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
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
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
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 435 30
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 435 31
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
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 435 33