Chapter 3. Electric Potential

Transcription

1 Chapt 3 Elctic Potntial 3.1 Potntial and Potntial Engy Elctic Potntial in a Unifom Fild Elctic Potntial du to Point Chags Potntial Engy in a Systm of Chags Continuous Chag Distibution Diving Elctic Fild fom th Elctic Potntial Gadint and Equipotntials Exampl 3.1: Unifomly Chagd Rod Exampl 3.: Unifomly Chagd Ring Exampl 3.3: Unifomly Chagd Disk Exampl 3.4: Calculating Elctic Fild fom Elctic Potntial Summay Poblm-Solving Statgy: Calculating Elctic Potntial Solvd Poblms Elctic Potntial Du to a Systm of Two Chags Elctic Dipol Potntial Elctic Potntial of an nnulus Chag Moving Na a Chagd Wi Concptual Qustions dditional Poblms Cub Th Chags Wok Don on Chags Calculating E fom V Elctic Potntial of a Rod Elctic Potntial Calculating Elctic Fild fom th Elctic Potntial Elctic Potntial and Elctic Potntial Engy Elctic Fild, Potntial and Engy

2 Elctic Potntial 3.1 Potntial and Potntial Engy In th intoductoy mchanics cous, w hav sn that gavitational foc fom th Eath on a paticl of mass m locatd at a distanc fom Eath s cnt has an invssqua fom: F g Mm = G ˆ (3.1.1) 11 wh G = N m /kg is th gavitational constant and ˆ is a unit vcto pointing adially outwad. Th Eath is assumd to b a unifom sph of mass M. Th cosponding gavitational fild g, dfind as th gavitational foc p unit mass, is givn by Fg GM g = = m ˆ (3.1.) Notic that g only dpnds on M, th mass which cats th fild, and, th distanc fom M. Figu Consid moving a paticl of mass m und th influnc of gavity (Figu 3.1.1). Th wok don by gavity in moving m fom to B is B B GMm GMm 1 1 Wg = g d = d = F s GMm = B (3.1.3) Th sult shows that W g is indpndnt of th path takn; it dpnds only on th ndpoints and B. It is impotant to daw distinction btwn Wg, th wok don by th 3-

3 fild and W xt ngativ sign: W = W., th wok don by an xtnal agnt such as you. Thy simply diff by a g xt Na Eath s sufac, th gavitational fild g is appoximatly constant, with a magnitud g = GM / E 9.8m/ s, wh E is th adius of Eath. Th wok don by gavity in moving an objct fom hight y to y B (Figu 3.1.) is B B yb W = F ds = mg cosθ ds = mg cos φ ds = mg dy = mg( y y ) g g y B (3.1.4) Figu 3.1. Moving a mass m fom to B. Th sult again is indpndnt of th path, and is only a function of th chang in vtical hight y y. B In th xampls abov, if th path foms a closd loop, so that th objct movs aound and thn tuns to wh it stats off, th nt wok don by th gavitational fild would b zo, and w say that th gavitational foc is consvativ. Mo gnally, a foc F is said to b consvativ if its lin intgal aound a closd loop vanishs: F d s = (3.1.5) Whn daling with a consvativ foc, it is oftn convnint to intoduc th concpt of potntial ngy U. Th chang in potntial ngy associatd with a consvativ foc F acting on an objct as it movs fom to B is dfind as: B U = UB U = F d s = W (3.1.6) wh W is th wok don by th foc on th objct. In th cas of gavity, W fom Eq. (3.1.3), th potntial ngy can b wittn as = W and g U g GMm = + U (3.1.7) 3-3

