# Answer, Key Homework 8 David McIntyre 1

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1 Answer, Key Homework 8 Dvid McIntyre 1 This print-out should hve 17 questions, check tht it is complete. Multiple-choice questions my continue on the net column or pge: find ll choices before mking your selection. The due time is Centrl time. Chpter problems..8 µc 7. m 001 (prt 1 of 1) 0 points A solid conducting sphere is given positive chrge Q. How is the chrge Q distributed in or on the sphere? 1. It is uniformly distributed throughout the sphere.. Its density decreses rdilly outwrd from the center.. It is concentrted t the center of the sphere. 4. Its density increses rdilly outwrd from the center. 5. It is uniformly distributed on the surfce of the sphere only. correct q µc 5.1 m Find q. Correct nswer: µc. Given : Q 1.8 µc, Q q, Q q, Q 4 µc, 7. m, nd d 5.1 m. Q 1 q The chrge distribution on conductors cn only be on the surfce, nd since on sphericl surfce every point is like ny other surfce point, the chrge distribution is uniform. The electric field is norml to the surfce of conductor. The conductor is symmetricl (since it is sphericl), so the chrge must be uniform. 00 (prt 1 of 1) 10 points Three point chrges re locted t the vertices of n equilterl tringle. The chrge t the top verte of the tringle is given in the figure. The two chrges q t the bottom vertices of the tringle re equl. A fourth chrge µc is plced below the tringle on its symmetryis, nd eperiences zero net force from the other three chrges, s shown in the figure below. Q Q d µc The force on Q 4 is due to the Coulomb forces from Q 1, Q, nd Q. Becuse Q nd Q hve equl chrge, the -components of their forces cncel out (by symmetry). Thus we only need to consider the y-components of the forces. Coulomb s lw tells us F i k Q i Q 4 is the i th force on Q 4 from the ith chrge. The forces from Q nd Q re equl to ech other, nd opposite the direction of the force from Q 1, since otherwise they could not cncel. Totl r

2 Answer, Key Homework 8 Dvid McIntyre force on Q 4 is F Q4 k Q 1 Q 4 r 14 k Q Q 4 r 4 sin with r 14 the distnce between Q 4 nd Q 1, r 4 r 4 the distnce between Q 4 nd either Q or Q, nd indicted in the sketch bove. Remember tht this force F Q4 will be set equl to zero since the problem tells us the forces re in equilibrium. Becuse Q 1, Q, nd Q form n equilterl tringle, of sides of length, it cn be seen tht r 14 Also, ( d sin d r 4 ) nd r Our force eqution becomes d. d 4 Q 1 0 k Q 4 ( ) d Q d ( ). 4 d 4 d 4 4 d. test chrge Q 4, the nswer is lso independent of the sign or mgnitude of the chrge Q (prt 1 of 1) 0 points Two identicl smll chrged spheres hng in equilibrium with the msses shown in the figure. The length of the strings re equl nd the ngle (shown in the figure) with the verticl is identicl m kg 0.0 kg Find the mgnitude of the chrge on ech sphere. Correct nswer: C. Given : L 0.14 m, m 0.0 kg, 6. nd Rerrnging, we get Q Q 1 d [ 4 d] / [ ] d (.8 µc) (5.1 m) [ (7. m) 4 ( 5.1 m) ] / [ ] (7. m) (5.1 m) µc. Note: Neither the sign nor the mgnitude of the chrge Q 4 (given in the problem s µc) enter into this eqution. Actully, the resultnt force on Q 4 is zero mens tht the resultnt electric field is zero. Becuse the electric filed is independent of the q q m m From the right tringle in the figure bove, we see tht sin L. Therefore, L sin (0.14 m) sin(6 ) m. The seprtion of the spheres is r m. The forces cting on one of the spheres re shown in the figure below. L

