Chapte Univesity Physics (PHY 6) Lectue lectostatics lectic field (cont.) Conductos in electostatic euilibium The oscilloscope lectic flux and Gauss s law /6/5
Discuss a techniue intoduced by Kal F. Gauss (777-855) to calculate electic fields. euies symmetic chage distibutions. Techniue based on the notion of electical flux. Flux Numbe of objects passing though a suface /6/5
To intoduce the notion of flux, conside a situation whee the electic field is unifom in magnitude and diection. Conside also that the field lines coss a suface of aea A which is pependicula to the field. The numbe of field lines pe unit of aea is constant. The flux, F, is defined as the poduct of the field magnitude by the aea cossed by the field lines. FA Aea=A /6/5
Units: Nm /C in SI units. Find the electic flux though the aea A = m, which is pependicula to an electic field = N/C FA Answe: F = 44 Nm/C. /6/5 4
If the suface is not pependicula to the field, the expession of the field becomes: FAcos Whee is the angle between the field and a nomal to the suface. N /6/5 5
emak: When an aea is constucted such that a closed suface is fomed, we shall adopt the convention that the flux lines passing into the inteio of the volume ae negative and those passing out of the inteio of the volume ae positive. /6/5 6
Question: Calculate the flux of a constant field (along x) though a cube of side L. y x z /6/5 7
Question: Calculate the flux of a constant field (along x) though a cube of side L. easoning: Dealing with a composite, closed suface. Sum of the fluxes though all sufaces. Flux of field going in is negative Flux of field going out is positive. is paallel to all sufaces except sufaces labeled and. So only those suface contibute to the flux. y x z /6/5 8
Question: Calculate the flux of a constant field (along x) though a cube of side L. easoning: Dealing with a composite, closed suface. Sum of the fluxes though all sufaces. Flux of field going in is negative Flux of field going out is positive. is paallel to all sufaces except sufaces labeled and. So only those suface contibute to the flux. Solution: F A F A cos L cos L Fnet L L y x z /6/5 9
The net flux passing though a closed suface suounding a chage Q is popotional to the magnitude of Q: F Acos Q net In fee space, the constant of popotionality is /e o whee e o is called the pemittivity of of fee space. eo 9 4 k 4 8.99 Nm / C e /6/5
The net flux passing though any closed suface is eual to the net chage inside the suface divided by e o. F Acos net Q e o /6/5
The field lines emitted by a chage ae popotional to the size of the chage. Theefoe, the electic field must be popotional to the size of the chage In ode to count the field lines, we must enclose the chages in some geometical suface (one that we choose) /6/5
Mathematically F F Chage enclosed within bounding limits of this closed suface integal da enclosed e e /6/5
Fluxes, Fluxes, Fluxes /6/5 4
Shapes Sphee Cylinde Pillbox /6/5 5
Sphee When to use: aound spheical objects (duh!) and point chages Hey! What if an object is not one of these objects? Closed suface integal yields: da (4 is the adius of the geometical object that you ae ceating ) /6/5 6
/6/5 7 Sphee xample 5 5 4 ) (4 ) (4 4 ) 4 )( ( : A A da A A sphee Inside the enclosed enclosed e e What if you had a sphee of adius, b, which contained a mateial whose chage density depend on the adius, fo example, =A whee A is a constant with appopiate units? 5 5 5 4 ) (4 ) (4 4 ) 4 )( ( : b A b A da b A b Ab sphee Outside the enclosed total e e At =b, both of these expessions should be eual
When to use: aound cylindical objects and line chages Closed suface integal yields: da ( L) L is the adius of the geometical object that you ae ceating and L is the length of the cylinde 8
What if you had an infinitely long line of chage with a linea chage density, l? (L) da (L) enclosed l e ll ll e o l e ˆ 9
When to use: aound flat sufaces and sheets of chage Closed suface integal yields: da A A is the aea of the pillbox
If excess chage is placed on an isolated conducto, the chage esides on the suface. Why? If thee is an -field inside the conducto then it would exet foces on the fee electons which would then be in motion. This is NOT electostatic. Theefoe, if thee is no -field inside, then, by Gauss s Law, the chage enclosed inside must be zeo If the chages ae not on the outside, you ae only left with the suface A caveat to this is that -field lines must be pependicula to the suface else fee chages would move.