4 wh U is an abitay constant which dpnds on a fnc point. It is oftn convnint to choos a fnc point wh is qual to zo. In th gavitational cas, w choos infinity to b th fnc point, with U ( = ) =. Sinc U g dpnds on th fnc point chosn, it is only th potntial ngy diffnc U g that has physical impotanc. Na Eath s sufac wh th gavitational fild g is appoximatly constant, as an objct movs fom th gound to a hight h, th chang in potntial ngy is Ug =+ mgh, and th wok don by gavity is Wg = mgh. U concpt which is closly latd to potntial ngy is potntial. Fom U gavitational potntial can b obtaind as, th U B g B Vg = = ( g / m) d = m F s d g s (3.1.8) Physically Vg psnts th ngativ of th wok don p unit mass by gavity to mov a paticl fom to B. Ou tatmnt of lctostatics is makably simila to gavitation. Th lctostatic foc F givn by Coulomb s law also has an invs-squa fom. In addition, it is also consvativ. In th psnc of an lctic fild E, in analogy to th gavitational fild g, w dfin th lctic potntial diffnc btwn two points and Bas B V = ( F / q ) d s = B E d s (3.1.9) wh q is a tst chag. Th potntial diffnc V psnts th amount of wok don p unit chag to mov a tst chag q fom point to B, without changing its kintic ngy. gain, lctic potntial should not b confusd with lctic potntial ngy. Th two quantitis a latd by Th SI unit of lctic potntial is volt (V): U = q V (3.1.1) 1volt = 1 joul/coulomb (1 V= 1 J/C) (3.1.11) Whn daling with systms at th atomic o molcula scal, a joul (J) oftn tuns out to b too lag as an ngy unit. mo usful scal is lcton volt (V), which is dfind as th ngy an lcton acquis (o loss) whn moving though a potntial diffnc of on volt: 3-4

5 = = V (1.6 1 C)(1V) J (3.1.1) 3. Elctic Potntial in a Unifom Fild Consid a chag + q moving in th diction of a unifom lctic fild E = E ˆ ( j), as shown in Figu 3..1(a). (a) (b) Figu 3..1 (a) chag q which movs in th diction of a constant lctic fild E. (b) mass m that movs in th diction of a constant gavitational fild g. Sinc th path takn is paalll to E, th potntial diffnc btwn points and B is givn by V V V B d E B = = E s = ds = E d < (3..1) B implying that point B is at a low potntial compad to. In fact, lctic fild lins always point fom high potntial to low. Th chang in potntial ngy is U = UB U = qed. Sinc q >, w hav U <, which implis that th potntial ngy of a positiv chag dcass as it movs along th diction of th lctic fild. Th cosponding gavitational analogy, dpictd in Figu 3..1(b), is that a mass m loss potntial ngy ( U = mgd ) as it movs in th diction of th gavitational fild g. Figu 3.. Potntial diffnc du to a unifom lctic fild What happns if th path fom to B is not paalll to E, but instad at an angl θ, as shown in Figu 3..? In that cas, th potntial diffnc bcoms 3-5

6 B V = VB V = E d s = E s = Escosθ = Ey (3..) Not that y incas downwad in F igu 3... H w s onc mo that moving along th diction of th lctic fild E lads to a low lctic potntial. What would th chang in potntial b if th path w C B? In this cas, th potntial diffnc consists of two contibutions, on fo ach sgmnt of th path: V = V + V (3..3) C BC Whn moving fom to C, th chang in potntial is VC = E y. On th oth hand, whn going fom C to B, V BC = sinc th path is ppndicula to th diction of E. Thus, th sam sult is obtaind ispctiv of th path takn, consistnt with th fact that E is consvativ. Notic that fo th path C B, wok is don by th fild only along th sgmnt C which is paalll to th fild lins. Points B and C a at th sam lctic potntial, i.., VB = V C. Sinc U = q V, this mans that no wok is quid in moving a chag fom B to C. In fact, all points along th staight lin conncting B and C a on th sam quipotntial lin. mo complt discussion of quipotntial will b givn in Sction Elctic Potntial du to Point Chags Nxt, lt s comput th potntial diffnc btwn two points and B du to a chag +Q. Th lctic fild poducd by Q is E = ( Q/4 πε ) ˆ, wh ˆ is a unit vcto pointing towad th fild point. Figu Potntial diffnc btwn two points du to a point chag Q. Fom Figu 3.3.1, w s that ˆ d s = dscosθ = d, which givs B Q B Q Q 1 1 V = V ˆ B V = d s = d = B (3.3.1) 3-6

7 Onc again, th potntial diffnc V dpnds only on th ndpoints, indpndnt of th choic of path takn. s in th cas of gavity, only th diffnc in lctical potntial is physically maningful, and on may choos a fnc point and st th potntial th to b zo. In pactic, it is oftn convnint to choos th fnc point to b at infinity, so that th lctic potntial at a point P bcoms V P P = E d s (3.3.) With this fnc, th lctic potntial at a distanc away fom a point chag Q bcoms 1 Q V() = (3.3.3) Whn mo than on point chag is psnt, by applying th supposition pincipl, th total lctic potntial is simply th sum of potntials du to individual chags: 1 qi V() = k = q i (3.3.4) i i i i summay of compaison btwn gavitation and lctostatics is tabulatd blow: Gavitation Elctostatics Mass m Chag q Mm Gavitational foc F ˆ g = G Gavitational fild g= F / m Potntial ngy chang Gavitational potntial V g B U = F g d s g B = g d s Qq Coulomb foc F ˆ = k Elctic fild E= F / q B Potntial ngy chang U = F d s B Elctic Potntial V = E d s GM Q Fo a souc M: Vg = Fo a souc Q: V = k U = mgd (constant g ) U = qed (constant E ) g 3-7

