X1.4a 115PHYS-115(4) (Kaldon-26364) Name __________S O L U T I O N_________WMU - Spring 2005Exam 1A - 100,000 points Book Title ________________________________________________Rev. 02/02/05 We.A5State Any Assumptions You Need To Make – Show All Work – Circle Any Final AnswersUse Your Time Wisely – Work on What You Can – Be Sure to Write Down EquationsShort Answers Should Be Short! – Feel Free to Ask Any QuestionsEXAM 1 [FORM - A]PHYS-115 (KALDON-4)SPRING 2005WMUPresidential Inaugurations, State of the Union Speech, BigWinter Storms, and Super Bowl Sunday…First 115 Physics exam? (with cookies).Priceless.And It’s Only Just February!Physics 115 / Exam 1 Spring 2004 Page 2This Story Has To Have a Point (50,000 points)1.) Three identical 15.0 nC (= 15.0 × 10-9 C) charges, q1 , q2 and q3 , are equallyspaced apart by a distance of 5.00 cm each. Find the total electric force, rFE,acting on q2 in the center.By Symmetry, rrrFFFE=+=12 320(b) Find the total electric field, rE, acting on q2 in the center from the other two charges.By Symmetry, rr rEE E=+=130(c) Find the total electric potential V acting on q2 in the center from the other two charges.dcm mVkqrVVVkqdkqdkqdCmvoltsNmC====+= + ==´´=×-5 00 0 0500228988 10 150 10005005393139922.....chchbg(d) Find the number of electrons added or subtracted to make q2.QNeqNeNCCelectrons subtracted=± =+=´´=--;..,,,15 0 101602 1093 630 000 000919Physics 115 / Exam 1 Spring 2004 Page 3(e) An electron is located a long way (1.00 meter to the right) from q2 in the center from the other two charges.Find the initial magnitude and direction of the electron’s acceleration. me = 9.11 × 10 -31 kgFkq qrCCmNFmaaFmNkgmsENmCEE==´´´=´===´´=´×-----1229919217173113 28 988 10 3 15 0 10 1602 101006 480 106480 10911 107113 1022......../chbgchc hbgNonpareil Parallels (50,000 points)2.) A parallel plate capacitor consists of two metal plates, each 5.00 cm wide by 10.0 cm long, with a gap d =2.00 mm. The space between the plates is filled with silicone oil (dielectric constant k = 2.50 and dielectricstrength 15.0 × 106 V/m). (a) Calculate the capacitance, C, of this capacitor. e0 = 8.85×10-12 F/mAmm mCAdFmmmFpF==== ´=´ =--0 0500 0100 0 00500250 885 100 005000 002005531 10 55 312012211.. ... /....bgbgchbgchchbgke(b) Show that you can safely charge this capacitor with DV = 25,000 volts.DDVEdEVdvoltsmVm Vm E====<=25 0000 0020012 500 000 15 000 000,.,, / ,, /max– OR –DVEdVEd Vm mvoltsSo volts V====<max maxmax,, / .,... ,15 000 000 0 0020030 00025 000bgbg(c) Find the charge ±Q on the plates with DV = 25,000 volts. If you didn’t get an answer to (a), use C = 621 pF.Physics 115 / Exam 1 Spring 2004 Page 4CQVQCV F volts C pC=== ´ = ´ =--5531 10 25 000 1383 10 1 383 00011 6.,.,,chbg(d) Use Gauss’ Law to find the magnitude of the electric field between the plates. Because of the dielectricsilicone oil between the plates, you’ll have to multiply e0 by k in the Gauss’ Law equation.Note: You do NOT get a factor of 2, because E-field is onlybetween the plates, not on the outside.skeskeske==´==´== ===´=---×QACmCm CmEAqAECmNCEinsideCNm1383 100 0050000 0002766 2 766 100 00027662 50 8 85 1012 500 0006224200021222.../. /./..,, /bgbgchFThis is the SAME answer as in part (b).(e) Four of our capacitors are connected together as shown. Find the equivalentcapacitance, Ceq , of this arrangement.If you didn’t get an answer to (a), use C = 621 pF.SeriesCCCCCCParallel CCCCSo for thecircuit C C pFeqeqeqeq111222255 31=+=