A Let n o A A So A A A A da enclosed ˆ ) )( ( e e e +++++++++++++++++++++++++++++++++++++++++++++ A
Let A +++++++++++++++++++++++++++++++++++++++++++++ A So A da enclosed A e e A A o A nˆ e So a conducto has x the electic field stength as the infinite sheet of chage
ecall the Divegence of a field of vectos Div( v) v Div=+lage Div= How much the vecto diveges aound a given point 4
v d A place of high divegence is like a faucet v Bounded suface of some egion da Suspiciously like LHS of Gauss s Law Sum of the faucets in a volume = Sum of the wate going thu the suface 5
so da e enclosed e d enclosed e and enclosed d So how the -field speads out fom a point depends on the amount of chage density at that point
Poblem A chage, +, is suounded by a thin, spheical shell of adius, a, which has a chage density of on its suface. This shell is, in tun, suounded by anothe thin shell of adius, b, which has a suface chage of +. Find the electic fields in egion : <a egion : a<= <=b egion : > b b + a - + /6/5 7
I choose spheical! So da (4 ) 8
enclosed = ) ( ˆ 4 ) ( 4 4 a outwad adially enclosed e e e e 9 + a b - +
enclosed =+(4a )*(-) 4 4 a enclosed 4e ( adially 4e e 4a e outwad) 4a ˆ ( a b) b + a - +
enclosed =+(4a )*(- )+ (4b )*() enclosed 4 4a 4b 4e ( adially 4e e 4a outwad) 4b 4a 4b ˆ ( a b) e b + a - +
An electic filed given by =4i-(y +)j pieces the Gaussian cube shown below. ( is in newtons/coulomb and y is metes). What net chage is enclosed by the Gaussian cube? y x z X=. m X=. m
Planes y 4 6.y-z: Nomal to +x.x-z: Nomal to y.y-z: Nomal to x 4.x-z: Nomal to +y 5.x-y: Nomal to +z x 6.x-y: Nomal to -z z X=. m 5 X=. m
6 5 4 da da da da da da da da enclosed enclosed enclosed e e e 4
but da 4xˆ da y dz xdydz ˆ yˆ 4dy 5 4** 6 egion, in which the nomal vecto points in the opposite diection, will have a value of -6
4 Since is pependicula to sides 5 & 6, the esult is zeo. yˆ dxdz dz yˆ dxdz dz enclosed ( y ( y y y ) dx ) dx 6()() 4 6(4 )() 7 N m da 6 4 6 7 48 C 48e 48(8.85 ) 4. C 6
The figue below shows a coss-section of two thin concentic cylindes with adii of a and b whee b>a. The cylindes eual and opposite chages pe unit length of l. a) Pove that = fo >a b) Pove that = fo >b c) Pove that, fo a<<b, a l b l e l 7
I choose cylindical! So da ( L) 8
enclosed = da (L) l l (L) a b 9
enclosed =ll (L) da (L) e l ll e a l b l 4
enclosed =ll-ll= da (L) l l (L) a b This is the pinciple of a coaxial cable 4
A vey long, solid insulating cylinde with adius has a cylindical hole with adius, a, boed along its entie length. The axis of the hole is a distance b fom the axis of the cylinde, whee a<b<. The solid mateial of the cylinde has a unifom chage density, p. Find the magnitude and diection of the electic field inside the hole and show that is unifom ove the entie hole. b a 4
L L L L da enclosed e e e e ˆ ) ( ) ( 4
we can tanslate coodinates by =-b Whee b is the diection and distance of the cente of the off-axis cylinde is a vecto fom the oigin b da ( ' L) enclosed (L) e e e ' ' e ' L ' ˆ e b L ' 44
b b b b hole hole axis off solid hole axis off solid e e e e e e e e 45 All of these ae constants and do not depend on.