8 3.3.1 Potntial Engy in a Systm of Chags If a systm of chags is assmbld by an xtnal agnt, thn U = W =+ Wxt. That is, th chang in potntial ngy of th systm is th wok that must b put in by an xtnal agnt to assmbl th configuation. simpl xampl is lifting a mass m though a hight h. Th wok don by an xtnal agnt you, is + mgh (Th gavitational fild dos wok mgh ). Th chags a bought in fom infinity without acclation i.. thy a at st at th nd of th pocss. Lt s stat with just two chags and q. Lt th potntial du to q1 at a point P b V1 (Figu 3.3.). q1 Figu 3.3. Two point chags spaatd by a distanc 1. Th wok don by an agnt in binging th scond chag q fom infinity to P is W thn W = qv1. (No wok is quid to st up th fist chag and W 1 = ). Sinc V = q /4 πε, wh is th distanc masud fom q to P, w hav U W 1 qq 1 1 = = (3.3.5) 1 If q1 and q hav th sam sign, positiv wok must b don to ovcom th lctostatic pulsion and th potntial ngy of th systm is positiv, U 1 >. On th oth hand, if th signs a opposit, thn U 1 < du to th attactiv foc btwn th chags. Figu systm of th point chags. To add a thid chag q 3 to th systm (Figu 3.3.3), th wok quid is 3-8

9 q 3 q1 q W3 = q3( V1+ V ) = + 13 (3.3.6) 3 Th potntial ngy of this configuation is thn U = W + W = 1 qq qq + qq + = U + U + U (3.3.7) Th quation shows that th total potntial ngy is simply th sum of th contibutions fom distinct pais. Gnalizing to a systm of N chags, w hav U N N 1 qq i j πε i= 1 j= 1 ij j> i = 4 (3.3.8) wh th constaint j > i is placd to avoid doubl counting ach pai. ltnativly, on may count ach pai twic and divid th sult by. This lads to U 1 qq 1 1 q 1 = = = 8πε 4 N N N N N i j j q i i 1 j 1 ij i 1 πε j 1 = = = = ij i= 1 j i j i i i qv( ) (3.3.9) wh V( i ), th quantity in th panthsis, is th potntial at th oth chags. i (location of q i ) du to all 3.4 Continuous Chag Distibution If th chag distibution is continuous, th potntial at a point P can b found by summing ov th contibutions fom individual diffntial lmnts of chag. dq Figu Continuous chag distibution 3-9

10 Consid th chag distibution shown in Figu Taking infinity as ou fnc point with zo potntial, th lctic potntial at P du to dq is dv 1 dq = (3.4.1) Summing ov contibutions fom all diffntial lmnts, w hav V 1 dq = (3.4.) 3.5 Diving Elctic Fild fom th Elctic Potntial In Eq. (3.1.9) w stablishd th lation btwn E and V. If w consid two points which a spaatd by a small distanc ds, th following diffntial fom is obtaind: x y z dv = E ds In Catsian coodinats, E= E ˆi+ E ˆj + E kˆ and ds = dx ˆ i+ dy ˆ j+ dzk ˆ, w hav which implis ( ˆ ˆ ˆ ) ( ˆ ˆ ˆ ) x y z x y z (3.5.1) dv = E i+ E j+ E k dxi+ dyj+ dzk = E dx+ E dy+ E dz (3.5.) V V V Ex =, Ey =, Ez = x y z (3.5.3) By intoducing a diffntial quantity calld th dl (gadint) opato ˆ ˆ i + j+ k ˆ (3.5.4) x y z th lctic fild can b wittn as E ˆ ˆ ˆ V ˆ V ˆ V E ˆ ˆ ˆ ˆ = i x + E j y + E k z = + = + V = V x i+ y j z k x i+ y j z k E = V (3.5.5) Notic that opats on a scala quantity (lctic potntial) and sults in a vcto quantity (lctic fild). Mathmatically, w can think of E as th ngativ of th gadint of th lctic potntial V. Physically, th ngativ sign implis that if 3-1

