Chapter 2: Electric Field

Transcription

1 P 6 Genel Phsics II Lectue Outline. The Definition of lectic ield. lectic ield Lines 3. The lectic ield Due to Point Chges 4. The lectic ield Due to Continuous Chge Distibutions 5. The oce on Chges in lectic ields lectic field A chge cn expeience n electosttic foce due to the pesence of othe chges. The ide of field is equied in ode to explin the "ction t distnce." Recll the ide of gvittionl field. m g g Chpte : lectic ield In this view, th cetes foce on the mss m. This is insne. th isn t even touching the mss. So we intoduce the ide of gvittionl field. Now we tke the view tht the field due to th, g, is exeting the gvittionl foce on the mss. Since the foce is g mg, the gvittionl field is defined s, electic foce. q g g. Let s tke the sme ppoch with the m Insted of thinking of q exeting the foce on q, we think of q ceting field nd the field exeting the foce on q. Mthemticll, we cn wite q e which is the definition of the electic field. The electic field is vecto, nd its diection is the sme s the diection of the foce on positive test chge. It hs units Newton pe coulomb (N/C) Note: -The suounding chges (not q) is the ones tht cete n electic field. -Simill between electic field nd gvittionl field q e q e Gzintep Univesit cult of ngineeing Deptment of ngineeing Phsics

2 P 6 Genel Phsics II The electic field due to point chge kq Definition of the electic field The electic field due to N chged pticles n i kq i i i xmple: A point chge Q +. C is t the oigin, nd second chge Q -5. C is plced on the -xis t. m. Wht is the totl electic field t the point P with coodintes x. m,. m? We use Coulomb lws to find the field due to ech chge, nd then we find the vecto sum. ist, we found the field due to Q Q 9 N. m (.C) ( i + ) j k (9. ) [.6 i +.8 ] j C [(.m) + (.m) ] 5 The field t P due to the negtive chge Q is in the x diection: Q 9 N. m ( 5.C) (9. ) k i i(. N / C) C (.m) i.8 + [.6 + j.] i N / C [.5 i +.8 ] j N / C o (.94x )N/C, in diection.5 i +.8 j N / C lectic field lines We need moe desciptive imge of the field. The most useful ide is lectic ield Lines. ield Lines o Lines of oce e used to visulize the field. The ules fo dwing them e:. The tngent to the field line points in the diection of the foce on positive test chge.. The densit of lines is popotionl to the stength of the field (bigge chges mke moe lines). One w to visulize the electic field is dwing connecting lines of vecto (electic field) nd these connecting lines e clled electic field lines. Gzintep Univesit cult of ngineeing Deptment of ngineeing Phsics

3 P 6 Genel Phsics II Clcultion of lectic field due to point chges In this section we will demonstte clcultion of electic field on illusttive exmples. xmple A poton nd n electon fom two cones of n equiltel tingle with sides of length 3 x -6 m. Wht is the mgnitude of thei net electic field t the thid cone? Answe: o this poblem we dw s digm of the chges nd find t point P(x,). Known: n q p +.6 x -9 C; kq i q e x -9 C; i i i 3 x -6 m Red line: electic field vecto of poton Blue line: electic field vecto of electon. Since electic field is vecto quntit, we cn wite e+ p cos 6 sin 6 e e x e cos 6 x+ sin 6 p p p q 9 nd mgnitude of e p k ; q.6 C x ( N / C) xmple : Two chges e t the cones of sque of side cm. The net electic field t C due to the chges is long diection nd foce cting on the chge Q due to Q is long + diection. If the mgnitude of foce is N, detemine: ) mgnitude of chges Q nd Q, b) Mgnitude of electic field. Gzintep Univesit cult of ngineeing Deptment of ngineeing Phsics 3

