. • Thus, summing over 2-D surface S: • Now, replace water with electric field, i.e. there is no physical motion. • The electric flux through a surface of area A is: • 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.

  25. The Gaussian Surface • The total electric flux (number of field lines) passing through this surface is: • 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! dA E

  26. Flux lines Gaussian surface 3. Gauss’ Law • Consider a point charge surrounded by a Gaussian sphere. • The electric field is: • where e0 = permittivity of free space = 8.85 x 10-12 C2 N-1 m-2 • The electric flux through the surface is then: Radial field lines are always normal to sphere

  27. Gauss’ Law in General • 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.

  28. Coulomb from Gauss • Consider surface elements dA • If E is along dA then: E.dA = E dA cos(0º) = E dA • Hence: • From Gauss’ Law: • Rearranging: • Which, since F = qE, gives Coulomb’s Law, where we put E radially outward from the charge q. E +q r dA Gaussian surface

  29. + + + + + + + + + + + + P Applications of Gauss’ Law • Use Gauss’ law to find electric flux or field in a symmetrical situation. • Shape of the Gaussiansurface 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.

  30. Electric Field of Long Rod dA h + + + + + + + + + + dA dA r • 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: • But, @ top & bottom EdA E.dA=0 • For the side E is parallel to dA E.dA=E dA

  31. Field due to a Long Rod Gauss’ Law Rod Charge Density Compare this with our previous result. E varies as 1/R

  32. r R E 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)

  33. E 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 • Inside the shell the field is zero: a physically important result. No field inside the shell Faraday Cage!!

  34. + 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 As before

  35. r R Inside the Solid Sphere Inside the sphere • r < R • Charge enclosed by a Gaussian sphere of radius r<R is:

  36. Field Inside Solid Charged Sphere • Hence, from Gauss’ Law: • The same behaviour is found for other forces, e.g. gravity. E r R

  37. 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.

  38. + + + + + + + + + E = 0 + + En + + + + + + + + dA + Why Doesn’t My Radio Work • The electric field outside a charged conductor is: • where: Proof • Consider a Gaussian cylinder straddling the conductor’s surface. • Closed hollow conductors admit no electric field • EM shielding  “Faraday Cages” • Car Radios and biomagnetics

  39. 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

  40. 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.

  41. Mechanical Analogue In mechanics • Work done in moving from point a  b • results in a change in potential energy: W a  b= Ua - Ub • When W a  b > 0 Ua > Ub • e.g. a mass falling under gravity

  42. 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: • When integrated along the path and thus: • This is the change in electric potential energy, for a charge q0 moving from a  b.

  43. Electric Potential Energy • Since: • By definition, a charge infinitely far away has zero potential energy. • The electric potential energy between 2 charges is then: • Since this is a scalar the total potential energy for a system of charges is:

  44. Uranium Nucleus Example Worked Example • Calculate the electrostatic potential energy between 2 protons in a Uranium nucleus separated by 2 x 10-15 m.

  45. Electric Potential Definition: • Electric potential is potential energy per unit charge: • 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

  46. Potential & Charge Distribution • For a single point charge; q, a distance r away, the electric potential is: • Potential is zero if r =  • For a collection of charges: • For a charge distribution:

  47. Electric Potential Difference • Difference in electric potential for a charge q between points a and b. • 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. For a uniform field, d || E

  48. 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.

  49. e - p+ r Worked Examples A very simplistic picture • Electric potential due to proton: • Electrostatic p.e. is given by:

  50. 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.

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