11 V incass as a positiv chag movs along som diction, say x, with V / x>, thn th is a non-vanishing componnt of E in th opposit diction ( E x ). In th cas of gavity, if th gavitational potntial incass whn a mass is liftd a distanc h, th gavitational foc must b downwad. If th chag distibution posssss sphical symmty, thn th sulting lctic fild is a function of th adial distanc, i.., E= E ˆ. In this cas, dv = E d. If V( ) is known, thn E may b obtaind as dv E ˆ E= = ˆ d (3.5.6) Fo xampl, th lctic potntial du to a point chag q is V() = q/. Using th abov fomula, th lctic fild is simply E = ( q/ 4 πε ). ˆ Gadint and Equipotntials Suppos a systm in two dimnsions has an lctic potntial V( x, y). Th cuvs chaactizd by constant V( x, y) a calld quipotntial cuvs. Exampls of quipotntial cuvs a dpictd in Figu blow. Figu Equipotntial cuvs In th dimnsions w hav quipotntial sufacs and thy a dscibd by V( x, y, z ) =constant. Sinc E = V, w can show that th diction of E is always ppndicula to th quipotntial though th point. Blow w giv a poof in two dimnsions. Gnalization to th dimnsions is staightfowad. Poof: Rfing to Figu 3.5., lt th potntial at a point Pxy (, ) b V( x, y). How much is V changd at a nighboing point Px ( + dxy, + dy)? Lt th diffnc b wittn as 3-11

12 dv = V( x+ dx, y+ dy) V( x, y) V V V V = V( x, y) + dx+ dy+ V( x, y) dx+ dy x y x y (3.5.7) Figu 3.5. Chang in V whn moving fom on quipotntial cuv to anoth With th displacmnt vcto givn by ds = dxˆi+ dyĵ, w can wit dv as V ˆ V ˆ dv = ( dxˆ dyˆ + + ) = ( V) d = d x i y j i j s E s (3.5.8) If th displacmnt d s is along th tangnt to th quipotntial cuv though P(x,y), thn dv = bcaus V is constant vywh on th cuv. This implis that E d s along th quipotntial cuv. That is, E is ppndicula to th quipotntial. In Figu w illustat som xampls of quipotntial cuvs. In th dimnsions thy bcom quipotntial sufacs. Fom Eq. (3.5.8), w also s that th chang in potntial dv attains a maximum whn th gadint V is paalll to d s : dv max ds = V (3.5.9) Physically, this mans that V always points in th diction of maximum at of chang of V with spct to th displacmnt s. Figu Equipotntial cuvs and lctic fild lins fo (a) a constant E fild, (b) a point chag, and (c) an lctic dipol. 3-1

13 Th poptis of quipotntial sufacs can b summaizd as follows: (i) (ii) Th lctic fild lins a ppndicula to th quipotntials and point fom high to low potntials. By symmty, th quipotntial sufacs poducd by a point chag fom a family of concntic sphs, and fo constant lctic fild, a family of plans ppndicula to th fild lins. (iii) Th tangntial componnt of th lctic fild along th quipotntial sufac is zo, othwis non-vanishing wok would b don to mov a chag fom on point on th sufac to th oth. (iv) No wok is quid to mov a paticl along an quipotntial sufac. usful analogy fo quipotntial cuvs is a topogaphic map (Figu 3.5.4). Each contou lin on th map psnts a fixd lvation abov sa lvl. Mathmatically it is xpssd as z = f( x, y) = constant. Sinc th gavitational potntial na th sufac of Eath is = gz, ths cuvs cospond to gavitational quipotntials. Vg Figu topogaphic map Exampl 3.1: Unifomly Chagd Rod Consid a non-conducting od of lngth having a unifom chag dnsity λ. Find th lctic potntial at P, a ppndicula distanc y abov th midpoint of th od. Figu non-conducting od of lngth and unifom chag dnsity λ. 3-13

