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Electricity and Magnetism

Electricity and Magnetism. Electric Field. F = q E. Or E = F / q. where q is a test charge. Electric Field of a Charge. Since F= kQq / r 2 We have E=F/ q = kQ/r 2. Gravitational. +. A.

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Electricity and Magnetism

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  1. Electricity and Magnetism

  2. Electric Field F=qE Or E=F/q where q is a test charge

  3. Electric Field of a Charge Since F=kQq/r2 We have E=F/q=kQ/r2

  4. Gravitational + A m F=mg Q b F=QE - B Earth + A PE=mgh (KE=0) F=mg W=PE W=QEd(Charge) W=mgh(mass) PE=QEd (KE=0) F=QE h c d PE=0 PE=0 - B Earth + A d PE=0 KE=mgh PE=0 KE=QEd - B Earth Electric + A g a E - B Earth

  5. Potential difference is the work done to move the charge from one point to another. Only potential difference between two pointsis physically meaningful DV=Vab=Vb-Va=Wab/Q Equipotential lines a E d b Vab=Ed(uniform electric field)

  6. V V r r + Electric Potential due to a Point Charge V=kQ/r(V=0 at r=∞) r r Electric Potential due to Uniform Electric Field Vab=Ed

  7. Electric force:F=kQAQB/r2, • Electric Field: E=F/Q • Electric potential:V=PE/q • Potential difference: • Vab=Vb-Va=DPE/q=Wba/q or Wba=qVab • (a) For uniform electric field:Vab=Ed • (b) For electric field due to a point charge: • V=kQ/r, (V=0 at r=∞)

  8. (e) X

  9. Q -Q + + + + + + + + - - - - - - - - d - + Capacitance C=Q/V Unit:farad, F. 1 F=1 coulomb/volt

  10. Dielectric Constant, Ka measure of how effective it is in reducing an electric field across the plates Q -Q Vo Co=Q/Vo Q -Q + + V=Vo/K C=Q/V=KQ/Vo=KCo + +

  11. + + + + + + + + - - - - - - - - d • Example:Imagine moving a proton from the negative plate to the positive plate. • The plates are 1.5 cm apart. The electric field is uniform with a magnitude of 1500 N/C. • What is the change in the proton’s electric potential energy? • What is the electric potential difference between the plates (voltage)? • If the proton is released from rest at the positive plate, what speed will it have • just before it hits the negative plate? • Solution: Given E=1500 N/C, q=+1.6x10-19 C, m=1.67X10-27 kg, d=1.5 cm=1.5X10-2 m • (a) DU=qEd=(1.6x10-19 C)(1500 N/C)(1.50x10-2 m)=3.60x10-18 J • (b) DV=DU/q=3.6x10-18 J/1.6x10-19 C=22.5 V • (c) DKE+DU=0 or DKE=-DU=mv2/2 • v=(2DU/m)1/2=6.57x104 m/s

  12. Ohm’s Law(German Physicist, Georg Ohm, 1787-1854) I=V/R Definition of Resistance R=V/I Theunit for R is Ohm, W 1W=1V/1A

  13. R=rL/A • ris the resistivity, and depends on the material used • r reflects intrinsic properties of the material • r is a function of temperature • The unit of r is Wm

  14. The work that must be done to take Q through V W=QV Since I=Q/t or Q=It, W=IVt(Electric work) Recall Power P=work done/time interval=W/t, P=IV (Electric Power) Using Ohm’s law I=V/R, P=I2R=V2/R The unit for Power is Watt, 1W=1J/s

  15. Ohm’s law V=IR Resistivity r=R•A/L (R=rL/A) Power P=IV=I2R=V2/R

  16. I R1 I R V Resistors in Series I I R2 R3 V=V1+V2+V3 I=I1=I2=I3 (Resistors in series have the same current) V=IR=IR1+IR2+IR3=I(R1+R2+R3) R=R1+R2+R3

  17. R3 I R2 I R1 I R V Resistors in Parallel I3 I2 I1 I=I1+I2+I3 V=V1=V2=V3 (Resistors in parallel have the same voltage) I=V/R=V/R1+V/R2+V/R3=V/(R1+R2+R3) 1/R=1/R1+1/R2+1/R3

  18. R3 I R2 I R1 IncreasingA1/R=1/R1+1/R2+1/R3+… R=rL/A > > R2 R3 R1 IncreasingL R=R1+R2+R3+…

  19. R’ =R’=R2R3/ (R2+R3); R2 = 1/R’=1/R2+1/R3=(R2+R3)/R2R3 R3 R’’ R’’=R1+R’ = R’ R1 R =R=R”R4/ (R”+R4) R” = R4 I2 I1 R1 R2 R’ R3 I3 I R” I > > R4 I4