4 P 6 Genel Phsics II c) Wht chge must be plced t D so tht the electic field t C will be zeo? Solution: Remembe electic field is vecto quntit. ) Vecto sum of the fields t point C: + (It is es! to see tht the fist chge will be positive nd second chge will be negtive) Q Q k x; k ( x ) Since hs no x component, then Q Q /. On the othe hnd ccoding to the coulomb lw QQ k ; N then QQ / k I thin ou cn solve the emining. xmple 3 lectic field of dipole moment. An electic dipole is sstem of two equl nd opposite point chges septed b smll distnce. The extended stight line joining the two point chges in dipole is clled the dipole xis. The pependicul bisecto of the dipole xis is clled the neutl xis. Dipole Axis A B Q Q + C D - +Q / / -Q θ Neutl Axis x The question is: clculte electic field t distnce fom the cente of the dipole. The nswe of this question is ve es. om the definition of electic field, we clculte Q + + In pctice is ve smll quntit when it comped to. Distnce is in tomic scle. Theefoe it is necess to mke n ppoximtion ( ). The clcultion is tedious. Let us wite the vectos: ; +. Mgnitude of the vectos cn be clculted fom the dot poduct popet: + cos θ; + + cosθ Gzintep Univesit cult of ngineeing Deptment of ngineeing Phsics 4

5 P 6 Genel Phsics II nd the unit vectos: ; The electic field tkes the fom: Q + 3/ 3/ + cosθ + + cosθ The clcultion cn be completed lte. (Potentil section). Hee we clculte specific fom of the field. lectic field on the dipole xis: To clculte electic field on the dipole xis we set θ nd, is the distnce fom the oigin to point on the positive -xis. The electic field tkes the fom: Q using the identit ( + s) n + ns when s we obtin: Q 3 πε We define the dipole moment p Q ^which is diected fom negtive to positive chge. p Then the electic field:. 3 πε lectic field on the neutl xis: To clculte electic field on the neutl xis we set θ π/ nd xx, x is the distnce fom the oigin to point on the positive x-xis. The electic field tkes the fom: xx xx Q + Q 3/ 3/ 3/ 4 πε x + x + x + using the identit ( + s) n + ns when s we obtin: Q 3 πεx We define the dipole moment p Q ^which is diected fom negtive to positive chge. p Then the electic field:. 3 πεx lectic field of continuous chge distibutions Wht s bout continuous chge? Gzintep Univesit cult of ngineeing Deptment of ngineeing Phsics 5

6 P 6 Genel Phsics II When we del with continuous chge distibutions, it is most convenient to expess the chge on n object s chge densit the thn totl chge. q q The totl electic field is just the sum of the fields of the smll (point) chges q's. i k q i o mximum ccuc we wnt the q's to become smlle nd smlle. In this limit, the sum becomes n integl: dq continuous chge distibution Intepettion of dq o line of chge, fo exmple, we would epot the line chge densit (o chge pe length) λ (C/m). Tble below shows the othe chge densities we shll be using: Nme Smbol SI unit chge q C Line chge densit λ C/m Sufce chge densit σ C/m Volume chge densit ρ C/m 3 xmple: lectic field fo lines of unifom chge ind the electic field t distnce m s in figue, fom the line of chge with chge densit λ C/m, length L. L/ Step dq I: let d be the length of n element II:Relte the chge dq of the element to the length of the element with dqλd III:xpess the field poduced t given point b dq using field eqution fo chge θ x distibution. IV:If the chge on the line is positive, then t given point dw vecto tht points diectl w fom dq. If the chge is negtive, dw the vecto pointing diectl towd dq. The distnce fom dq to the point is! V:xpess in othe foms fo integtion (dx) -L/ e.g. + i i Gzintep Univesit cult of ngineeing Deptment of ngineeing Phsics 6

7 P 6 Genel Phsics II VI:xpess in components e.g., x, VII:Look fo cnceling nd dding components VIII:Tnsfom (x,) to (, θ) o (, θ) to (x,) IX: inlize ou nswe dq λd + λd x cosθ + λd sin θ + substituting cos θ / nd sin θ / we obtin λd x + using the identities: d + ( ) 3/ + ( ) 3/ λd ( + ) 3/ nd nd tking the limits of the integls fom L/ to L/ we obtin: λl x 4 πε + L /4 (s expected) Wht is the electic field s L tends to infinit! xmple d ( ) 3/ + + ind the field on the xis of ing of chge, q, (distibuted unifoml ove the ing) nd dius,, s function of the distnce fom the cente, x. dq θ d x Using the field due to the point chge dq, d k dq. B the smmet of this poblem the field will onl point in the x-diection so we onl hve to wo bout the x- components, d x k dq cosθ.(you cn find the electic field fo component, this will be zeo). Notice tht is the sme fo ll the dq's nd cosθ is lso the sme fo ll dq's so the integl is stight fowd. x k dq cosθ k cosθ dq k q cosθ Gzintep Univesit cult of ngineeing Deptment of ngineeing Phsics 7