14 Solution: Consid a diffntial lmnt of lngth dx which cais a chag dq = λ dx, as shown in Figu Th souc lmnt is locatd at ( x, ), whil th fild point P is locatd 1/ on th y-axis at (,y). Th distanc fom dx to P is = ( x + y ). Its contibution to th potntial is givn by dv 1 dq 1 λ dx = = 4 πε ( x + y ) 1/ Taking V to b zo at infinity, th total potntial du to th nti od is V λ / dx λ = ln x = + x + y x + y / / / λ ( /) + ( /) + y = ln 4 πε ( / ) + ( / ) + y (3.5.1) wh w hav usd th intgation fomula dx x + y ln ( x x ) y = + + plot of V( y)/ V, wh = λ /, as a function of y / is shown in Figu V Figu Elctic potntial along th axis that passs though th midpoint of a nonconducting od. In th limit y, th potntial bcoms 3-14

15 V λ ( /) + / 1 + ( y/ ) λ ( y/ ) = ln = ln ( /) / 1 ( y/ ) ( y/ ) λ λ ln ln = y / y λ = ln πε y (3.5.11) Th cosponding lctic fild can b obtaind as E y V λ / = = y πε y + y ( /) in complt agmnt with th sult obtaind in Eq. (.1.9). Exampl 3.: Unifomly Chagd Ring Consid a unifomly chagd ing of adius R and chag dnsity λ (Figu 3.5.7). What is th lctic potntial at a distanc z fom th cntal axis? Figu non-conducting ing of adius R with unifom chag dnsity λ. Solution: Consid a small diffntial lmnt d = Rdφ on th ing. Th lmnt cais a chag dq = λ d = λr dφ, and its contibution to th lctic potntial at P is dv 1 dq 1 λrdφ = = R + z Th lctic potntial at P du to th nti ing is 3-15

16 1 λr 1 πλr 1 Q V = dv = dφ = = R + z R + z R + z (3.5.1) wh w hav substitutd Q= π Rλ fo th total chag on th ing. In th limit z R, th potntial appoachs its point-chag limit: V 1 Q z Fom Eq. (3.5.1), th z-componnt of th lctic fild may b obtaind as E z V 1 Q 1 Qz = = = z z R + z 4 πε ( R + z ) 3/ (3.5.13) in agmnt with Eq. (.1.14). Exampl 3.3: Unifomly Chagd Disk Consid a unifomly chagd disk of adius R and chag dnsityσ lying in th xyplan. What is th lctic potntial at a distanc z fom th cntal axis? Figu non-conducting disk of adius R and unifom chag dnsity σ. Solution: Consid a cicula ing of adius and width d. Th chag on th ing is dq = σ d = σ( π d ). Th fild point P is locatd along th z -axis a distanc z fom th plan of th disk. Fom th figu, w also s that th distanc fom a point on 1/ th ing to P is = ( + z ). Thfo, th contibution to th lctic potntial at P is dv 1 dq 1 σ (π d ) = = + z 3-16

17 By summing ov all th ings that mak up th disk, w hav R σ R πd σ σ V = = + z = R + z z + z ε ε (3.5.14) In th limit z R, 1/ R R R z z z + = 1 + = 1 + +, z z and th potntial simplifis to th point-chag limit: V σ R 1 σ( πr ) 1 Q = = ε z 4 πε z 4 πε z s xpctd, at lag distanc, th potntial du to a non-conducting chagd disk is th sam as that of a point chag Q. compaison of th lctic potntials of th disk and a point chag is shown in Figu Figu Compaison of th lctic potntials of a non-conducting disk and a point chag. Th lctic potntial is masud in tms of V = Q/ R. Not that th lctic potntial at th cnt of th disk ( z = ) is finit, and its valu is V σ R Q R 1 Q = = = = V (3.5.15) R c ε πr ε This is th amount of wok that nds to b don to bing a unit chag fom infinity and plac it at th cnt of th disk. Th cosponding lctic fild at P can b obtaind as: E z V σ z z = = + z ε z R z (3.5.16) 3-17

18 which ags with Eq. (.1.18). In th limit R z, th abov quation bcoms E = σ /ε, which is th lctic fild fo an infinitly lag non-conducting sht. z Exampl 3.4: Calculating Elctic Fild fom Elctic Potntial Suppos th lctic potntial du to a ctain chag distibution can b wittn in Catsian Coodinats as V( x, y, z) = x y + Bxyz wh, B and C a constants. What is th associatd lctic fild? Solution: Th lctic fild can b found by using Eq. (3.5.3): Thfo, th lctic fild is E E E x = = xy y = = x y z V x V y V = = Bxy z Byz Bxz ˆ E= ( xy Byz) i ( x y + Bxz) ˆj Bxy kˆ. 3.6 Summay foc F is consvativ if th lin intgal of th foc aound a closd loop vanishs: F d s = Th chang in potntial ngy associatd with a consvativ foc F acting on an objct as it movs fom to B is B U = UB U = F d s 3-18

