= ρ. Since this equation is applied to an arbitrary point in space, we can use it to determine the charge density once we know the field.

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1 Gauss s Law In diffeentia fom D = ρ. ince this equation is appied to an abita point in space, we can use it to detemine the chage densit once we know the fied. (We can use this equation to ve fo the fied if we know the chage densit. Thee must be me smmet that we can eiminate two of the vecto components of D. Howeve, duing this couse, we don t utiize this technique.) In intega fom E d = 1 ε V ρ dv V. This equation is used to detemine the tota fied (fux) passing though a suface due to the chage encosed. We can app this intega if we know the chage distibution and we can make assumptions about the behavio of E. Note, the vecto D and E ae eated b the pemittivit, ε. In this couse, we imit ou fied distibutions and geometies to the foowing smmetic pobems: 1. Pana chage densit the chage is an infinite sheet o sab and has a constant densit on a pane. B smmet, the eectic fied on an pane paae to the pane of the chage densit wi be constant and pependicua to that pane. Exampe: ab of chage chage densit ρ v pane paae to chage densit x The fied wi point in the xˆ (o - xˆ ) diection fo this geomet. Fo smmet to app, the chage densit can on be a function of x.

2 . Cindica chage densit the chage is a cindica coumn o she and has a constant densit on a suface = constant. Note, on that suface the chage is constant but that does not imp it is a suface chage. B smmet the eectica fied wi on point in the adia diection, E E Exampe: Cindica voume chage The fied wi point in the ˆ (o - ˆ ) diection fo this geomet. Fo smmet to app, the chage densit can on be a function of.

3 3. pheica chage densit the chage is a spheica voume o she and has a constant densit on a suface = constant. Note, on that suface the chage is constant but that does not imp it is a suface chage. B smmet the eectica fied wi on point in the adia diection, E E Exampe: pheica voume chage ρ v chage densit uface = constant The fied wi point in the ˆ (o - ˆ ) diection fo this geomet. Fo smmet to app, the chage densit can on be a function of. An impotant point to ecognize is that these tpes of pobems ae pesented since the ae simpe geometies that we ae capabe of ving in cass. These smmetic pobems povide us with a simpification of the eft side of the Gauss s Law equation, which is ve difficut to achieve in pactice. It is ae that an inteesting pobem wi have an exact anatic ution. Occasiona, we ma be abe to simpif the appoach when consideing a ution in a given egion, but that ution ma not be vaid outside of that domain. To ve an intega fom of Gauss s Law, we need to pefom the foowing steps 1.) Recognize the coodinate sstem.) Using smmet, detemine which components of the fied exist 3.) Ceate a Gaussian suface such that the sides of the suface ae eithe paae o pependicua to the diection of the fied in step emembe, the Gaussian suface is abita in size 4.) Detemine the tota chage inside that suface. The chage distibution can be a voume, suface, ine and/o point chage. 5.) Evauate the fux passing though that suface. If the fied is paae to the suface, then E d =. If the fied is pependicua to the suface, then E d E i d, whee Ei is the fied in the component diection. Again, it is impotant to ecognize that this simpification on exists fo the smmetic pobems we use in cass. Rea pobems ae much moe compex. 6.) You can now equate the esuts fom step 4 and step 5 to detemine the fied in the egion ou define the Gaussian suface.

4 As an exampe, conside a sheet of chage that is ocated on the pane = and has a chage densit ρ. A sheet of chage is a suface chage densit. In ode fo smmet to app, the sheet must have a unifom chage densit. The coodinates ae given in the owe eft. The sheet is infinite in the z-diection (pependicua to the pape) and in the x-diection. Gaussian suface ρ s = ρ z x The Gaussian suface is dawn as an ovehead view. In actuait, it has six sides: 6 back, z = constant 3 top, = constant 4 eft, x = constant ight, x = constant x 1 bottom, = constant z 5 font, z = constant

5 tep 1) The geomet is Catesian tep ) B smmet, the fied is in the diection. It is impotant to note that the fied is positive fo > and negative fo < since the chage distibution is positive. The diection of the fied ines indicates positive o negative fied. E E ρ s = ρ sp tep 3) The Gaussian suface is in the pevious figues. The box has sides of ength,. A, the box is smmetic about the suface chage distibution. We know that the fied on a pane paae to the suface chage is constant, the magnitude of the fied at me ocation =C must be the same as that at the ocation = C. Howeve, we do not know how the fied behaves as a function of. tep 4) To detemine the tota chage inside the box, we must integate acoss the section of suface chage encosed b ou Gaussian suface. Q enc = ρ dxdz The imits of integation indicate that the chage encosed is independent of the oigin. Again, a unifom chage distibution is necessa fo this assumption. Now, Q = ρ enc The tota chage encosed is dependent on the dimensions of the Gaussian suface. tep 5) To detemine the tota fux passing though the Gaussian suface, we must integate acoss a six sides that make up that suface, 6 E d = n= 1 n E d

6 This equation can be simpified b noticing that the fux passing though sides, 4, 5, and 6 is zeo. The fied is paae to those sufaces and theefoe the dot poduct between the fied and the unit vecto noma to the suface is zeo, E d =. In othe wods, the fux passing though those sufaces is zeo. Now, E d = E d E d On side 1, d dxdz ˆ and the fied points in the negative diection. On side 3, d dxdz ˆ and the fied points in the positive diection. Fo both sides E d = E dxdz E d = E dxdz The indicates that both integas ied the same esuts. Again, the imits of integation indicate an abita sized box. tep 6) Equate sides to find the fied. E d = E E = ρ E 1 = We have the magnitude of the fied. Note, the magnitude of the fied is independent of and that the dimensions of the Gaussian suface vanish fom the ution. We sti need to incude the diection, a compete ution woud ook ike. ρ 1 ρ E = 1 ρ ˆ ˆ > <

Jackson 3.3 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

Jackson 3.3 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell Jackson 3.3 Homewok Pobem Soution D. Chistophe S. Baid Univesity of Massachusetts Lowe POBLEM: A thin, fat, conducting, cicua disc of adius is ocated in the x-y pane with its cente at the oigin, and is