3 Answer, Key Homework 8 Dvid McIntyre T cos F e mg T T sin Becuse the sphere is in equilibrium, the resultnt of the forces in the horizontl nd verticl directions must seprtely dd up to zero: F T sin F e 0 Fy T cos m g 0. From the second eqution in the system bove, we see tht T m g, so T cn be cos eliminted from the first eqution if we mke this substitution. This gives vlue F e m g tn (0.0 kg) ( 9.8 m/s ) tn(6 ) N, for the electric force. From Coulomb s lw, the electric force between the chrges hs mgnitude F e k e q r, where q is the mgnitude of the chrge on ech sphere. Note: The term q rises here becuse the chrge is the sme on both spheres. This eqution cn be solved for q to give F e r q k e ( N) ( m) ( N m /C ) C. 004 (prt 1 of 1) 0 points Two identicl conducting spheres, A nd B, crry equl chrge. They re seprted by distnce much lrger thn their dimeters. A third identicl conducting sphere, C, is unchrged. Sphere C is first touched to A, then to B, nd finlly removed. As result the electrosttic force between A nd B, which ws originlly F, becomes 1. 1 F. 4 F. 8 F correct F F F 8. F F Since the two conducting spheres re identicl (i.e. sme rdius), when the spheres touch the chrges redistribute themselves eqully between the two spheres. Let spheres A nd B hve n initil chrge Q. When n identicl unchrged sphere C comes in contct with sphere A nd removed, then by conservtion of chrge, ech sphere will crry chrge Q C1 Q A Q/ When sphere C touches sphere B, then ech sphere will crry chrge Q C Q B Q C 1 Q B Q/ Q 4 Q Hence if the initil force is given by Q F i k e then the finl force is d F f k e (/4 Q)(1/ Q) d 8 F i

4 Answer, Key Homework 8 Dvid McIntyre (prt 1 of 1) 0 points Two lrge, prllel, insulting pltes re chrged uniformly with the sme positive rel chrge density σ, which is the chrge per unit re. The permittivity of free spce 1 4πk e. The mgnitude of the resultnt electric field E (where outside stnds for bove nd below the two pltes) is 1. σ between the pltes, zero outside. σ σ. between the pltes, outside.. σ between the pltes, zero outside. 4. Zero between the pltes, σ outside. correct 5. Zero between the pltes, σ outside. 6. σ everywhere. 7. Zero everywhere. σ 8. Zero between the pltes, outside. σ 9. between the pltes, zero outside. 10. σ between the pltes, σ outside. Ech plte produces constnt electric field of E σ directed wy from the plte for positive chrge density, nd towrd the plte for negtive chrge density. Between the two pltes, the two fields cncel ech other so tht E net 0. Outside the two pltes, the fields dd together, so tht E net σ. 006 (prt 1 of ) 4 points Given two chrges q 1.01 µc t the origin, q 5.71 µc, nd 10 cm in the figure below. Identify the direction of E in the region II (0 < <, long -is). I II III O q 1 q 1. ll possibilities: right, left, or zero. right correct. down 4. up 5. left 6. none of these The direction of the electric field t point P is the direction tht positive chrge would move if plced t P. A positive chrge plced in region II would be ttrcted to q nd repelled by q 1. Thus the direction is to the right. 007 (prt of ) points Identify the direction of E in region III. ( >, long -is). 1. left correct. none of these. up 4. right 5. down 6. ll possibilities: right, left, or zero In region III, positive chrge would be forced to the left since q > q 1 nd q is closer to region III. The effect of q domintes nd the direction of the electric field is to the left.

5 008 (prt of ) points Locte the coordinte such tht E 0. (Note tht the origin O is t q 1.) Correct nswer: m. We hve lredy seen tht the electric field is nonzero in regions II nd III. Thus the only cndidte is region I (negtive -is). The point where E 0 is the point where the mgnitudes re equivlent E 1 E. Cll the point where this hppens c. Then k q 1 c k q (c ) Answer, Key Homework 8 Dvid McIntyre 5 y y A II I r III IV O ( ) c c q q 1 1 c q q 1 q is given by 1. q Q π. q Q. q Q π 4. q π Q 5. q Q π B Solving for c, c q q m c m C C 1 We find tht E 0 t m I II III c q 1 q 009 (prt O 1 of ) 4 points Consider the setup shown in the figure below, where the rc is semicircle with rdius r. The totl chrge Q is negtive, nd distributed uniformly on the semicircle. The chrge on smll segment with ngle is lbeled q. 6. None of these. 7. q Q π 8. q Q π 9. q Q π 10. q π Q correct The ngle of semicircle is π, thus the chrge on smll segment with ngle is q Q π 010 (prt of ) points The mgnitude of the -component of the electric field t the center, due to q, is given by 1. E k q (cos ) r. E k q cos r. E k q (sin ) r. correct