19 Th lctic potntial diffnc V btwn points and B in an lctic fild E is givn by U B V = VB V = = d q E s Th quantity psnts th amount of wok don p unit chag to mov a tst chag fom point to B, without changing its kintic ngy. q Th lctic potntial du to a point chag Q at a distanc away fom th chag is V 1 Q = Fo a collction of chags, using th supposition pincipl, th lctic potntial is V 1 Qi = i i Th potntial ngy associatd with two point chags and q spaatd by a distanc 1 is q1 U 1 qq = 1 1 Fom th lctic potntial V, th lctic fild may b obtaind by taking th gadint of V : E = V In Catsian coodinats, th componnts may b wittn as V V V Ex =, Ey =, Ez = x y z Th lctic potntial du to a continuous chag distibution is V 1 = dq 3-19

20 3.7 Poblm-Solving Statgy: Calculating Elctic Potntial In this chapt, w showd how lctic potntial can b calculatd fo both th disct and continuous chag distibutions. Unlik lctic fild, lctic potntial is a scala quantity. Fo th disct distibution, w apply th supposition pincipl and sum ov individual contibutions: V = k i qi i Fo th continuous distibution, w must valuat th intgal V = k dq In analogy to th cas of computing th lctic fild, w us th following stps to complt th intgation: dq (1) Stat with dv = k. () Rwit th chag lmnt dq as λ dl dq = σ d ρ dv (lngth) (aa) (volum) dpnding on whth th chag is distibutd ov a lngth, an aa, o a volum. (3) Substitut dq into th xpssion fo dv. (4) Spcify an appopiat coodinat systm and xpss th diffntial lmnt (dl, d o dv ) and in tms of th coodinats (s Tabl.1.) (5) Rwit dv in tms of th intgation vaiabl. (6) Complt th intgation to obtain V. Using th sult obtaind fo V, on may calculat th lctic fild by E = V. Futhmo, th accuacy of th sult can b adily chckd by choosing a point P which lis sufficintly fa away fom th chag distibution. In this limit, if th chag distibution is of finit xtnt, th fild should bhav as if th distibution w a point chag, and falls off as 1/. 3-

21 Blow w illustat how th abov mthodologis can b mployd to comput th lctic potntial fo a lin of chag, a ing of chag and a unifomly chagd disk. Chagd Rod Chagd Ring Chagd disk Figu () Expss dq in tms of chag dnsity (3) Substitut dq into xpssion fo dv dq = λ dx dq = λ dl dq = σ d dv k λ dx dv k λ dl dv = = = k σ d (4) Rwit and th diffntial lmnt in tms of th appopiat coodinats dx = x + y dl = R dφ = R + z d = π d = + z λ dx (5) Rwit dv dv = k 1/ ( x + y ) dv λrdφ = k ( R ) dv = k πσ d ( + z ) 1/ 1/ + z (6) Intgat to gt V λ V = / / dx x + y λ ( /) + ( /) + y = ln 4 πε ( /) + ( /) + y Rλ V = k dφ 1/ ( R + z ) = k = k ( πrλ) R + z Q R + z V= k πσ R d ( + z ) 1/ ( ) ( ) = kπσ z + R z kq = z + R z R Div E fom V V Ey = y = λ / πε y ( /) + y E z V kqz = = z ( R + z ) 3/ V kq z z Ez = = z R z z + R Point-chag limit kq Ey fo E y y E z kq z kq z z R E z z R 3-1

22 3.8 Solvd Poblms Elctic Potntial Du to a Systm of Two Chags Consid a systm of two chags shown in Figu Figu Elctic dipol Find th lctic potntial at an abitay point on th x axis and mak a plot. Solution: Th lctic potntial can b found by th supposition pincipl. t a point on th x axis, w hav 1 q 1 ( q) q 1 1 V( x) = + = 4 πε x a 4 πε x+ a 4 πε x a x+ a Th abov xpssion may b wittn as V( x) 1 1 = V x/ a 1 x/ a+ 1 wh V = q/ a. Th plot of th dimnsionlss lctic potntial as a function of x/a. is dpictd in Figu Figu

