An o5en- confusing point:

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Percival Henry
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An o5en- confusing point: Recall this example fom last lectue: E due to a unifom spheical suface chage, density = σ. Let s calculate the pessue on the suface.

1 An o5en- confusing point: Recall this example fom last lectue: E due to a unifom spheical suface chage, density = σ. Let s calculate the pessue on the suface. Due to the epulsive foces, thee is an outwad pessue. The foce pe unit aea on a chaged suface is F = σe. Howeve, E = Q outside, and E = 0 inside. What value of E should we use? 4πε o The way to deal with this is to eliminate the disconpnuity by consideing a vey thin shell, analyze the poblem, and then take the limit as the thickness goes to zeo. If we get a plausible esult, then we e good; othewise, we e in touble. So, conside a spheical shell between 1 and, with an abitay (but spheically symmetic) chage density, ρ(). The foce on an infinitesimal shell, d, at adius ( 1 < < ) is: df = E() ( 4π 1 )ρ( )d I wite 1 (i.e., a constant), because, fo a thin shell, the suface aea does not change much between 1 and. x z R E 1 This is discussed in secpon 1.14 of Pucell and Moin, Electicity and Magne3sm. y Physics 435 7

2 We have: df = E() ( 4π 1 )ρ( )d. To integate this, (to find the total foce) we fist need to elate E() and ρ(). Remembe Gauss s law. It tells us that as one passes though a slab of chage, the electic field changes: de = σ = ρd ε 0 ε 0 E + de Thus, df = E( 4π 1 )ε 0 de Now we can do the integal: F = 4πε 0 1 E( ) 1 E de = 4πε 0 1 E ( ) E ( 1 ) E( 1 ) ( ) E( )+E( 1 ) = 4πε 0 1 E( ) E( 1 ) = 4π 1 σ E( )+E( 1 ) E( = Q )+E( 1 ) tot ( ) E ρ d So, the ight answe is to use the aveage of the fields on the inside and outside. This is a eliable esult as the shell appoaches zeo thickness, because it does not depend on the thickness (as long as the shell is easonably thin). Physics 435 8

3 The Electic Potential: Cul of E Conside ou tusty point chage: Let s calculate the wok done on anothe chage as it moves fom a to b. We need to integate: W = qq 4πε 0 E = This integal is most easily done in spheical coodinates. It simplifies fom 3- D to 1- D, because ˆ = ( 1,0,0 ). Theefoe we only need. d dl = d Thus, W = qq b. 4πε 0 = qq 1 1 a ε 0 a b b a Q 4πε 0 ˆ d l This is nice: V = Q. But that s not my main point. 4πε 0 The impotant point is that W does not depend on the path taken! Q a wiggly path b Fo completeness: dl θ = dθ dl Φ = sinθ dφ Physics 435 9

4 On to the cul! (all you sufe dudes) SupeposiPon tells us that path independence holds fo any distibupon of chage. Path independence tells us that the loop integal of E is zeo fo any loop: E dl = 0 Conside two paths between a and b. The two path integals ae equal. Thus, the loop integal is zeo, because one path will be tavesed backwads. Now, put in a likle math: Stokes theoem (discoveed by Kelvin). E d l = loop E suface( ) n da No physics hee! The loop is the bounday of the suface. Evey suface has a bounday, and the loop integal on evey bounday is zeo. Theefoe, we have E fo evey suface. suface( ) n da = 0 Theefoe, E = 0 eveywhee. loop This is anothe pat of Maxwell s equapons. a Path 1 loop b Path ˆn dl Physics

5 Comment: Thee is an inpmate connecpon between E = 0 and the existence of the electic potenpal. Basically, if E 0, then enegy is not conseved. Example: Paallel plate capacito We usually model the capacito like this: E has only a z- component, and it is independent of x, except at the edge of the capacito. The disconpnuity doesn t make sense, so let s suppose that E z goes linealy to zeo. (This is not the exact solupon, but close.) Howeve, this implies that the y- component of E is not zeo: E ( ) y = E x z E z x - a To keep it zeo, we need E x = - az. So, the field looks like this: This is called the finge field of the capacito. E = 0,0, E 0z E = 0 Note: This can t be the ight answe, because E x is disconpnuous at x = 0 and z 0, changing fom 0 inside to az outside. We ll lean how to solve this kind of poblem soon. E 0z ( ) inside outside E z Fo stapc fields. We ll deal with Pme vaying fields late. z E z = E 0z ax x x Puzzle: Why do the field lines cuve diffeently as one moves to the ight? Physics

6 E = 0 Electic Potential An impotant implicapon: We can define V(), a funcpon of posipon only, such that: E tot = KE + qv() is a constant of the mopon. (Enegy is conseved.) Because E = 0, this math idenpty: lets us wite E = V V. Calculate the wok done on a chage in a field: W = q E dl 0 = q V dl 0 ( ( ) V ( 0) ) = q V Ref. pt. = 0 So, W + qδv = 0, o KE + qv is constant. ( ( )) = 0 Wok Note: U = qv. U is the potenpal enegy. V is the electic potenpal. F = U E = V Physics 435 3

7 A poblem solving stategy: If you have a choice between using E and V to solve a poblem (e.g., What is the papcle s enegy when it is at? ), you ae usually beke off with V. If you use E, you ll pobably end up doing the E. dl integal anyway. Don t do unnecessay wok. Cul example: Conside this vecto field (in cylindical coodinates): A = ϕˆ It ciculates aound, and its magnitude is popoponal to. The cul only has a z- component (out of the page): ( A ) = 1 ( A ϕ ) = 1 =, i.e., it s constant. z As a check, conside a cicle of adius R. The integal of the cul ove the aea of the cicle is (πr ). The loop integal of A aound the cicle is (πr)r, the same esult. Puzzle: What about this field: B = ϕ ˆ /? (a magnepc field nea a wie). It ciculates, but has no cul. Does that make sense? Note: In mechanics, field A descibes otapon of a solid, and field B descibes the votex mopon of a fluid (a whilpool). End 8/30/13 Physics

Gauss Law. Physics 231 Lecture 2-1

Gauss Law. Physics 231 Lecture 2-1 Gauss Law Physics 31 Lectue -1 lectic Field Lines The numbe of field lines, also known as lines of foce, ae elated to stength of the electic field Moe appopiately it is the numbe of field lines cossing