Today in Physics 122: work, energy and potential in electrostatics

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Tody in Physics 1: work, energy nd potentil in electrosttics Leftovers Perfect conductors Fields from chrges distriuted on perfect conductors Guss s

1 Tody in Physics 1: work, energy nd potentil in electrosttics Leftovers Perfect conductors Fields from chrges distriuted on perfect conductors Guss s lw for grvity Work nd energy Electrosttic potentil energy, potentil, nd voltge Potentil s shortcut in E clcultions 0 Septemer 01 Physics 1, Fll 01 1

2 How to do tht integrl It s simple exercise in integrtion y prts: u First, let x = u, dy = e du dx = udu, y = e u u u e u du = u e + e udu Now, in lst term, let x = u, dy = e du dx = du, y = e u u u u e u du = u e + ue + e du u u u u e ue e = + + u ( = e u + u+ ) 0 Septemer 01 Leftovers Physics 1, Fll 01 u xdy = xy ydx. u u

3 Nonuniformly chrged sphere (continued) So using this enclosed chrge in Guss s Lw gives 4πr E = 4πkQ encl. 3 kqencl. 4πkρ ˆ 0R rr r r E = r = e + + rˆ r r R R Although tht my hve hd non-trivil integrl, it ws still dole. Now imgine trying to do the sme clcultion with Coulom s lw! 0 Septemer 01 Leftovers Physics 1, Fll 01 3

4 Perfect electricl conductors So fr we ve een considering chrge distriutions in which the chrges re stuck in predetermined positions: tht is, perfect insultors. In contrst re mterils in which the chrges re free to move in response to field. If chrges cn move without ny impediment, the mterils re clled perfect conductors. Perfect conductors re not to e confused with superconductors, which hve other interesting properties in ddition to perfect conductivity. Of course most conductors re not perfect, ut perfect is good pproximtion for mny metls if the chrge distriution is sttic (constnt in time). 0 Septemer 01 Leftovers Physics 1, Fll 01 4

5 Properties of perfect conductors Electric chrge cn only reside on the surfce of perfect conductor. Since like chrges repel ech other, ny free chrges within it will move s fr s they cn wy from other ones, nd the surfce is s fr s they re free to go. Ner perfectly-conducting surfce, E is perpendiculr to tht surfce. If it weren t, then chrge would flow long the surfce until ny component of E to the surfce were cncelled, leving only the component. Aside: This is why lightning rods work. Prticulrly lrge E 0 Septemer 01 Leftovers Physics 1, Fll 01 5

6 Exmple 5: Guss s Lw nd conducting plnes Spce t z > 0 is empty, ut t z = 0 lies n infinite plnr surfce of conductor, which crries uniform chrge per unit re σ. Clculte E everywhere. E, da We use the sme setup s Exmple 1. Becuse E hs to e to the plne (cosθ = 0), the flux is still zero z through ll of the verticl, z fces. σ For the upper fce, E nd da still point long +z, so its flux is still E. z But this time the lower fce hs E = 0: no flux threds it. E = 0 0 Septemer 01 Leftovers Physics 1, Fll 01 6

7 Guss s Lw nd conducting plnes (continued) So Guss s Lw for this Gussin surfce ecomes E da = E da = E = 4πkQencl. = 4πkσ 4 πkσzˆ, z> 0 E = 0, z 0 Tht is, the field t z > 0 is twice s strong s in Exmple 1. The chrge produces the sme flux s efore, ut insted of dividing etween up nd down, it ll goes up. E = 0 E = 4πkσzˆ σ 0 Septemer 01 Leftovers Physics 1, Fll 01 7

8 Exmple 6: point chrge nd conducting shell A neutrl, conducting sphericl shell inner nd outer rdii R 1 nd R respectively hs point chrge Q t its center. Clculte the chrge, nd chrge per unit re on ech surfce. r To do this methodiclly, follow σ the setup of Exmples 3 nd 4. However, I don t think you ll σ mind the following shortcuts: 1 r 1 Consider sphericl Q Gussin surfce with rdius R 1 r1, R1 < r1 < R. It lies within the conductor, so E = 0 R everywhere on its surfce. 0 Septemer 01 Leftovers Physics 1, Fll 01 8