  20. Capacitors in Combination Capacitors in Series -Q +Q -Q +Q -Q +Q C1 C2 C3 Q=CV or V=Q/C V=V1+V2+V3 >>> Q/C=Q/C1+Q/C2+Q/C3 1/C=1/C1+1/C2+1/C3 (Q=Q1=Q2=Q3) Resistors in series R=R1+R2+R3 (I=I1=I2=I3)

  21. Capacitors in Parallel -Q1 +Q1 C1 -Q2 +Q2 C2 C=εA/d -Q3 +Q3 C3 Q1=C1V, Q2=C2V, Q3=C3V Q=Q1+Q2+Q3 CV=C1V+C2V+C3V C=C1+C2+C3 Resistors in parallel 1/R=1/R1+1/R2+1/R3 (V=V1=V2=V3)

  22. x (a) 6.8 W

  23. Magnetic Force Magnetic forces arise from the interactions of moving charges in magnetic fields. (The electrical force is between two charges at rest)

  24. A Moving Charge in Magnetic Field Magnetic field, B, is defined in terms of the magnetic force F on a moving charge: Magnetic field magnitude: B=F/QvsinQ orF=BQvsinQ (electric force: F=EQ) Fmax=BQvif Q=90o or sinQ=1 Direction of B: Right-hand rule F v Q Q B Unit: 1 tesla=1 T=1 N/A·m (1T=104 gauss (G))

  25. An Electric Current in a Magnetic Field A magnet exerts a magnetic force on a current-carrying wire: B=F/IlsinQ or F=IlBsinQ (for moving charge F=qvBsinQ) Direction: Right-hand rule F=0 when Q=0o Fmax=IlB when Q=90o I B Q

  26. F=BQvsinQ for a moving charge F=BIlsinQ for a current

  27. Which one of the following statements concerning the magnetic force on a charged particle in a magnetic field is true? (a) It is a maximum if the particle is stationary. (b) It is zero if the particle moves perpendicular to the field. (c) It is a maximum if the particle moves parallel to the field. (d) It acts in the direction of motion for a positively charged particle. (e) It depends on the component of the particle's velocity that is perpendicular to the field. X

  28. Example: Alpha particles of charge q=+2e and mass m=6.6x10-27 kg and emitted from a radioactive source at a speed of 1.6x107 m/s. What magnetic field strength would be required to bend these into a circular path of radius r=0.25 m? Solution: To have a circular path, the magnetic field has to be perpendicular to the velocity, i.e. Q=90o or sinQ=1. The magnetic force provides the centripetal acceleration: qvB=mv2/r (rotational motion a=w2r=v2/r) thus B= mv2/rqv =(6.6x10-27 kg)(1.6x107 m/s)2 /(0.25m)(2x1.6x10-19 C)(1.6x107 m/s) =1.3 T v r ×× ×× B F

  29. 2r ••••• ••••• FE=qE, FM=qvB Stage 1: qE=qvB thus v=E/B Stage 2: qvB’=mv2/r so m=qB’r/v=qBB’r/v

  30. Example: If E=40,000 V/m, B=B’=0.08T and r=39.0 cm, what is the mass of the ions? Solution:E=vB (stage 1) or v=4.0x104 V/m/8.0x10-2 T=5.0x105 m/s qvB’=mv2/r (stage 2) so m=qB’r/v=(1.6x10-19 C)(8.0x10-2 T)(0.39m)/5.0x105 m/s =9.98x10-27 kg

  31. B=moI/2r > B > < > < r I Magnetic field at the center of a current loop B=moI/2pr I × r B

  32. Faraday’s Law EMF= - DFB/Dt

  33. EMF Induced in Moving Conductor Motional EMF=DFB/Dt=BDA/Dt=BlvDt/Dt=Blv ×××××××××××××××××××× ×××××××××××××××××××× ×××××××××××××××××××× ×××××××××××××××××××× ×××××××××××××××××××× ×××××××××××××××××××× I

  34. I2 I1 R1 R2 R’ R1=3W, R2=12 W, R3=8W, R4=6 W, R’=4.8W, R”=7.8 W R3 I3 I I” R” I > > R4 I4 A potential difference V=12 V is applied across the set of resistors shown. Find the current that flows through each resistor. Solution:V=Vupper=Vlower=12 V I4=V/R4=12V/6W=2 A I1=I”=V/R’’=12V/7.8W=1.54 A V’=I1R’=(1.54A)(4.8W)=7.4 V or V1=I1R1=(1.54V)(3W)=4.6V, V’=V-V1=12V-4.6V=7.4V I2=V’/R2=7.4V/12W=0.62 A I3=V’/R3=7.4V/8W=0.92 A

  35. At which of the labeled points will the electric field have the greatest magnitude? (a) G (c) A (e) D (b) I (d) H x At which of the labeled points will an electron have the greatest potential energy? (a) A (c) G (e) I (b) D (d) H x What is the potential difference between points B and E? (a) 10 V (c) 40 V (e) 60 V (b) 30 V (d) 50 V x

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