Exam 1 September 21, 2012 Instructor: Timothy Martin

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1 PHY 232 Exm 1 Sept 21, 212 Exm 1 September 21, 212 Instructor: Timothy Mrtin Stuent Informtion Nme n section: UK Stuent ID: Set #: Instructions Answer the questions in the spce provie. On the long form questions, show your work! Answers with work shown will not be gre. Points will be given bse on correctness n coherency if it is ifficult to follow your work, you will likely receive less prtil creit. If you run out of room for n nswer, continue on the bck of the pge. You re not llowe use of clcultor or ny other electronic evice uring this test. Question Points Score Totl: 1

2 PHY 232 Exm 1, Pge 2 of 7 Sept 21, Multiple choice Here re two fixe point chrges q 1 =2e n q 2 = e; there is istnce 2 between them. I q 1 II q 2 III () (3 points) The irection of the electric force between the two chrges is A. To the left B. To the right C. Attrctive D. Repulsive Solution: C, becuse opposite chrges ttrct. (b) (3 points) In which region oes the electric fiel equl? A. Region I B. Region II C. Region III D. Nowhere Solution: C; the two chrges contribute oppositely, n we re closer to the smller one. (c) (3 points) If chrge q 3 = e were plce hlfwy between them, wht woul the force on it be? e A. k 2 e 2 to the right 3e B. k 2 e 2 to the right e C. k 2 e 2 to the left 3e D. k 2 e 2 to the left Solution: B. Both exert force to the right, n F 1 + F 2 = k e ( 2e2 2 + e2 2 ) () (3 points) If sphericl shell of imeter 4 ws centere hlfwy between the two chrges, wht woul the net flux through it be? A. B. C. e 2e 3e D. Solution: A; Use Guss s Lw: the net chrge within the sphere is +e. Pge 2

3 PHY 232 Exm 1, Pge 3 of 7 Sept 21, 212 (e) (3 points) Which of the following fiel igrms is impossible, whether chrge is present or not? A) C) B) D) Solution: B; These lines coul not begin/en on chrges. Aitionlly, n integrl roun loop must give E s =, which wouln t work here. (f) (3 points) For conuctor in electrosttic equilibrium, which of the following is not necessrily true A. Insie the conuctor, V =. B. The chrge ensity ρ insie the conuctor is. C. Just outsie the conuctor, the electric fiel is proportionl to the chrge ensity on the surfce. D. The chrge ensity on the surfce epens on its curvture Solution: A; the voltge V is constnt, but generlly not. (g) (3 points) If positively chrge prticle moves prllel to the electric fiel (i.e. long electric fiel lines), its potentil energy A. Increses B. Decreses C. Stys the sme D. Who knows? Not enough informtion is given. Solution: B; E s woul be negtive. The electric force is oing positive work. (h) (3 points) A cyliner of rius n length L hs chrge ensity ρ r. The totl chrge on the cyliner is A. π 2 Lρ B. 2π 2 Lρ C. 4π 3 Lρ D. 2πLρ Solution: B; ρv = ρ rθz Pge 3

4 PHY 232 Exm 1, Pge 4 of 7 Sept 21, (3 points) A line of chrge of length L extens long the x xis, with the left en istnce from the origin. It hs chrge ensity λ = αx. x = λ = αx Fin Solution: All of these require integrting over the chrge istribution. Use x s the prmeter. Then little bit of chrge q = λx = αx x. An integrl woul go from x = to x = + L. () (1 points) The electric fiel t the origin. Solution: We use E = q k e r. The istnce from the origin to point x will just be x. Using q 2 from bove, +L αx E = k e x 2 x = k eα ln( + L) ln() = k e α ln( + L ) (b) (1 points) The potentil t the origin. Solution: Sme pproch s bove, but V = k e q r. V = +L k e αx x x = k eαl (c) (1 points) The totl chrge of the istribution. Solution: Or equivlently Q = q = +L αxx = 1 2 α ( + L) 2 2 Q = αl 2 (L +2) =1 2 αl2 + αl Pge 4

5 PHY 232 Exm 1, Pge 5 of 7 Sept 21, (21 points) We hve 3 ienticl point chrges Q rrnge in n equilterl tringle, with sie length. Fin: () (1 point) The electric fiel t the center of the tringle. Solution: This will be, since the forces from ll three cncel. (b) (1 points) The electric fiel istnce z irectly bove the center of the tringle Solution: For convenience, efine b s the istnce from ech prticle to the center. Then this works just like the exmple problem from worksheet 2; sum the contribution from ech prticle, n by symmetry only the z component will remin. The nswer is 3Qz E = k 2 ˆk (b 2 + z 2 ) 3/2 For more etile pproch see the solution to ws2. (c) (1 points) The totl potentil energy of the system of three prticles. Q Solution: Between ny pir U = k 2 e. So between the three pirs, Q 2 U tot =3k e Pge 5

6 PHY 232 Exm 1, Pge 6 of 7 Sept 21, (25 points) Consier set of three slbs, ech infinite in extent long the y n z xis, n ech with thickness t long the x xis. The mile slb hs constnt chrge ensity of ρ. To its left n right re metllic slbs which o not crry net chrge. Use Guss s lw to fin Conuctor Chrge slb ŷ Conuctor () (5 points) The electric fiel in region I I II III IV (b) (5 points) The electric fiel in region II (c) (1 points) The chrge ensity on the surfce of the conuctor, between regions II n III. () (5 points) The electric fiel in region IV ρ = ρ ˆx t Solution: In ech cse, imgine gussin cyliner or prism with cross-sectionl re A, the ens sitting t ±x s pproprite for the region consiere. The flux through the encps will be E A = 2EA, n Guss s Lw then gives us 2EA = Q enc () The E fiel will vry s function of x. If we plce the ens of the prism t ±x, then the mount of chrge enclose is Q = ρ V = ρ A(2x). So E(x) = ρ x î (b) In region II, Q = ρ V = ρ At. So E(x) = ρ t 2 î (c) We re looking for σ t the conuctor. Imgine gussin prism with encps insie the conuctor. In region III, we know tht E = Q enc =. The chrge enclose will be the sme s in (b), plus the contribution from the surfce chrge on both conuctors. Therefore Q enc = ρ At +2σA = σ = ρ t 2 () Since the conuctor is overll neutrl, the nswer here is the sme s in (b). Pge 6

7 PHY 232 Exm 1, Pge 7 of 7 Sept 21, 212 Formul sheet Constnts Ner point chrge, Between two chrge prticles: k e = 1 4π E = k e q r 2 ˆr V = k e q r q 1 q 2 F = k e r 2 ˆr U = k e q 1 q 2 r When chrge q is in n electric fiel, the electric force on it is F = q E If chrge q moves through potentil, the chnge in potentil energy is U = q V Reltionships between electric fiel n potentil V = E s Guss s Lw Line chrge: q = λl Surfce chrge: q = σa Volume chrge: q = ρv For cyliner, V = z r r θ E x = V x ; E y = V y ; E z = V z E A = Q enc Pge 7

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