9 Point chrge nd conducting shell (continued) For E to e zero in the conductor there must e chrge induced on the inner surfce, to cncel the point chrge s field. So Guss Lw for this surfce (1) nd the volume it encloses is σ r E da = 0 = 4π kq 1 nd Q σ encl. ( Q ) = 4 π kq+, = Q = Q 4πR1 Q σ 1 R r 1 R 1 0 Septemer 01 Leftovers Physics 1, Fll 01 9

10 Point chrge nd conducting shell (continued) But the conductor is neutrl, so if chrge Q is induced on its inner surfce, there must e chrge +Q on its outer surfce. r And thus the outer-surfce chrge σ density is σ = Q 4 πr. We weren t sked, ut: since the outer surfce cts s if ll of its chrge lies t its center, the field outside the conductor is E = kqrˆ r, just s if the conductor weren t there t ll. Q σ 1 R r 1 R 1 0 Septemer 01 Leftovers Physics 1, Fll 01 10

11 Your turn. Suppose tht the center of the inner surfce nd the point chrge were displced from the center of the outer surfce, s shown. Which quntities differ from their vlues in the previous exmple? (Check ll tht pply.) o Q 1 o σ 1 o Q o σ o E(r) o None of these hs chnged. 0 Septemer 01 Leftovers Physics 1, Fll Q σ 1 σ E( r)

12 Your turn gin. Now suppose the inner nd outer surfces of the shell still hve the sme dimensions, ut the point chrge nd the centers of oth surfces re ll displced from ech other. Which quntities differ from their vlues in Exmple 6? (Check ll tht pply.) o Q 1 o σ 1 o Q o σ o E(r) o None of these hs chnged. 0 Septemer 01 Leftovers Physics 1, Fll 01 1 Q σ σ 1 E( r)

13 Guss s Lw nd grvity Guss s Lw works for grvity too. If one repets the resoning which leds from Coulom s to Guss s lw (13 Septemer, esp. pp. 3-9), nd sustitutes msses nd grvittionl field g for chrges nd electric field E: kqq GmM FQ q = rˆ F M m = rˆ r r F = qe F = mg kq GM E = rˆ g = rˆ r r one otins the grvittionl nlogue of Guss s Lw: E da = 4πkQ g da = 4πGM encl. encl. 0 Septemer 01 Leftovers Physics 1, Fll 01 13

Guss s Lw nd grvity (continued) Since most physicists nd mny engineers never del with grvity, their lives re not drsticlly improved y Guss s Lw for grvity.

14 Guss s Lw nd grvity (continued) Since most physicists nd mny engineers never del with grvity, their lives re not drsticlly improved y Guss s Lw for grvity. Astrophysicists nd geophysicists use it, though, wherever grvity is the dominnt force nd symmetricl ut nonuniform mss densities re encountered. Milky Wy mss model: Tom Jrrett, Cltech 0 Septemer 01 Leftovers Physics 1, Fll 01 14

15 Work in electrosttics Electric chrges exert forces on one nother, so to move chrges round in ech other s presence requires work. Consider point chrge Q, niled down t the coordinte origin, nd nother point chrge q we wish to move from one plce to nother plce. The work involved in this move is expressed in the usul wy: W = Fd = kqq rd ˆ r F q d Q 0 Septemer 01 Physics 1, Fll 01 15

16 Work in electrosttics (continued) But, s ws the cse for grvity in PHY 11, the dot product ensures tht only the rdil prt of the pth mtters nd the integrl hs the sme vlue no mtter wht the pth, s long s it hs the sme rdil prt. Thus we cn write the integrl in terms of just the endpoints of the pth: r ( ) dr W = kqq r ( ) r r() q q q Q Q r() Q 0 Septemer 01 Physics 1, Fll 01 16

17 Work in electrosttics (continued) This is esily solved: r ( ) r ( ) dr W = kqq = kqq kqq r = ( ) r ( ) r( ) r( ) r r In the sme nlogy s tht of force nd field, we cn define the work per unit (movle) chrge: r ( ) dr 1 1 V = Ed = kq = kq r ( ) r ( ) r ( ) r V is clled the voltge etween the two points nd. 0 Septemer 01 Physics 1, Fll 01 17

