ElectricElectricFieldsFieldsandandDipoleGauss’s LawJanuary 14, 2014Physics for Scientists & Engineers 2, Chapter 221
Announcements! Homework set 1 is due 1/20! Lecture notes: linked from lon-capa, or directly athttp://www.pa.msu.edu/ schwier/courses/2014SpringPhy184/! Section 1 lecture notes:http://www.pa.msu.edu/ nagy t/phy184/lecturenotes.html! Strosacker learning center schedule – BPS 1248 Mo: 10am – noon, 1pm – 9pmTue: noon – 6pmWe: noon – 2pmTh: 10am – 1pm, 2pm – 9pm! Clicker: If you didn’t enter it on lon-capa, then email me your clickernumber Clickers only work for the section in which you are registeredJanuary 14, 2014Physics for Scientists & Engineers 2, Chapter 212
The Electric Field! The electric field is defined at any point in space as the netelectric force on a charge, divided by that charge F (r )E (r ) q! Electric field linesstart at positivecharges and end atnegative charges! Electric field line density: Test charge q High density of field lines – large field Low density of field lines – small fieldJanuary 14, 2014Physics for Scientists & Engineers 2, Chapter 224
The Electric Field and force! The electric force on a charge is parallel or antiparallel to theelectric field at that pointEFpositive chargeEFnegative charge! The electric force is F ( r ) qE ( r )January 14, 2014F qEPhysics for Scientists & Engineers 2, Chapter 225
Superposition of electric fields! The total electric field at any point is the vector sum of allindividual electric fields Enet Eii! Note that this means adding the x-, y-, and z- componentsseparatelyEnet ,X Ei ,XiEnet ,Y Ei ,YiEnet ,Z Ei ,ZiJanuary 14, 2014Physics for Scientists & Engineers 2, Chapter 216
Electric Field from 4 Point ChargesPROBLEM! We have four charges q1 10.0 nCq2 -20.0 nCq3 20.0 nCq4 -10.0 nC! These charges form a square ofedge length 5.00 cm! What electric field do the particlesproduce at the square center?January 14, 2014Physics for Scientists & Engineers 2, Chapter 228
Problem solving strategy! Use this approach to solve problems, in particular if at first you have no clue.Step 1ThinkStep 2SketchRecognize the problemWhat’s going on?ThinkDescribe the problemin terms of the fieldWhat does this have to do with ?SketchStep 3ResearchStep 4ExecuteStep 5Double-check1/14/14!!!!Plan a solutionHow do I get out of this?!Research!!Execute the planLet’s get an answer!Simplify, Calculate, RoundEvaluate the solutionCan this be true?!Draw a picturePhrase the question in your ownwordsRelate the question to somethingyou just learnedIdentify physics quantities, forces,fields, potentials, Find a physics principle(symmetry, conservation, )Write down the equationsSolve equations, starting withintermediate stepsCheck units, order-of-magnitude,insert into original question, Double-checkPhysics for Scientists & Engineers 29
Electric Field from 4 Point ChargesPROBLEM! We have four charges q1 10.0 nCq2 -20.0 nCq3 20.0 nCq4 -10.0 nC! These charges form a square ofedge length 5.00 cm! What electric field do the particlesproduce at the square center?SOLUTIONTHINK! Each of the four charges produces a field at the center! We can use the principle of superposition The electric field at the center of the square is the vector sum of theelectric field from each chargeJanuary 14, 2014Physics for Scientists & Engineers 2, Chapter 2210
Electric Field from 4 Point ChargesSKETCH! We define an x-y coordinate system We place q2 and q3 on the x axis We place q1 and q4 on the y axis The center of the square is atx y 0 The sides of the square area 5.00 cm The distance from each charge tothe center is rRESEARCH! The electric field at the center of the square is given by theprinciple of superposition Ecenter E1 E2 E3 E4January 14, 2014Physics for Scientists & Engineers 2, Chapter 2211
Electric Field from 4 Point Charges! We can write the field in terms of the x- and y-componentsEcenter,x E1,x E2,x E3,x E4,xEcenter,y E1,y E2,y E3,y E4,y! All four charges are a distance r from the center2a ar 22! The electric field component from each charge at the centerof the square is given byqir2SIMPLIFYEi k! Let’s start with the x-componentsEcenter,x E1,x E2,x E3,x E4,xJanuary 14, 2014q2q3 k k 2 k 2 2 ( q2 q3 )rrrPhysics for Scientists & Engineers 2, Chapter 2212