23 s can b sn fom th gaph, V( x ) divgs at x/ a = ± 1, wh th chags a locatd Elctic Dipol Potntial Consid an lctic dipol along th y-axis, as shown in th Figu Find th lctic potntial V at a point P in th x-y plan, and us V to div th cosponding lctic fild. By supposition pincipl, th potntial at P is givn by Figu V 1 q q = Vi = i 4 πε + wh a acosθ ± = +. If w tak th limit wh a, thn 1 1 1/ 1 1 = 1 ( a/ ) ( a/ )cosθ 1 ( a/ ) ( a/ )cosθ + = ± + ± and th dipol potntial can b appoximatd as q 1 1 V = 1 ( a/ ) ( a/ )cosθ 1 ( a/ ) ( a/ )cosθ q acosθ pcosθ ˆ = = p wh p = aqˆj is th lctic dipol momnt. In sphical pola coodinats, th gadint opato is 1 ˆ 1 = ˆ+ θ+ φ ˆ θ sinθ φ 3-3

24 Sinc th potntial is now a function of both and θ, th lctic fild will hav componnts along th ˆ and ˆθ dictions. Using E = V, w hav E V pcosθ 1 V psinθ = =, Eθ = = 3 4, E φ = πε θ πε Elctic Potntial of an nnulus Consid an annulus of unifom chag dnsity σ, as shown in Figu Find th lctic potntial at a point P along th symmtic axis. Solution: Figu n annulus of unifom chag dnsity. Consid a small diffntial lmnt d at a distanc away fom point P. Th amount of chag containd in d is givn by dq = σ d = σ( ' dθ) d ' Its contibution to th lctic potntial at P is dv 1 dq 1 σ dd ' ' θ = = 4 πε ' + z Intgating ov th nti annulus, w obtain V σ b π dd ' ' θ πσ b ds ' σ b z a z πε a a πε ε = = = ' + z 4 ' + z wh w hav mad usd of th intgal 3-4

25 s ds s + z = s + z Notic that in th limit a and b R, th potntial bcoms = σ + ε V R z z which coincids with th sult of a non-conducting disk of adius R shown in Eq. (3.5.14) Chag Moving Na a Chagd Wi thin od xtnds along th z-axis fom z = d to z = d. Th od cais a positiv chag Q unifomly distibutd along its lngth d with chag dnsity λ = Q/d. (a) Calculat th lctic potntial at a point z > d along th z-axis. (b) What is th chang in potntial ngy if an lcton movs fom z = 4d to z = 3d? (c) If th lcton statd out at st at th point z = 4d, what is its vlocity at z = 3d? Solutions: (a) Fo simplicity, lt s st th potntial to b zo at infinity, V ( ) =. Consid an infinitsimal chag lmnt dq = λ dz locatd at a distanc z ' along th z-axis. Its contibution to th lctic potntial at a point z > d is dv λ dz ' = 4 πε z z' Intgating ov th nti lngth of th od, w obtain λ z d dz' λ z + d V() z = ln = z z' z d z+ d (b) Using th sult divd in (a), th lctical potntial at z = 4d is Similaly, th lctical potntial at z λ 4d+ d λ 5 V( z= 4 d) = ln = ln 4d d 3 = 3d is 3-5

26 λ 3d+ d λ V( z= 3 d) = ln = ln 3d d Th lctic potntial diffnc btwn th two points is λ 6 V = V( z = 3 d) V( z = 4 d) = ln > 5 Using th fact that th lctic potntial diffnc V is qual to th chang in potntial ngy p unit chag, w hav wh q= is th chag of th lcton. λ 6 U = q V = ln < 5 (c) If th lcton stats out at st at z = 4d thn th chang in kintic ngy is 1 K = mv f By consvation of ngy, th chang in kintic ngy is λ 6 K = U = ln > 5 Thus, th magnitud of th vlocity at z = 3d is v f λ 6 = ln m Concptual Qustions 1. What is th diffnc btwn lctic potntial and lctic potntial ngy?. unifom lctic fild is paalll to th x-axis. In what diction can a chag b displacd in this fild without any xtnal wok bing don on th chag? 3. Is it saf to stay in an automobil with a mtal body duing sv thundstom? Explain. 3-6

