8/30/16 Quiz 2 8/25/16 A positive test chage qo is eleased fom est at a distance away fom a chage of Q and a distance 2 away fom a chage of 2Q. How will the test chage move immediately afte being eleased? qo Q A. B. C. D. E. 2Q 2 To the left To the ight It doesn t move at all Moves upwad Moves downwad Continuous Chage Distibutions: Electic Field and Electic Flux Lectue 3 Last Week Discete Chages Magnitude of the electic field due to a point chage: E 1 q E P E Electic field due to a dipole at point P: E qd p 2πε o z 3 2πε o z 3 z - d q Dipole - q- cente 1
Today Continuous Chage Distibutions Two appoaches to calculating electic field Extension of point chage methods Coulomb s Law E kq d E k dq ˆ Focus on this today Integate Gauss Law Flux (E A) enclosed chage E n da 1 S ε 0 Text Refeence: Chapte2.1 Many useful examples 22-1 to 8, etc. Q inside Electic Field fom Coulomb s Law Discete chages vs. continuous chage distibution Pinciples emain the same Coulomb s Law Law of Supeposition Only change: - - - - - Electic Field fom Coulomb s Law Bunch of Chages P i - q - i - - - k q i 2 i ˆ i i E 1 Summation ove discete chages Continuous Chage Distibution P dq Chage density E ρdv dq σ da λdl 1 dq " ˆ de volume chage suface chage line chage Integal ove continuous chage distibution http://www.falstad.com/vecto3de/ Good visualization 2
Chage Densities How do we epesent the chage Q on an extended object? small pieces total chage of chage Q dq Line of chage: λ chage pe dq λ dx unit length [C/m] Suface of chage: σ chage pe unit aea [C/m 2 ] dq σ da σ dx dy (Catesian coodinates) Volume of Chage: ρ chage pe unit volume [C/m 3 ] dq ρ dv ρ dx dy dz (Catesian coodinates) Continuous Chage Distibutions Teminology used in the text Souce point, x s, whee chage ΔQ(x s ) is located Field point, x p, whee field ΔE(x p ) is evaluated Vecto fom x s to x p : x p - x s Supeposition Add up all the ΔE ceated by all chage elements ΔQ Limiting case: eplace sum by integal ove chage distibution 1 dq " E ˆ de Need to ewite befoe we ae able to evaluate this Some examples Chage Distibution Poblems 1) Undestand the geomety 2) Choose dq 3) Evaluate de contibution fom the infinitesimal chage element 4) Exploit symmety as appopiate 5) Set up the integal 6) Solve the integal 7) The esult 8) Check limiting cases 3
Field Due to Chage on a Line Calculate the electic field at point P due to a thin od of length L of unifom positive chage, with density λq/l Steps 1-3 E î E y ĵ d de cosθ de y de sinθ Pick any infinitesimal segment of the od dx s Chage of the segment is then dq λdx s Solve fist fo de k dq cosθ x s d k λ dx s cosθ d k λ x 2 x 1 cosθ dx s Field Due to Chage on a Line Calculate the electic field at point P due to a thin od of length L of unifom positive chage, with density λq/l Steps 4-7 Change integation vaiable fom x s to θ tanθ y p x s x s y p cotθ x 2 cosθ dx s x 1 sinθ y p y p sinθ d cotθ dx s y p y dθ p csc 2 θdθ θ 2 cosθ y p csc 2 θdθ 1 θ 2 θ 1 y p 2 / sin 2 θ y p θ 1 cosθ dθ Field Due to Chage on a Line Calculate the electic field at point P due to a thin od of length L of unifom positive chage, with density λq/l Steps 4-7 Evaluate integal, solve fo kλ θ 2 cosθ dθ kλ ( sinθ y 2 sinθ 1 ) p y p θ 1 kλ y p y p Similaly fo E y y p kλ 1 1 1 >0, 1 >0 1 E y kλ cotθ 2 cotθ 1 1 E y 0 y p 0 y p 0 4
Field Due to Chage on a Ring Calculate the electic field along z-axis due to a cicula ing of adius R of unifom positive chage, with density λ Steps 1-3 Pick any infinitesimal segment of the ing Total chage of the segment is then dq λds The electic field due to this (point) chage de 1 dq Symmety: diection of the field at point P must be along the positive z diection Field Due to Chage on a Ring Step 4 Exploit symmety The z-component of the field: de z de cosθ 1 dq cosθ 1 λds ( z 2 R 2 ) z z 2 R 2 zλ z 2 R 2 ( ) ( ) 1/2 3/2 ds Field Due to Chage on a Ring Steps 5-7 Integate ove the entie ing with λ unifom z and R fixed E E z de z zλ(2πr) z 2 R 2 ( ) 3/2 qz E z 2 R 2 ( ) 3/2 zλ ds ( ) 3/2 0 z 2 R 2 2πR q: the total chage of the ing 2λπR 5
Limiting Cases Step 8 qz E z 2 R 2 ( ) 3/2 Two limiting cases: The electic field vanishes as z appoaches zeo (o at the cente of the ing) The electic field appoaches that of a point chage at lage distances (i.e., lage z) qz E z 2 R 2 ( ) 3/2 1 q z 2 Field Fom Chaged Disk Poblem: calculate the electic field along z- axis due to a cicula disk of adius R of unifom positive chage, with density σ. Pick any ing element (of infinitesimal width d) of the disk The chage of the element is dqσ daσ (2π d) The electic field due to the ing element (σ 2πd)z de z 2 ( ) 3/2 E de σ z 4ε 0 z 2 R 2 z 2 Field Fom Chaged Disk Integate ove the entie disk R 0 σ z 4ε 0 dx X 3/2 2d ( z 2 ) 3/2 σ z 4ε 0 σ z 2X 1/2 z ( ) 2 R 2 4ε z 2 0 R d ( z 2 ) ( z 2 ) 3/2 0 whee we have set Xz 2 and dx2d z > 0 E σ 1 2ε 0 z z 2 R 2 z < 0 E σ 1 2ε 0 z z 2 R 2 6
Limiting Cases z > 0 E σ z 1 2ε 0 z 2 R 2 Small distance z (o lage disk) z 0 (o R ), E σ 2ε 0 E z Discontinuity at z 0 z Limiting Cases z > 0 E σ z 1 2ε 0 z 2 R 2 Lage distance z E σ 1 2ε 0 1 1 R 2 σ R 2 2ε 0 2z 2 z 2 1 σ (πr 2 ) 1 q z 2 z 2 Infinite Sheet of Positive Chage y σ E 2ε x σ 0 2ε 0 x Unifom electic fields geneated on both sides of sheet Discontinuity at x 0 7
y 90 Ac of Chage E x In this coodinate system you have to deal with both and E y Yʹ E In this coodinate system you only have to deal with the hoizontal component of E Xʹ unifom chage distibution total chage Q 90 Ac of Chage (cont d) dθ y λ Q 2Q 2π π 4 θ de x dq λ dθ d k dq kλ cosθ dθ cosθ kλ θ 2 cosθ dθ θ 1 kλ ( sinθ sinθ 1 ) Fo the 90 0 ac, θ 1 45 0 and θ 2 45 0, so kλ 1 1 2 2kλ 2k 2Q π 2 2kQ π 8