Electric_force_electric_field_examples

Electric_force_electric_field_examples - kr 8,7ax/0W/Z:...

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Unformatted text preview: kr 8,7ax/0W/Z: .— 6 Example: p 636, #5 /flC v” /0 C Three point charges are located at corners of an equilateral triangle as shown. Calculate the resultant force on the 7.00—pC charge. [/ng U/Vf/ ytny’U/‘g‘ /:/ : éwé (6956005: 99/2 00; “'2— Em)‘; , : mama/Vi » (9 436%; KM”? 3 ml)‘ 0 \pb“ 3 F, Example: p 636, #8 Two small beads having positive charges 3g and q are fixed at opposite ends of a horizontal, insulating rod, extending from the origin to the point x = d. A third small charged bead is free to slide on the rod. At what position is the third bead in equilibrium? Can it be in stable equilibrium? l d 62 61/va 15 ' r: 7 j" """""""""""""""""""""""""" " “"— 6 "' F6? 0 0 +30 +q ’ ' 7/ , 7 I f‘ , 6/? 56% 7/6 0 2 16? O/Ikcfier UWMM 625% fly»? Q X? Example: (a) Show that the electric field at point P a distancey above the midpoint of an electric dipole aligned along the x-axis is: !Xflfii’zi : (a) f/eclm': Fz’e/J 450/! MI'J/Na’ P4404 " Podyf 7 I; (fiqr'jlhf flat-‘1 +5 J‘g, _ .__..——-’ r II; )0 / ‘ A 3 I’ : ,5; £7: [.7 + 5,, :0 I I ‘r 1; 60:11:49 E ‘E — A“ a? lea—916a»! M “' a'q’ a“ "Er Etr‘gzx' iii—3 Coséo azty’ a. gal cos a: W 312.3 a 2L6“ a "I{ d.yl/( £0" '1? E“, z a" (A, Ii”; } mud/k Dl‘a { “(A'r Alva. ,7Lt- é¢CQ.¢$( /,\(//;;’5 fl" {6% /,v( «3 paw F116 , anal. (b) Show that the electric field at point P a distance x along the x-axis from the midpoint of an electric dipole aligned along the x-axis is: ,7 1121—1 (C) Using (I‘HC) “1+”x+ (2 )x2+"',showthatfor x>>a the answer in part (b) becomes ~ 4keqa EP 2 — 3 i x (éavle-a-d A P ii ® 0 fl f 1% -1 I é—-—--———-""'_ gr _——-—————-————--—-"‘ é- : 0 ) - lit—3 A ' (my a; min /5/) f = i4. ) - w A I £2 ' (AI-(q): (A .6 5 I -: fl: 5—.- ._—-— g. / 2.4 2 “‘ H3!) C/"ZQ’ ] t 0 (v 4. / _ .. (1%,)! 0‘2) “A” /'.Z§.’3 .1. '2 (#53:); ' “a” "V" a; 5‘0 £15" 5 A‘s 2’ 7%?» {I 273[(”?O’{"‘2;q)j (:(.’Q,¢X) 44¢ q i 1 Example: An electron (g: -e) starts with Vi = 0 at x = d and accelerates toward x20. What is vwhen x20? What is the kinetic energy of the electron at x = 0 ? < Bleéém: 6:»: W: 7/,‘v‘0 a} at“! acqhwfif bun/J *pa. (’1’. Offlifilk Dr“ J g Ax: w" a*‘ "‘2’; M ’0; - o 2. User vzr’v, + ‘24,, AX (consulachn-flau) [lit/inc; [4/52 4y: ,é: gag/Z .Zéfof}:a)rlo7 0265}! :eEA’ ‘2 will 92!. /¢fl" Ht [‘5 fig a P“ adhqt. r‘n [/cofifvz, pa [in . (Aw ’ an? ‘3 any, a E/eJ. 41-597,), A charge moving across E field lines is deflected: o For an electron projected into a uniform field ~ ' E 2 E} with ‘71 = Vhl‘ , acceleration is: v=vfi + + + + o This is like projectile motion EXCEPT that E field only exists for range of x. Example: An electron is projected into a uniform field E = with ‘71 = ijf . The field exists between x = O and x :1 . (a) What is the electron velocity when it leaves the region of electric field? (b) Relative to its starting level, what is the vertical displacement of the electron when it leaves the region of electric field? Application: Cathode ray tube: __" 0 can accelerate and steer electrons to a I pOint on the screen. 8 "I I. ' " '" a .. — "‘ -. .. + “" Example: An electron is projected into a uniform field E = Ej with 13} = inf . The field exists between x = 0 and x =1 . (a) What is the electron velocity when it leaves the region of electric field? (b) Relative to its starting level, what is the vertical displacement of the electron when it leaves the region of electric field? " ' ~ “ ‘f— we - Q 7" "' ~ " {~11 -—[... ‘1‘}4¥ L a, "‘ W \ 5 \ I F :20 x:( fl"; 1’; //ll Flo jet}! 7: low flow, e/Lcfirm {'5 6! pa“ X30 «aekf‘g. - e E M (Vd}¢ Iol‘m am flc’. Jury: O//¢‘ré 2"» J g ) 4x30 a]: (a) 0: /oc/ A: ’U,; = ’14," 7/4”" 7/," Ca, 0":0) - - eE ’ _ e52 ' UN ' mu"- 'o (a; ’ W“ 2‘ - 5%. If 6"” 0 Example: A cube with faces of area A is oriented with edges parallel to E in a region of uniform electric field. What is the total electric flux through the surface of the cube? / —’ A: . .. £0» H2 4 gi‘oéé; Tin?” h; 4,23 =3 (#5 70 ) «- $9» H-L lap} 31%: 613: po-‘Js “(f w 095‘ = 5/} cos “'0‘ ‘ «Fl? 9.» M 9908: 4957' _. £9!“ [a I‘I\J"' I’Aflkllvu (pg: 5",; :[fltos a» : +529 Bo» M We ' [a («5? ¢ 0 : “5’4 *[4’ 0 (a! ma 6‘ 7°“ ’V r7 as come; 041/!) Example: Use Gauss’s law to find the electric field around a point charge q. @0911”) 7 7 @ Cd 1m" [5 gym/72397 % g {I} =5? 77mm aw? pram$ @ CA00§€ ‘Gaaggx‘m gar/Mi Ppck gf/ré're way/‘45: f [a ré,/L/a/?) ‘ OLVL Zn % r \r“ ‘ L ,3 , A .' 77W /:,/ « EN) «V ‘f’ ‘.‘ vEn 1;; QMZfiwAe/e (944 §M7flw¢€‘ L// \-V En gym/mm” PM £1547: Hm ~ 3 g” : ékflngQ/C BUT 649’wa A/M/ S4 we» ' v A / glxflpvég QC/MX %raw7A SM #085 €09 5] xe‘ffzy fifii‘aé? {4/ (Mfg/70): % x A? W720 r7 f? CU; éfiflpeaw flh, 0W ...
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This note was uploaded on 03/08/2012 for the course PHYSICS 1051 taught by Professor Michaelmorrow during the Winter '12 term at Memorial University.

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Electric_force_electric_field_examples - kr 8,7ax/0W/Z:...

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