8 P 6 Genel Phsics II Note tht + x nd cosθ x so we cn wite + x x k q x ( + x ) 3. Check the nswe b exmining the limits s x goes to zeo nd to infinit. Wht hppens if goes to zeo! xmple ind the electic field distnce, x, w fom disc of dius R nd chge densit, σ(c/m). Solution: In the pevious exmple we hve obtined electic field of ing. We cn use the esult to obtin field of the disc. ield of the ing of dius is given b x k q x + x. x ( ) 3 We think tht disc contins lge numbe of ings of dius chnging fom zeo to R. lectic field of ech ing is diected long x- xis. Hee let us obtin the electic field b diect integtion. Becuse of smmet the electic field diected long x-xis. We cn wite: xdq d x ; dq σρdρdφ ρ + x ( ) 3/ dρ is thickness of ech ing. Integtion ields: σx R πρdρ σ x x. ( ) 3/ ρ + x ε R + x When R tends to infinit we obtin electic field of lge conducto with sufce chge densit σ(c/m). σ x ε Motion of the Chges in lectic ields This section cn be ecognized s the ppliction of the Newton s lws. Plese eview the kinemtics, foce, eneg etc. sections in P 5. The foce on single chge, s shown t the ight, cn now be witten in tems of the field it feels. Accoding to the definition of electic field, e e q. q xmple igue shows n electon of mss m nd chge e pojected with speed v t ight ngles to unifom field. Descibe its motion. q e Gzintep Univesit cult of ngineeing Deptment of ngineeing Phsics 8

9 P 6 Genel Phsics II nd e m since x vt t t m e v Remembe pojectile motion!!! x liminting t ields eqution of tjecto. e x fo the mv xmple: The th hs n electic field of bout 5N/C pointed downwd. A.µm dius wte doplet is suspended in clm i. ind ()the mss of the wte doplet, (b)the chge on the wte doplet nd (c)the numbe of excess electons on the doplet. ()Use the definition of densit nd the volume of sphee, ρ m vol m ρ(vol) ρ4 3 π 3 () 4 3 π(.x 6 ) 3 4.9x 5 kg e g (b)the foces on the doplet e its weight nd the electic foce. Using Newton s Second Lw, Σ m e g e g Using the definitions of the electic nd gvittionl fields, q mg q mg (4.9x 5 )(9.8).74x 6 C 5 (c)since chge is quntized, q Ne N q e xmple p θ 6.74x 7 electons 9.6x A dipole in constnt electic field s shown t the left. It feels no net foce becuse the two foces on it cused b the field e equl nd opposite. The dipole does feel net toque, howeve. This toque tends to lign it with the field. τ ( sin θ) ( q)+ ( sinθ) q ( ) qsinθ p sinθ This toque points into the ppe so we cn wite the toque on the dipole s, τ p The Toque on Dipole The potentil eneg of the dipole cn be found fom the definition of potentil eneg, U W c. The wok done s the dipole ottes though n ngle, dθ, is, dw τ dθ p sinθ dθ. θ The totl wok done s the ngle goes fom π/ to θ is, W psin θ dθ p cosθ. The potentil eneg is U U(θ) U( π ) p cosθ U p, whee the zeo fo potentil eneg is θ π. U p The Potentil neg of Dipole π Gzintep Univesit cult of ngineeing Deptment of ngineeing Phsics 9

10 P 6 Genel Phsics II xmple Wte molecules hve dipole moment of 6.x -3 C m. ind ()the mximum toque on wte molecule in the -field of th nd (b)the potentil eneg lost s the molecule moves fom the position of mximum toque until it ligns with the field. ()The toque on dipole in constnt field is τ p. The mximum occus when the dipole moment is pependicul to the field, τp (6.x 3 )(5) 9.3x 8 N m (b)the potentil eneg of dipole is U p. When the moment is pependicul to the field this is zeo. When the moment is ligned with the field U p 9.3x 8 J. This then is the eneg tht is lost U lost 9.3x 8 J. Gzintep Univesit cult of ngineeing Deptment of ngineeing Phsics