27 4. Why a quipotntial sufacs always ppndicula to lctic fild lins? 5. Th lctic fild insid a hollow, unifomly chagd sph is zo. Dos this imply that th potntial is zo insid th sph? 3.1 dditional Poblms Cub How much wok is don to assmbl ight idntical point chags, ach of magnitud q, at th cons of a cub of sid a? 3.1. Th Chags 18 6 Th chags with q = 3. 1 C and q 1 = 6 1 C a placd on th x-axis, as shown in th figu Th distanc btwn q and q1 is a =.6 m. Figu (a) What is th nt foc xtd on q by th oth two chags q 1? (b) What is th lctic fild at th oigin du to th two chags q 1? (c) What is th lctic potntial at th oigin du to th two chags q 1? Wok Don on Chags Two chags q 1 = 3. µ C and q = 4. µ C initially a spaatd by a distanc =.cm. n xtnal agnt movs th chags until thy a = 5.cm apat. (a) How much wok is don by th lctic fild in moving th chags fom to? Is th wok positiv o ngativ? (b) How much wok is don by th xtnal agnt in moving th chags fom to? Is th wok positiv o ngativ? f f f 3-7

28 (c) What is th potntial ngy of th initial stat wh th chags a apat? =. cm (d) What is th potntial ngy of th final stat wh th chags a = 5.cm f apat? () What is th chang in potntial ngy fom th initial stat to th final stat? Calculating E fom V Suppos in som gion of spac th lctic potntial is givn by 3 Eaz 3/ V( x, y, z) = V E z+ ( x + y + z ) wh a is a constant with dimnsions of lngth. Find th x, y, and th z-componnts of th associatd lctic fild Elctic Potntial of a Rod od of lngth L lis along th x-axis with its lft nd at th oigin and has a nonunifom chag dnsity λ = α x,wh α is a positiv constant. (a) What a th dimnsions of α? (b) Calculat th lctic potntial at. Figu 3.1. (c) Calculat th lctic potntial at point B that lis along th ppndicula biscto of th od a distanc b abov th x-axis. 3-8

29 3.1.6 Elctic Potntial Suppos that th lctic potntial in som gion of spac is givn by V( x, y, z) = V xp( k z )cos kx. Find th lctic fild vywh. Sktch th lctic fild lins in th x z plan Calculating Elctic Fild fom th Elctic Potntial Suppos that th lctic potntial vais along th x-axis as shown in Figu blow. Figu Th potntial dos not vay in th y- o z -diction. Of th intvals shown (igno th bhavio at th nd points of th intvals), dtmin th intvals in which has (a) its gatst absolut valu. [ns: 5 V/m in intval ab.] (b) its last. [ns: (b) V/m in intval cd.] (c) Plot E x as a function of x. (d) What sot of chag distibutions would poduc ths kinds of changs in th potntial? Wh a thy locatd? [ns: shts of chag xtnding in th yz diction locatd at points b, c, d, tc. along th x-axis. Not again that a sht of chag with chag p unit aa σ will always poduc a jump in th nomal componnt of th lctic fild of magnitud σ / ε ]. E x 3-9

30 3.1.8 Elctic Potntial and Elctic Potntial Engy ight isoscls tiangl of sid a has chags q, +q and q aangd on its vtics, as shown in Figu Figu (a) What is th lctic potntial at point P, midway btwn th lin conncting th +q and q chags, assuming that V = at infinity? [ns: q/ πε o a.] (b) What is th potntial ngy U of this configuation of th chags? What is th significanc of th sign of you answ? [ns: q /4 πε o a, th ngativ sign mans that wok was don on th agnt who assmbld ths chags in moving thm in fom infinity.] (c) fouth chag with chag +3q is slowly movd in fom infinity to point P. How much wok must b don in this pocss? What is th significanc of th sign of you answ? [ns: +3q / πε o a, th positiv sign mans that wok was don by th agnt who movd this chag in fom infinity.] Elctic Fild, Potntial and Engy Th chags, +5Q, 5Q, and +3Q a locatd on th y-axis at y = +4a, y =, and y = 4a, spctivly. Th point P is on th x-axis at x = 3a. (a) How much ngy did it tak to assmbl ths chags? (b) What a th x, y, and z componnts of th lctic fild E at P? (c) What is th lctic potntial V at point P, taking V = at infinity? (d) fouth chag of +Q is bought to P fom infinity. What a th x, y, and z componnts of th foc F that is xtd on it by th oth th chags? () How much wok was don (by th xtnal agnt) in moving th fouth chag +Q fom infinity to P? 3-3