University Physics 227N/232N Old Dominion University. More Electrostatics 20:3 Electric Field and Forces

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University Physics 227N/232N Old Dominion University More Electrostatics 20:3 Electric Field and Forces Dr. Todd Satogata (ODU/Jefferson Lab) satogata@jlab.org http://www.toddsatogata.net/2014-odu Wednesday, January 15 2014 Happy Birthday to Edward Teller, Gene Krupa, Dove Cameron, Skrillex, and Drew Brees! Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 1

Review: Electric Charge Electric charge is a fundamental property of matter. Many particles, including the electron and proton, carry electric charge. Charge comes in two varieties, positive and negative. Most charged particles carry exactly one elementary charge, e, either positive or negative. The proton carries exactly charge +e, the electron exactly charge e. The quarks, which make up protons, neutrons, and other subatomic particles, carry ±1/3 e or ±2/3 e. But they re never observed in isolation. The charge in a closed system is conserved The algebraic sum of charges remains unchanged. This is true even if new particles are created or destroyed. The SI unit of charge is the coulomb (C), equal to approximately 6.25 10 18 elementary charges. Thus e is approximately 1.602 10 19 C. Charges interact with electromagnetic radiation (light, radio, ) Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 2

Electric Charges and Atoms Most atoms are roughly electrically neutral Same number of electrons and protons Reality: many electrons are promiscuous and move among nuclei quite a lot. Macroscopic objects have over ~10 20 atoms If electrons and protons did not have very nearly exactly equal and opposite charges, large objects would experience unusual electrical forces (contrary to observation) Proton and electron electric charges are equal and opposite to the best precision we have every measured them Best measurement: 1.602176565(35) 10 19 C About 1 part in 46 billion! (Most accurate physical measurement: hyperfine rubidium transition frequency: 1 part in 417 quadrillion) (That s about the same relative accuracy as the thickness of a hair to the distance to the sun!) Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 3

Review: Coulomb s Law and the Electric Force Like charges repel, and opposite charges attract, with a force that depends on The product of the two charges The inverse square of the distance between them Mathematically, the electric force is described by Coulomb s law (much like the force of gravity): ~F 12 = Gm 1m 2 Gravity r 2 ˆr ~F 12 = kq 1q 2 Coulomb r 2 ˆr k 9.0 10 9 Nm 2 /C 2 ~F 12 Here (Coulomb) is the force q 1 exerts on q 2 and is a unit vector pointing from q 1 toward q 2. Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 4

That ˆr thing is kinda important ˆr(force on q 1 from q 2 ) q 1 q 2 ˆr(force on q 2 from q 1 ) ~F 12 = kq 1q 2 Coulomb r 2 ˆr Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 5

That ˆr thing is kinda important ˆr(force on q 1 from q 2 ) q 1 q 2 Newton s Third Law If object A exerts a force on object B, then object B exerts and oppositely directed force of equal magnitude on A. ˆr(force on q 2 from q 1 ) ~F 12 = kq 1q 2 Coulomb r 2 ˆr Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 6

The Superposition Principle The electric force obeys the superposition principle. That means the force two charges exert on a third force is just the vector sum of the forces from the two charges, each treated without regard to the other charge. The superposition principle makes it mathematically straightforward to calculate the electric forces exerted by distributions of electric charge. The net electric force is the sum of the individual forces. Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 7

Example: Coulomb Force from Multiple Charges What is the force (magnitude and direction; force is a vector!) on a proton at the origin from the two other charges shown below? ĵ q 1 = e at (1 cm, 1 cm) q 2 = e at ( 1 cm, 0 cm) proton,q =+e î Draw the picture, including vectors for forces Calculate magnitudes of forces Calculate and add components Find final total force and direction; evaluate Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 8

Example: Continued ĵ q 1 = e at (1 cm, 1 cm) r 2 =1cm ~F 1 r 1 = p 2 cm F 1 F 1y ~F 2 î proton,q =+e F 1x q 2 = e at ( 1 cm, 0 cm) F 2y =0 F 2x = F 2 sin(45 )=F 1y /F 1 cos(45 )=F 1x /F 1 F 2 = kqq 2 /r 2 2 F 1 = kqq 1 /r 2 1 Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 9

Example: Continued ĵ q 1 = e at (1 cm, 1 cm) r 2 =1cm ~F 1 r 1 = p 2 cm F 1 F 1y ~F 2 î proton,q =+e F 1x q 2 = e at ( 1 cm, 0 cm) F 2y =0 sin(45 )=F 1y /F 1 F tot,x = F 1x + F F 2x 2x = F 2 cos(45 )=F 1x /F 1 F 2 = kqq 2 /r2 2 = F 1 cos(45 ) F 2 F 1 = kqq 1 /r1 2 (be careful!) = kq q1 r 2 1 cos(45 ) 1 = ke 2 1 p r1 2 2 r2 2 q 2 r2 2 = 1.5 10 24 N=F tot,x Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 10