18 Electric potentil In the cse of grvity (in PHY 11) it ws found useful to define the potentil energy: the work done ringing two odies from infinity to given seprtion. In our present terms this is r r dr 1 kqq U ( r) = kqq = kqq. = r r r So in nlogy we define the electric potentil s the potentil energy per unit chrge: r r dr kq V ( r) = Ed = kq = r r The potentil t point is the voltge etween nd point very very ( infinitely) fr wy. 0 Septemer 01 Physics 1, Fll 01 18

19 Electric potentil (continued) Just s the work moving chrges etween two seprtions is the sme s the potentil energy difference etween strt nd finish: r ( ) r ( ) dr dr dr W = kqq = kqq kqq r ( ) r r ( ) r r ( [ ]) [ ] ( ) = U r + U r = U U the voltge etween two points is the potentil difference: r ( ) dr dr V = Ed = kq kq r ( ) r r ( [ ]) [ ] ( ) = Vr + Vr = V V 0 Septemer 01 Physics 1, Fll 01 19,

20 Electric potentil (continued) Now, why is it good to inflict nother physicl quntity on the concept-overurdened students? Becuse the electric potentil, s we hve defined it, is sclr which oeys the principle of superposition: kq1 kq V ( r ) = r r r r 1 esily extended to continuous distriution of chrge: ( r ) kdq dv V( ) k ρ r = = r r r r. r r r r dq = ( ) ρ r dv ρ ( r ) 0 Septemer 01 Physics 1, Fll 01 0

21 Electric potentil (continued) Sclrs re usully esier to clculte thn vectors! Wht s more, this sclr cn give the vector one is usully interested in E y differentition. Note tht In one dimension, V ( r) ( ) V x = nd, y differentiting with respect to the integrl s upper ound, nd invoking the fundmentl theorem of clculus, dv ( x ) = E ( x ) dx 0 Septemer 01 Physics 1, Fll 01 1 r x Ed = Edx

22 Electric potentil (continued) In three dimensions, V r V V V x y z ( r) = Ed E( r) = xˆ yˆ zˆ = V, where is the grdient opertor the 3-D version of the derivtive expressed in Crtesin coordintes s = xˆ + yˆ + zˆ. x y z It is usully esier to clculte sclr nd tke its derivtive, thn it is to integrte vector quntities! Thus, like Guss s Lw, the electric potentil will sometimes offer shortcut in electric-field clcultions. 0 Septemer 01 Physics 1, Fll 01

23 Exmple: field long the xis of chrged disk Now for n illustrtion of getting E y getting V first nd tking its derivtive: A circulr disk with rdius R crries chrge Q which is uniformly distriuted on the disk. Clculte the electric field distnce z ove the center of the disk. E =? It s -D chrge distriution: σ = Q π R nd we cn plce the coordinte origin t the center of the disk with the z xis pointing up. Q z 0 Septemer 01 Physics 1, Fll 01 3

24 Field long the xis of chrged disk (continued) Since we re only concerned with the xis of the disk, we cn choose n infinitesiml chrge element tht s symmetric out the xis: dq = σ da = πσ r dr r r = r + z kdq r dr V( z) = = πkσ r r 0 r + z Sustitute: u = r + z du = r dr r = 0 R u= z z + R R dr z r + z r 0 Septemer 01 Physics 1, Fll 01 4

25 Field long the xis of chrged disk (continued) After sustituting, we hve the potentil: z + R du u V( z) = πkσ = πkσ u 1 z = πkσ z + R z And we get the field y differentition: dv d E( z) = zˆ = πkσzˆ z + R z dz dz 0 Septemer 01 Physics 1, Fll 01 5 z z + R Compre to lecture on 6 Septemer 1 1 z = πkσzˆ z 1 = πkσzˆ 1 z + R z + R

26 Units of electric potentil The MKS unit of electric potentil is the volt: volt = 1 joule coulom = 1 kg m sec coul The units of electric field re often expressed in voltge terms: N coulom = 1 joule m coulom = 1 volt m Since electrons nd protons re often ccelerted cross regions with given voltge, convenient energy unit lso comes from the volt: 19 1 electron volt ( ev) = e 1 volt = joule. 0 Septemer 01 Physics 1, Fll 01 6

Lecture 13 - Linking E, ϕ, and ρ

Lecture 13 - Linking E, ϕ, and ρ Lecture 13 - Linking E, ϕ, nd ρ A Puzzle... Inner-Surfce Chrge Density A positive point chrge q is locted off-center inside neutrl conducting sphericl shell. We know from Guss s lw tht the totl chrge on