Electric Field from 4 Point Charges! Now the y-componentsEcenter,y E1,y E2,y E3,y E4,y kq1q4 k k 2 ( q1 q4 )22rrr! The magnitude of the electric field at the center is22Ecenter Ecenter,x Ecenter ,yEcenterk 2r(q2 q3 ) (q1 q4 )222k 2a(q2 q3 ) (q1 q4 )22 2 a2 r 2 ! The angle is given by k q q()142 Ecenter,y 1 1 r 1 q1 q4 θ tan tan tan kq q Ecenter,x 23 2 ( q2 q3 ) rJanuary 14, 2014Physics for Scientists & Engineers 2, Chapter 2213
Electric Field from 4 Point ChargesCALCULATE! Putting in our numerical values we get2k22q q q q(( 1 4)23)2a(q2 q3 ) ( 20.0 nC ) ( 20.0 nC ) 40.0 nC 40.0 10 9 CEcenter (q1 q4 ) (10.0 nC ) ( 10.0 nC ) 20.0 nC 20.0 10 9 CEcenter 2 9 Nm 2 8.99 10C 2 (0.0500 m )2( 40.0 10 9C ) ( 20.0 10 C )2 92Ecenter 321636.0179 N/CJanuary 14, 2014Physics for Scientists & Engineers 2, Chapter 2214
Electric Field from 4 Point Charges! For the angle we get q1 q4 θ tan q q 123 10.0 ( 10.0) 1θ tan tan 0.500 )( 20.0 20.0 1θ 26.56505118 ROUND! We round our results to three significant figuresEcenter 3.22 10 5 N/Cθ 26.6 January 14, 2014Physics for Scientists & Engineers 2, Chapter 2215
Electric Field from 4 Point ChargesDOUBLE-CHECK! What about the angle? E3 E4January 14, 2014x E2 E1θ 26.6 Physics for Scientists & Engineers 2, Chapter 22 Ecenter16
Electric Field from an Electric Dipole! A system of two oppositely charged point particles is calledan electric dipole! The vector sum of the electric field from the two chargesgives the electric field of the dipole! We have shown the electric field lines from a dipole! We will derive a generalexpression goodanywhere along thedashed line and then getan expression for theelectric field a longdistance away from thedipoleJanuary 14, 2014Physics for Scientists & Engineers 2, Chapter 2218
Electric Field from an Electric Dipole! Start with two charges on the x-axis a distance d apart Put -q at x -d/2 Put q at x d/2! Calculate the electric field at a point P a distance x from theoriginJanuary 14, 2014Physics for Scientists & Engineers 2, Chapter 2219
Electric Field from an Electric Dipole! The electric field at any point x is the sum ofthe electric fields from q and –q1 q1 qE E E 24πε 0 r 4πε 0 r 2! Replacing r and r- we get the electric fieldeverywhere on the x-axis (except for x d/2) q 11E 2 2 114πε 0 ( x 2 d ) ( x 2 d ) ! Interesting limit far away along the positive xaxis (x d)January 14, 2014Physics for Scientists & Engineers 2, Chapter 2220
Electric Field from an Electric Dipole! We can rewrite our result as 2 2 qd d E 1 1 2 2x 4πε 0 x 2x ! We can use a binomial expansion or a Taylor expansiondnα , (1 α ) 1 nα .2xd 2(1 α ) 1 ( 2)α . 1 2 .2xd 2(1 α ) 1 ( 2)α . 1 2 .2xf (z) (1 z)n f (z) n 0f (n) (z 0)(z 0)nn!For small z d/2x, stop after n 1! So we can writeqE 4πε 0 x 2January 14, 2014q d d 1 x 1 x 4πε x 2 0qd 2d x 2πε x 30Physics for Scientists & Engineers 2, Chapter 2221
Definition of Electric Dipole Moment! We define the electric dipole moment as a vector thatpoints from the negative charge to the positive charge p qd p is the magnitude of the dipole moment q is the magnitude of one of the opposite charges d is the distance between the charges! Using this definition we can write the electric field far awayfrom an electric dipole aspE 2πε 0 x 3January 14, 2014Physics for Scientists & Engineers 2, Chapter 2222
Electric Dipole Moment of Water! Chemistry reminder the H2O molecule! The distribution of electric charge inan H2O molecule is non-uniform! The more electronegative oxygenatom attracts electrons from thehydrogen atoms! Thus, the oxygen atom acquires a partial negative chargeand the hydrogen atoms acquire a partial positive charge! The water molecule is “polarized”January 14, 2014Physics for Scientists & Engineers 2, Chapter 2223
Electric Dipole Moment of WaterPROBLEM! Suppose we approximate the water molecules as twopositive charges located at the center of the hydrogen atomsand two negative charges located at the center of the oxygenatom! What is the electric dipole moment of a water molecule?SOLUTION! The center of charge of the two positivecharges is located exactly halfway between thecenters of the hydrogen atoms! The distance between the positive and negativecharge centers is θ d Δr cos (10 10 m ) cos(52.5 ) 0.6 10 10 m 2 January 14, 2014Physics for Scientists & Engineers 2, Chapter 2224