Example: Continued ĵ q 1 = e at (1 cm, 1 cm) r 2 =1cm ~F 1 r 1 = p 2 cm F 1 F 1y ~F 2 î proton,q =+e F 1x q 2 = e at ( 1 cm, 0 cm) F 2y =0 sin(45 )=F 1y /F 1 F 2x = F F tot,y = F 1y + F 2y 2 cos(45 )=F 1x /F 1 F 2 = kqq 2 /r2 2 = F 1 sin(45 )+0 F 1 = kqq 1 /r1 2 = kqq 1 r1 2 sin(45 ) = ke2 (be careful!) p =+8.2 10 25 N=F r1 2 tot,y 2 Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 11

Example: Continued ĵ ~F tot ~F 1 ~F 2 proton,q =+e î 1.5 10 24 N=F tot,x F tot = q F 2 tot,x + F 2 tot,y +8.2 10 25 N=F tot,y = tan 1 (F tot,y /F tot,x ) = 29 Makes sense more strongly attracted to closer charge! Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 12

Example: Balloon on the Ceiling You rub a 30g balloon on your head and stick it to the ceiling Assume the balloon just barely sticks to the ceiling and the charges on the balloon (negative) are 1mm from the same number of positive charges on the ceiling About how many excess electrons did you rub onto the balloon? F g = mg =(0.03 kg)(9.8 m/s 2 )=0.3 N=F electrostatic F electrostatic = kq2 r 2 Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 13

Example: Balloon on the Ceiling You rub a 30g balloon on your head and stick it to the ceiling Assume the balloon just barely sticks to the ceiling and the charges on the balloon (negative) are 1mm from the same number of positive charges on the ceiling About how many excess electrons did you rub onto the balloon? F g = mg =(0.03 kg)(9.8 m/s 2 )=0.3 N=F electrostatic q = r Felectrostatic r 2 k F electrostatic = kq2 = r 2 s (0.3 N)(10 3 m) 2 9 10 9 Nm 2 /C 2 =5.8 10 9 C q e = 5.8 10 9 C 1.6 10 19 C 3.6 1010 electrons Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 14

The Electric Field The electric field at a point in space is the force per unit charge that a charge q placed at that point would experience: ~E = ~ F q N C = V m The force on a charge q in an electric field ~E is ~F = q E ~ Electric field is analogous to the gravitational field, which gives force per unit mass. Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 15

Vector Field Examples http://hint.fm/wind/ Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 16

Vector Field Examples Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 17

Fields of Point Charges and Charge Distributions The field of a point charge is radial, outward for a positive charge and inward for a negative charge. ~E point charge = kq r 2 ˆr The superposition principle shows that the field due to a charge distribution is the vector sum of the fields of the individual charges. ~E total = X i ~E i = X i kq i r 2 i Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 18 ˆr i

Demo: Electric Field of Dreams (5-10 minutes) http://phet.colorado.edu/en/simulation/efield ~E total = X i ~E i = X i kq i r 2 i ˆr i Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 19

Demo: Electric Field Hockey http://phet.colorado.edu/en/simulation/electric-hockey Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 20

Electric Field Hockey (10 minutes) http://phet.colorado.edu/en/simulation/electric-hockey Click Download or Run Now! to run Java simulation Click and drag electric charges from box in upper right Click Start to watch how the black proton moves Click Field checkbox to see net electric field Remember that a positive charge moves in the direction of the electric field! Click Trace to draw the path of the proton as it moves Add more charges! Watch pretty pictures! You can see how this is very complicated, nonlinear motion This is not first semester motion with constant acceleration When you feel good, try Difficulties 1, 2, 3 to see if you can hit the goal on the right with your proton Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 21

DC Accelerating Gaps: Cockcroft-Walton Accelerates ions through successive electrostatic voltages First to get protons to >MeV Continuous HV applied through intermediate electrodes Rectifier-multipliers (voltage dividers) Limited by HV sparking/breakdown FNAL still uses a 750 kv C-W Also example of early ion source H gas ionized with HV current Provides high current DC beam Rectifiers Source Target ~1930 1.4 MeV Cavendish Lab Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 22

DC Accelerating Gaps: Van de Graaff How to increase voltage? R.J. Van de Graaff: charge transport Electrode (1) sprays HV charge onto insulated belt Carried up to spherical Faraday cage Removed by second electrode and distributed over sphere Limited by discharge breakdown ~2MV in air Up to 20+ MV in SF 6! Ancestors of Pelletrons (chains)/ Laddertrons (stripes) Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 23

Van de Graaff Popularity Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 24

The Dipole: an Important Charge Distribution An electric dipole consists of two point charges of equal magnitude but opposite signs, held a short distance apart. The dipole is electrically neutral, but the separation of its charges results in an electric field. Many charge distributions, especially molecules, behave like electric dipoles. The product of the charge and separation is the dipole moment: p = qd. Far from the dipole, its electric field falls off as the inverse cube of the distance. Prof. Satogata / Spring 2014 ODU University Physics 227N/232N 25