Problem Solving Session 1: Electric Dipoles and Torque

Transcription

1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Dpatmnt of Physics 8.02 Poblm Solving Sssion 1: Elctic Dipols and Toqu Sction Tabl (if applicabl) Goup Mmbs Intoduction: In th fist poblm you will lan to apply Coulomb s Law to find th lctic fild of an lctic dipol. You will invstigat th poptis of th lctic fild associatd with an lctic dipol. You will also xplo what happns to an lctic dipol whn it is placd in an unifom lctic fild. Radings: Cous Nots 8.02: Chapt 2 Coulomb s Law Sction Intoduction Elctic Dipol An lctic dipol consists of two qual but oppositly chagd point-lik objcts, q and q, spaatd by a distanc 2a, as shown in Figu 1. Figu 1 Elctic dipol 0

2 Th dipol momnt vcto p which points fom q to q (in th y- diction) is givn by p = 2qa ˆj (1) Th magnitud of th lctic dipol is p = 2qa, wh q > 0. Fo an ovall chagnutal systm having N chagd objcts, th lctic dipol vcto p is dfind as p i= N q i i (2) wh i is th position vcto of th chagd objct with chag q i. Poblm 1 Elctic fild of a Dipol Qustion 1: Consid th lctic dipol momnt shown in Figu 1. Find th x- and y- componnts of th lctic fild at a point with coodinats (x, y,0), using i=1 q E = k and q E = k. Solution: Fom th figu w hav that th vctos fom th chagd paticls to th fild point a ± = xî ( y a)ĵ Th spctiv distancs fom th positiv and ngativly chagd objcts to th fild point at (x, y,0) a givn by th xpssions ± = (x 2 ( y a) 2 ) 1/ 2 Th chags a q = q and q = q. Thfo th supposition of th two lctic filds yilds E = E E q = k q k xî = k q ( y a)ĵ (x 2 ( y a) 2 ) xî ( y a)ĵ / 2 (x 2 ( y a) 2 ) / 2 Th x -componnt of th lctic fild stngth at th point P is thn 1

3 x x. E x = k q x 2 ( y a)2 / 2 x 2 ( y a)2 / 2 Similaly, th y -componnt is givn by ya ya E y = k q x 2 ( y a)2 / 2 x 2 ( y a)2 / 2 W can show that th lctic fild of th dipol in th limit wh >> a is Ex = k p sin θ cosθ, Ey = k p (cos 2 θ 1) () wh sin θ = x / and cosθ = y /. S Chapt Elctic Fild of a Dipol. Qustion 2: By what pow of distanc dos th stngth of lctic fild fall off? How dos this compa to a singl point chag? Bifly xplain a ason fo th diffnc btwn ths two cass. Solution: Th lctic fild E du to a dipol dcass with as 1/, unlik th 1 / 2 bhavio fo a point-lik chagd objct. This is to b xpctd sinc th nt chag of a dipol is zo and thfo must fall off mo apidly than 1 / 2 at lag distanc. Th xact lctic fild lins du to th oppositly chagd objcts a shown in th figu on th lft in Figu 2. Th lctic fild lins cosponding to th lctic dipol givn by Eq. () a shown in th figu on th ight in Figu 2. Figu 2 Elctic fild lins fo (a) a finit dipol and (b) a point dipol. 2

4 Qustion Elctic Dipol Animation Opn up th applt, Two Point Chags, Th units fo chag a µc. Cat an lctic dipol with chags q 1 =.0 µc and q 2 =.0 µc. Click on th Elctic Filds: Gass Sds ba to s a psntation of th lctic fild. Not that this applt can shows th th diffnt psntations w us fo vcto filds: (1) vcto fild gid of aows, (2) fild lins, and () gass sds.

5 Poblm 2 Elctic Dipol in a Unifom Elctic Fild Plac an lctic dipol in a unifom fild E = E î, with th dipol momnt vcto p making an angl θ with th x-axis. Qustion 1: What is a vcto xpssion fo th dipol momnt p? Giv you answ in th fom p = p x î p y ĵ wh you dtmin th componnts ( p x, p y ). Solution: Fom Figu, w s that th unit vcto which points in th diction of p is cosθ î sinθ ĵ. Thus, w hav p = 2 qa(cosθ ˆi sin θ ˆj ) (1) Qustion 2: Find a vcto xpssion fo th toqu on th dipol. What point did you comput th toqu about? Is you answ indpndnt of that choic of point? Solution: As sn fom Figu abov, sinc ach point-lik chagd objct xpincs an qual but opposit foc du to th fild, th nt foc on th dipol is F nt = F F = 0. Evn though th nt foc vanishs, th fild xts a toqu a toqu on th dipol that is indpndnt of a choic of point. W slct th midpoint O of th dipol about which to calculat th toqu. τ = F F = (acosθ î asinθ ĵ) (F î) (acosθ î asinθ ĵ) (F î) = asinθ F ( ˆk) asinθ F ( ˆk) = 2aF sinθ( ˆk) wh w hav usd F = F = F. (2) 4

6 Qustion : What typ of motion dos th dipol undgo if it is lasd fom its position and is f to mov? Solution: Th diction of th toqu is ˆk, o into th pag. Th ffct of th toqu τ is to otat th dipol clockwis so that th dipol momnt p bcoms alignd with th lctic fild E. Qustion 4: Show that τ = p E. Solution: With F = qe, th magnitud of th toqu can b wittn as τ = 2 a( qe)sin θ = (2 aq) E sinθ = pe sinθ and th gnal xpssion fo toqu bcoms τ = p E. Qustion 5: Toqu on an Elctic Dipol Animation Opn up th applt, Toqu on an Elctic Dipol Th units fo th dipol momnt a 10 6 m C and th units fo lctic fild stngth a 10 N C -1 St th dipol momnt p = m C, th lctic fild stngth E = N C -1, and th damping to zo. What appoximatly is th piod of oscillation? Click on th Elctic Filds: Gass Sds ba to s a psntation of th lctic fild at vaious momnts in th oscillation. At what point in th cycl a th fild lins most contotd? At what point a th fild lins naly staight (xcpt clos to th dipol). Summay Two qual but opposit point-lik chagd objcts fom an lctic fild that fa fom th chagd objcts is an lctic dipol. Th lctic dipol momnt vcto p points fom th ngativ point-lik chagd objct to th positiv point-lik chagd objct, and has a magnitud p = 2aq Th toqu acting on an lctic dipol p placd in a unifom lctic fild E is τ = p E 5