Electric Dipole Moment of Water! This distance times the transferred charge, q 2e,is the magnitude of the dipole moment of waterp 2ed 2 (1.6 10 19 C )( 0.6 10 10 m ) 2 10 29 C m! This oversimplified calculation comes fairly closeto the measured value of the electric dipolemoment of water of 6.2·10-30 C m! The fact that the real dipole moment of water is less thanour calculated result indicates that the two electrons of thehydrogen atoms are not pulled all the way to the oxygen, butonly one-third of the wayJanuary 14, 2014Physics for Scientists & Engineers 2, Chapter 2225
Demo: electric force on dipole! Demo: electric force on conducting end of stick balanced atmid-point! Demo: electric force on water Force due to dipole moment of water! Demo: electric force on wooden end of stick balanced atmid-pointJanuary 14, 2014Physics for Scientists & Engineers 2, Chapter 2126
Explanation of the Demo! The presence of the charged stick polarizes the atoms evenon the wooden side of the stickJanuary 14, 2014Physics for Scientists & Engineers 2, Chapter 2127
January 14, 2014 Physics for Scientists & Engineers 2, Chapter 22 5 The Electric Field and force ! !e electric force on a charge is parallel or antiparallel to the electric "eld at that point ! !e electric force is F qE points toward charge 2, as shown in Figure 21.
Radiation by a Hertzian Dipole 253 Hertzian dipole. Figure 25.3: Schematics of a small Hertzian dipole. In (25.1.1), lis the e ective length of the dipole so that the dipole moment p ql. The charge qis varying in time harmonically because it is driven by the generator. Since dq dt I; we have Il dq dt l j!ql j!p (25.1.2) for a Hertzian dipole.
and an electric field is given to leading order by the Electric Dipole approximation: The dipole moment of a pure dipole: -Vector quantity -Points from - to . -Magnitude is charge _ distance For Hydrogen atom this gives: Electric Dipole Approximation d q(r r ) rrr V E DE(r CM) rr Semi-Classical' Approx .
Example: Force and torque on an electric dipole Consider an electric dipole in a uniform E-field with magnitude 5.0 x 105 N/C. The two charges are of magnitude 1.6 x 10-19 C and are separated by a distance of 0.125 nm. Find a) net force exerted by the field on the dipole b) the electric dipole
External electric fields are needed to polarize electric dipole moments, which mixes rotational eigenstates with opposite parities. However, the dipole moment of these mixed states is unquantized and, thus the electric dipole moment is a classic vector. When two dipole moments are aligned, say, along the
Cushcraft R-7 Vertical Antenna for 10M/15M/20M/40M Cushcraft R-5 Vertical Antenna for 10M/15M/20M (SARC has two of these.) 6M Dipole 10M Dipole 15M Dipole 20M Dipole 75M Dipole (second one) G5RV Antenna Finance and Review – Paul Guido, N5IUT, has not
Random Length Radiator Wire Antenna 6 6. Windom Antenna 6 7. Windom Antenna - Feed with coax cable 7 8. Quarter Wavelength Vertical Antenna 7 9. Folded Marconi Tee Antenna 8 10. Zeppelin Antenna 8 11. EWE Antenna 9 12. Dipole Antenna - Balun 9 13. Multiband Dipole Antenna 10 14. Inverted-Vee Antenna 10 15. Sloping Dipole Antenna 11 16. Vertical Dipole 12 17. Delta Fed Dipole Antenna 13 18. Bow .
Random Length Radiator Wire Antenna 6 6. Windom Antenna 6 7. Windom Antenna - Feed with coax cable 7 8. Quarter Wavelength Vertical Antenna 7 9. Folded Marconi Tee Antenna 8 10. Zeppelin Antenna 8 11. EWE Antenna 9 12. Dipole Antenna - Balun 9 13. Multiband Dipole Antenna 10 14. Inverted-Vee Antenna 10 15. Sloping Dipole Antenna 11 16. Vertical Dipole 12 17. Delta Fed Dipole Antenna 13 18. Bow .
A. Electric dipole moment The electric dipole moment is calculated using the total charge density r r 3 ionr d r i i e , 1 where the integration is over the supercell volume . ii e ef rn d3r 2 is the contribution to the total dipole moment from the elec-tron density n i r of the ith electronic state with spin and occupation f i, and .