Today in Physics 1: electostatics eview David Blaine takes the pactical potion of his electostatics midtem (Gawke). 11 Octobe 01 Physics 1, Fall 01 1 Electostatics As you have pobably noticed, electostatics is about calculating foces on electic chages, and wok done moving electic chages aound, in the pesence of fixed (static) distibutions of electic chage. To the static distibution of chage coesponds an electic field E(x,y,z) yz) and an electic potential V(x,y,z), yz) which can be calculated fo any point in space x,y,z. Usually we calculate the electic field E instead of the foce, and obtain the foce fom F = qe. And we usually calculate potential diffeence V instead of wok, and obtain wok fom W = qv. Potential enegy is the wok done assembling the chages. 11 Octobe 01 Physics 1, Fall 01 The had way to calculate E is to use Coulomb s law, de ˆ kdq. It always woks, but it involves vecto addition and/o vecto calculus. Use when one only has a few point chages, o continuous distibutions of chage with no paticula symmety. Steps: Detemine the numbe of dimensions of you chage distibution. t ib ti Choose a coodinate system that suits the geomety of the chage distibution and the location at which you need to know the field, and expess the chage distibution in that coodinate system. Usually this will still be a Catesian coodinate system. 11 Octobe 01 Physics 1, Fall 01 3 (c) Univesity of Rocheste 1
The had way to calculate E (continued) If the dimension is zeo point chages, o outside of spheical chages add the field components of the chages (positions = x,y,z ) at the location in space (position = x,y,z) whee you need to know the field: kq1 kq E... 1 1 1 kq 1 1 E x 1 1 x Q 1 kq... 1 Q x 11 Octobe 01 Physics 1, Fall 01 4 The had way to calculate E (continued) If the dimension is one o moe: expess the infinitesimal element of chage, and the distance fom the infinitesimal element to the test point, in that coodinate system. (-D shown.) Expess the components (x,y,z) of the unit vecto, again in that coodinate system. Do the integals, one vecto component at a time: fo example, kdq Ex dq bounds of x da chage distibution 11 Octobe 01 Physics 1, Fall 01 5 Shotcut to E #1: Gauss s Law The flux of E though a closed suface depends only on the chage the suface encloses: EdA 4kQencl 4kdv s v This is i a shotcut h t t to t E whenh one is given a distibution of electic chage, and the symmety and/o extent of the chage distibution makes it clea what the patten (diection!) of E is. E da s dv 11 Octobe 01 Physics 1, Fall 01 6 (c) Univesity of Rocheste
How to use Gauss s Law to get E Daw imaginay closed suface (Gaussian suface) though point at which you want to know the field. The Gaussian suface needs to be dawn so as to make the unknown E come out of the flux integal: EdA EdA. This usually means the symmety of suface matches symmety of the chage distibution. Then calculate Qencl. dv and solve fo E. 11 Octobe 01 Physics 1, Fall 01 7 Shotcut to E #: potential Electic potential is a scala athe than a vecto, and it still obeys the pinciple of supeposition. It is elated to the electic field by V V V V x y z Ed E V xˆ yˆ zˆ Because it is a scala, it can be easie to wok out than it is to wok out E fom Coulomb s law. And if you know V and you can take deivatives, you can get E. It is, of couse, easie to diffeentiate than to integate. 11 Octobe 01 Physics 1, Fall 01 8 How to use potential to get E As long as the chages do not extend to infinity, and as long as you don t need to know E within a continuous distibution of chages: Detemine the numbe of dimensions of the chage distibution. Choose a coodinate system that t suits the geomety of the chage distibution and the location at which you need to know the field, and expess the chage density in that coodinate system. (Usually this will still be a Catesian coodinate system.) Expess the infinitesimal element dq, and the distance - fom the infinitesimal element to the test point, in that coodinate system. 11 Octobe 01 Physics 1, Fall 01 9 (c) Univesity of Rocheste 3
How to use potential to get E (continued) Then do the integal: kdq dv V k and diffeentiate the esult to get E: V V V E V xˆ yˆ zˆ x y z, dq dv 11 Octobe 01 Physics 1, Fall 01 10 The had way to calculate V Of couse, sometimes one does want to know what the potential is, in the two cases chage distibution extends to infinity lies within the distibution of chage which ae fobidden fo the Coulomb s law vesion of V, V kdq. In these cases, Deive E fist, using Coulomb s o Gauss s Law. Most often Gauss s Law. Then integate the esult to get V: b V V a V b Ed o V Ed a 11 Octobe 01 Physics 1, Fall 01 11 Othe nuances of E o V calculations Supeposition. If you ae given a distibution of chages that beak down into a combination of symmetical chage distibutions: solve the poblem fo each symmetical distibution and supepose the solutions, e.g. = + Conductos. Pefect conductos ae equipotentials, have zeo electic field inside, and have sufaces pependicula to any extenal electic field. 11 Octobe 01 Physics 1, Fall 01 1 (c) Univesity of Rocheste 4
Quick: how does David Blaine (below) make lightning stike his hands, and how does he live though it? He s holding lightning ods in his hands: pointy conductos, which concentate E at thei points because E has to be pependicula to the conductos. He s weaing a conductive suit and helmet. 11 Octobe 01 Physics 1, Fall 01 13 Capacitance and esistance as examples of E and V deivations To calculate C o R between two conductos: Pesume electic chage to be pesent; say, Q if thee is only one conducto, o ±Q if thee ae two. Eithe: Calculate the electic field fom the chages often using Gauss s s Law and integate it to find the potential diffeence V between the conductos, o Solve fo the potential diffeence diectly, using V kdq -. Then C 0 = Q/V. If the space between the conductos is filled with dielectic with constant K, then C = KC 0 = KQ/V. 11 Octobe 01 Physics 1, Fall 01 14 Resistivity and esistance (continued) If the space between the conductos is filled with impefectlyconducting mateial with esistivity : At eithe electode, apply the micoscopic fom of Ohm s law: J E. Multiply though by electode aea A; use expessions fo E and V to substitute E out, and V in; and use I = JA. J The esult will be of the fom V = IR (i.e. Ohm s Law), whence R = V/I. This esisto also has capacitance C 0, which can be viewed as a C in paallel with its R. 11 Octobe 01 Physics 1, Fall 01 15 (c) Univesity of Rocheste 5
Capacitos and esistos as cicuit elements Capacitos stoe chage and enegy: 1 1 Q Q CV U CV C Resistos cay cuent and dissipate enegy: V IR V P I R R 11 Octobe 01 Physics 1, Fall 01 16 Combinations of Rs, Cs Paallel and seies pais of esistos and capacitos: 1 1 1 C 1 Ceq C1 C = CC C 1 Ceq C1 C R 1 R = Reqq R1 R 1 1 1 R R R eq 1 C1 C= Ceq C1 C R1 R = RR R 1 eq R1 R 11 Octobe 01 Physics 1, Fall 01 17 The midtem exam Midtem #1: 8-9:15 AM, tomoow: Fiday, 1 Octobe 01 Sunames A-K in Hoyt Auditoium Sunames L-Z in Lowe Stong Auditoium 11 Octobe 01 Physics 1, Fall 01 18 (c) Univesity of Rocheste 6