W Fo gavity : W U U ab b a U F ds U Gmem d U Gmem Gmem Gmem o U Gmem Gmm Gmem 46
F F G C Gm m k e ˆ ˆ so k U 47
Potential is actually potential enegy pe unit chage =U/ olts ()= Joule/Coulomb o J/C ecall -field units wee N/C N N J C C C N m J m Now we have a new unit fo electic field: /m So now we see that the electic field is expession of how the voltage is changing ove space 48
It is a SCALA! But it can tell us about the -field and Foces! 49
U k k Point chage (I put the minus sign in ) i k i i Collection of point chages k d Distibution of chage 5
Take and divide by : W ab W b a ab b a F ds U F ds U Wa b b a ds b ds Potential diffeence between point A and point B a 5
k/ k/ -field Lines lectic field lines ae always pependicula to euipotential sufaces 5
ecall that all the chage of an isolated conducto esides on the suface All excess chage placed on an isolated conducto will distibute itself so that the suface will come to same potential because: If thee is moe o less chage between two points on the suface, thee will be cuent flow until euilibium e-established Futhemoe, the electic field lines must be pependicula to the suface at evey point Because if thee wee components tangential to the suface, thee would be 5
ecall d d o d x ds : b a d d dx ds x dx d d d holdingy, z constant y dy x dx x d y dy z dz z dz 54
xˆ x yˆ y zˆ z 55
56
ecall: W ab b a F ds b a ds But the wok done is independent of path Just like in gavity 57
What if: W aa a a F ds a a ds Well, fom point a to point a seems facetious, what about to? Such an integal is called a closed path integal and is expessed ds 58
ecall Cul(v) Cul( v) v Cul= Cul= Cul=+, out of page (ight Hand ule) 59
vda v ds 6
If ds then Because da ds 6
Needless to say, Gauss s Law woks well fo shapes: Sphee Cylinde Big, Flat Sufaces But what if you have a funny shape? 6
ecall that What is this uantity? and so e e 6
64 ˆ ˆ ˆ ˆ ˆ ˆ z y x so x x x z z y y x x z z y y x x z z y y x x
o u k e This euation applies to many physical systems Gavitation Heat flow M Quantum Mechanics 65
A numeical method to solve Laplace s euation Method of elaxation Method of ecusion Finite lement Method Big Bothes Discete Odinate (eacto physics) Finite Diffeence Time Domain (F engineeing) euies Known bounday conditions The ability to beak geomety into small egula shaped volume elements Compute with sufficient memoy (PM) 66
Based on neaest neighbo aveaging 5 5 5??=(5+++)/4 =.5 This is the estimate of the voltage in this suae 67
Loop thu multiple times You save the new values in a tempoay aay The new values eplace the old values and you loop again Fo -d objects, thee ae 6 sufaces 68
By a simple aveage, you ae assuming pefect conduction Fo diffeent mateials, you can slow up the tansfe of voltage, heat, etc by using Heat: diffeent themal conductivity values lecticity: diffeent electical conductivity values Fluids: diffeent viscosities 69
......5.5..........5 4 4.5.......5.5 4 4 4 4.5.....5 4 4 4 4 4 4.5.....5 4 4 4 5 5 4.5......5.5 4 5 5 4.5.......5 4 4 4 4.5.......5.5.5.5.5.5................................................... 7
In the figue below, point P is at the cente of the ectangle. With = at infinity, what is the net electic potential at P due to the six chaged paticles? +5 d - d - d P d + d - d +5 7
d d +5 - - d/ s d P d d + d - d +5 7
7 6 5 4 i i d k s k 5 - + +5 - - +5 P d d d d d d s d d/ 4 5 6 s k 5 d k s k 5 6 s k 4
74 6 5 4 i i d k d k s d k s d s s d s k.94 8 5 4 4 5 5 6 5 4
A chage is distibuted unifomly thoughout a spheical volume of adius,. a) Setting = at infinity shown that the potential at a distance fom the cente, whee < is given by 8e b) What is the potential diffeence between a point on the suface and the sphee s cente? 75
76 da outside outside enclosed outside ˆ 4 4 e e e da inside inside inside enclosed inside ˆ 4 ˆ 4 4 4 4 e e e e e
77 d d ds f i i f outside 4 4 4 ˆ ˆ 4 e e e e
78 d ds in in in in inside 8 8 8 8 4 ˆ 4 ˆ e e e e e e e
79 ) ( 8 8 8 8 8 4 8 8 4 in e e e e e e e e e
8 at Cente Cente 8 8 4 8 ) ( 8 e e e e e
8 8 8 4 4 ˆ 4 But d d inside e e e e e at in 8e Same as pevious Same as pevious
The electic potential at points in a space ae given by =x -y +5z What is the magnitude and diection of the electic field at the point (,,-)? 8
8 C N z y x z y x z z y y x x z y x /.6 5 5ˆ ˆ ˆ ) (,, ˆ ) 5( ˆ 6() 4() ˆ ) (,, 5 6 4 5
(,, ) xˆ yˆ 5ˆ z tan tan y x ( ) 5 Diection w..t +x axis cos z whee x y z.6 cos 5.6 8.4 Diection w..t +z axis 84
Thee +. C chages fom an euilateal tiangle,.7 m on a side. Using enegy that is supplied at a ate of.8 kw, how many days would be euied to move one of the chages to the midpoint of the line joining the othe two chages? 85
.C Initially, this chage is.7 m fom the othe two chages.7 m.7 m.c.85 m.85 m.7 m.c Finally, this chage is.85 m fom the othe two chages 86
87 C m L whee L k L k L k L k L k L k L k L k L k i f f i i f..7 4 4
k U L 9 (9x )(.) U.7 whee L.7m.C 5x 6 J 88
P=W/t W=-U So.8 kw= 8 J/s And t= U/P=5x 6 /8 t=8,699 s o 5 hous o. days 89