6 Answer, Key Homework 8 Dvid McIntyre 6 4. E k q r 5. E k q r k q cos 6. E r k q sin 7. E r 8. E k q (sin ) r 9. E k q sin r 10. E k q (cos ) r Negtive chrge ttrcts positive test chrge. At O, E points towrd q. According to the sketch, the vector E is pointing long the negtive is. The mgnitude of the E is given by E E cos k q r cos. 011 (prt of ) points Given: Q 41 µc, r 86 cm, nd k N m /C. Determine the mgnitude of the electric field t O. Correct nswer: 1718 N/C. By symmetry of the semicircle, the y- component of the electric field t the center is E y 0. Combining prt 1 nd prt, E k q cos r k Q π r cos Therefore, the mgnitude of the electric field t the center is given by E E k Q π r. π/ π/ k Q π r cos d For the bove vlues, the mgnitude is given by E ( N m /C ) ( 41 µc) π (86 cm) 1718 N/C. The direction is long negtive is. y y A II I r E III IV O B 01 (prt 1 of 1) 10 points A line of chrge strts t 0, where 0 is positive, nd etends long the -is to positive infinity. If the liner chrge density is given by λ λ 0 0 /, where λ 0 is constnt, determine the electric field t the origin. (Here î denotes the unit vector in the positive direction.) 1. k λ 0 0 ( î) correct. k λ 0 0 (î). k λ 0 0 (î) 4. k λ 0 0 (î) 5. k λ 0 0 ( î) 6. k λ 0 (î) 0 First we relize tht we re deling with continuous distribution of chrge (s opposed to point chrges). We must divide the distribution into smll elements nd integrte.

7 Answer, Key Homework 8 Dvid McIntyre 7 Using Coulomb s lw, the electric field creted by ech smll element with chrge dq is de k dq where dq λ d λ 0 0 d Now we integrte over the entire distribution (i.e. from 0 to ) nd insert our dq: E k dq 0 k λ 0 0 k λ 0 0 k λ 0 0 d 1 0 Since the distribution is to the right of the point of interest, the electric field is directed long the is if λ 0 is positive. Tht is, positive chrge t the origin would eperience force in the direction of î from this chrge distribution. In fct, the direction of n electric field t point P in spce is defined s the direction in which the electric force cting on positive prticle t tht point P would point. So E k λ 0 0 ( î). 01 (prt 1 of 1) 0 points The digrms below depict three electric field ptterns. Some of these ptterns re physiclly impossible. Assume: These electric field ptterns re due to sttic electric chrges outside the regions shown. () (b) (c) Which electrosttic field ptterns re physiclly possible? 1. (b) only correct. (b) nd (c). () nd (c) 4. () only 5. () nd (b) 6. (c) only Electrosttic lines of force do not intersect one nother. Neither do they form closed circuit (unless there is chnging mgnetic field present). 014 (prt 1 of 1) 0 points A prticle of mss g nd chrge 6 mc moves in region of spce where the electric field is uniform nd is E 7. N/C, E y E z 0. If the initil velocity of the prticle is given by v y m/s, v v z 0, wht is the speed of the prticle t 0.7 s? Correct nswer: m/s.

8 Answer, Key Homework 8 Dvid McIntyre 8 Given : m g kg, E 7. N/C, E y E z 0, v y m/s, v v z 0, t 0.7 s. According to Newton s second lw nd the definition of n electric field, F m q E. Since the electric field hs only n component, the prticle ccelertes only in the direction q E m. To determine the component of the finl velocity, v f, use the kinemtic reltion v f v i (t f t i ) t f. Since t i 0 nd v i 0, v f q E t f m (0.06 C) (7. N/C)(0.7 s) v f ( kg) m/s. No eternl force cts on the prticle in the y direction so v yi v yf m/s. Hence the finl speed is given by v f vyf v f [ ( m/s ) ( m/s ) ] 1/ m/s. Note: This is nlogous to prticle in grvittionl field with the coordintes rotted clockwise by π (90 ). 015 (prt 1 of ) 0 points Given: q e C. An electron enters the region of uniform electric field of 18 N/C, s in the figure m ĵ 10 6 m/s Find the mgnitude of the ccelertion of the electron while in the electric field. Correct nswer: m/s. Given : q e C, m e kg, E 18 N/C. F m q E, q e E ĵ m e so ( C)(18 N/C) kg ( m/s ) ĵ, nd nd the mgnitude of the ccelertion of the electron is m/s. 016 (prt of ) 0 points Find the time it tkes the electron to trvel through the region of the electric field, ssuming it doesn t hit the side wlls. Correct nswer: 10 8 s. Given : l 0.09 m, v m/s. The horizontl distnce trveled is l v 0 t t l v m 10 6 m/s 10 8 s. ĵ î

9 Answer, Key Homework 8 Dvid McIntyre (prt of ) 0 points Wht is the mgnitude of the verticl displcement y of the electron while it is in the electric field? Correct nswer: m. Using the eqution for the displcement in the verticl direction nd the results from the first two prts of the problem, we find tht y 1 t ( m/s ) ( 10 8 s) m. which hs mgnitude of m.

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