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Transcript
Lecture 2 Question from Last Class Simple Answer: Yes! Chapter 13. Electric Field Complicated Answer: Depends! Pauli Exclusion Principle: No two electrons in an atom can be in the same quantum state (n, l, ml, ms) *Superposition of Electric Fields http://www.hydropole.ch/hydropole/hydrogen/about.htm Coulomb’s Law of Electro-static Force ! F= 1 Q1Q2 r̂ 4!" 0 r 2 r + ! F =F= + ! F21 2 Force on “2” by “1” 1 Force repulsive and the permittivity constant is How Strong is the Coulomb’s Force? 1 Q1Q2 4!" 0 r 2 r + Compare Coulomb and gravitational forces for a proton and electron in a hydrogen atom - ! F21 2 1 Force attractive ! o = 8.55x10 "12 C 2 / Nm 2 1 Electric Field Electric field due to a Point Charge Observation: The net Coulomb force on a given charge is always proportional to the strength of that charge. F1 - + q1 F q0 test charge For a Point Charge: ! E1 = 1 q1 rˆ 4!" 0 r 2 F2 q2 ! ! ! F = F1 + F2 ! # q &q q F = 0 $$ 21 r̂1 + 22 r̂2 !! 4'( 0 % r1 r2 " Define the electric field, which is independent of the test charge, q0, and depends only on position in space: ! ! F E= q0 Electric Field q1 and q2 in F are the sources of the electric field, q0 is the test charge The Superposition Principle With this concept, we can “map” the electric field anywhere in space produced by any arbitrary: Charge Distribution Bunch of Charges E= F 1 qi r̂i # 4!" 0 ri2 + + + + - - + + E= 1 4!" 0 dqi #r 2 r̂ + + + + + + + + ++ + “Net” E at origin - These charges or this charge distribution “source” the electric field throughout space 2 Electric Field due to an Electric Dipole ! E+ Electric Field due to an Electric Dipole ! E+ P P ! E! ! E! r+ r+ x r- qd 2!" o r 3 E= The electric dipole moment (C-m): x ! ! p = qs r- + q+ s - q- + Dipole center q+ s - q- + Dipole center ! p Direction of p is from negative to positive - Rewrite magnitude of electric field at P: E= Electric Field due to an Electric Dipole ! E+ ! E! r+ x E! q+ s - q- ! E= + Dipole center ! p 1 r3 1 q rˆ 4!" 0 r 2 ! $+ s s r+ = 0, y ,0 % , ,0,0 = % , , y ,0 !( 2 2 !( #* ' r+ = r, = ! s s r%, = 0, y ,0 % , % , ,0,0 = , y ,0 !( 2 2 "!)( In general: r- + E along the y-axis p E= 2!" o r 3 P p 2!" o r 3 &s# y2 + $ ! %2" 2 ! E+ = 1 4() 0 ! E' = 1 4() 0 - 2 &s# y2 + $ ! %2" 2 s , y ,0 2 'q &s# y2 + $ ! %2" 2 s ' , y ,0 2 q &s# y2 + $ ! %2" &s# y2 + $ ! %2" 2 &s# y2 + $ ! %2" 2 3 E along the y-axis &s# y2 + $ ! %2" E along the z-axis ! ! ! E2 = E + + E ! = 1 4./ 0 ! ! ! E2 = E + + E ! = 1 4./ 0 2 ! if r>>s, then E2 ! Other Locations q 3 ! s,0,0 ' 2 - s *2 $ 2 %y + + ( " , 2 ) #" &% qs 3 ! 1,0,0 ' 2 - s *2 $ 2 %y + + ( " , 2 ) #" &% "1 qs , 0, 0 4#$ 0 r 3 at <0,r,0> Due to the symmetry E along the z-axis must be the same as E along the y-axis! Electric field For a Point Charge: ! E1 = 1 q1 rˆ 4!" 0 r 2 Dipole: for r>>s : y s -q z +q x ! E= 1 2qs , 0, 0 4!" 0 r 3 at <r,0,0> ! E= !1 qs , 0, 0 4"# 0 r 3 at <0,r,0> ! E= !1 qs , 0, 0 4"# 0 r 3 at <0,0,r> 4 Electric Dipole Moment Example Problem y E=? A dipole is located at the origin, and is composed of particles with charges e and –e, separated by a distance 2×10-10 m along the xaxis. Calculate the magnitude of the E field at <0,2×10-8,0> m. Since r>>s: E1, x = 200Å 1 sq 4!" 0 r 3 s !" Dipole moment: p = q ! Ls s is the separation of the two charges 2Å The direction of the dipole moment is taken to be from the negative to the positive end of the dipole. x p≈ 6×10-30 C·m∼0.04 e·nm Electric Dipole in a uniform electric field Choice of System Multi-particle systems: Split into objects to include into system and objects to be considered as external. •the charges that are the sources of the field • the charge that is affected by that field F = qE 5 A Point Charge in a Uniform Field • DEMO An particle of charge q injected in a E field will experience a force given by F = q E. The resulting acceleration can be found from Newton's second law. ! ! F = ma • ! q ! a= E m In a region of uniform E field, the particle will experience a constant acceleration and the resulting parabolic trajectory. x -----------------P v0 E + + ++ + + + + + + ++ + 6