test2_solution

test2_solution - Problem 1(25 pts A thin plastic rod of...

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Unformatted text preview: Problem 1 (25 pts) A thin plastic rod, of length 4m, lies along the x—axis (only a portion of the rod is shown in the diagram). It has a charge of —300 nC uniformly distributed over its surface. A thin glass ring, or radius 6 cm, is centered on the y-axis, 10 cm above the rod. It has a uniform charge of +2 nC. y K a . 0‘ w ring radius 6 cm charge +2nC ———— ——+ / 0/9/68 7: 29' “a C' )/ 1 Chrged rod 4 In Charge _3OOHC x Z L (a: 23 pts) Calculate the force on a small ball with charge —5 nC placed at < 0, 0.05, 0> m. Your answer must be a vector. Clearly show all steps in your work. __ //_’_ : .2 ‘* ('05-) V /Em/6/—- Wee maﬁa,” % (ﬁe? ( a \$2)?“ 3 /897;; . (.o¢1+ 05' EM: <0) ~127/7)o> % a 2 cal/L) _ _ M /Ele°o/: 5/7760 4r 2 46; (2))i:0)oe 61): “27°00m 05 z? — .. V ' EMD— {0, 27000 ) 0774 15/57: €400+E3N6 ~ <Ol#‘2839?)0>% .A A V F: ivrf: (“52—7)C<O)”2839?,0>74 E :<O)/.s’5‘e-_‘/,o>'\/ (b: 2 pts ) Draw the force vector you calculated on the diagram. Problem 2 (20 pts): A particular diatomic molecule consists of a positive ion with charge +e and a negative ion with charge —e, separated by a dis- tance of 1.5X10— m, as shown in the diagram. Locations 1, 2, :1 > 3, and 4 are shown in the diagram. 3 :1 be. so (a: 15 pts) Location 1 is 2X10_8 m from the center of the mole— T A I cule, and location 2 is 3x10‘8m from the center of the mole- g E‘ cule.Calculate the potential difference V2 — V1 , both 1‘ magnitude and sign. Show all steps in your work. ‘ /E/: 9192.0 2%; - ........ .. 3 Vﬁ—V‘: —f2E.,a-‘: "fl/E’Jr‘ a: -jrz / {fgjafr I” WW; *3 mo", V A A {516M 0K) 5 5.04? me As AZ) (b: 5 pts) Location 3 is ZXIO—8 m from the center of the molecule, and location 4 is 3><10—8 m from the center of the mole- Icule.Calculate the potential difference V4 — V3 , both magnitude and sign. Show all steps in your work. 0N J. AXIS) E m “X p/K 5 A 4Q m +7 0172 §°A?=O Problem 3 (15 pts) An isolated large-plate capacitor (not connected to anything) with an air gap of 3.5 mm originally has a potential difference of VA — V5 = 900 V. Then a plastic slab 3 mm thick is inserted into the middle of the air gap as shown in the diagram. Now the potential difference VA — VB = 310 V. Locations A and B are on the inside faces of the metal plates, directly across from one another, as shown. (a: 12 pts) Calculate the dielectric constant of the plastic slab. Show all work! ‘- Iél Ofelémnccy) VA~VB: c(0C) \/ _ E T" 2 Too V - \$765- .! : é— ifi 3.§e —3 H” '2‘ “4 83\$? Aswan/6 "Fina/6e“ 1:161!) 0F FLasl-uc gum IS SMALL) WITH 5cm}: _) M1 Am) Ex: ~2.\$7€5'ri,‘/ _ :: "Z‘ " E'AL —- l 7 "I VA VD 3/0 V 2 MI folAerr) gum.” x: 35 6591—. ) 310 2 ’ («Ex(.2§e-3)~_€)‘((10e—3}— é} (Jae—3)) K ‘ K 3/0 : a? EX(.2§e-3l 4' fig(3.0e"3) K __ (2,57e5)(3.Oe-3) 3/0»Q(2.s’7e5—)(~25€_3> ‘ ﬂ 6?,5'7e f)(3.oe'3l k g /m~ /<_ 3/0v02(2.s*7es)[--25_e”3) /<: 9425' (b: 3 pts) Location C is on the outside face of the right metal plate, 4 cm below location B. What is the potential difference VA ~ VC? Explain brieﬂy. EMT m METAL PLATE (5 0) so VB’Vc10) 5° VA—VC : VA‘VB +VB—VC = VAFVBZZ“9 BIO? ...
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This note was uploaded on 02/08/2009 for the course PHYSICS 2212M taught by Professor Weisenstein during the Spring '08 term at Georgia Tech.

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test2_solution - Problem 1(25 pts A thin plastic rod of...

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