X1.5a 1150PHYS-1150(5) (Kaldon-40527) Name __________S O L U T I O N_________WMU - Fall 2006Exam 1A - 100,000 points Book Title ________________________________________________Rev. 09/27/06 Tu.6State 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-1150 (KALDON-5)FALL 2006WMUkNmCCNm=´×=´×-8 988 10885 1092201222..ee = 1.602 × 10 -19 CAn electron walks into a bar and asks for a drink. He getsit and is charged $5. A proton comes into the bar and hepays $5 for a drink. A neutron walks in, asks for a drink,but when the bartender gives it to him, never asks for anymoney. When the neutron walks out, the electron and theproton are upset. “How come he didn’t have to pay?”“Yeah!” “That guy? He’s a neutron – no charge.”I love this joke!Physics 1150 / Exam 1 Fall 2006 Page 2If Three Charges Are Good, Then Four Are Better (50,000 points)1.) Four identical 15.0 nC (= +15.0 × 10 -9 C) charges, q1 , q2 , q3 and q4 ,are equally spaced apart by a distance of 5.00 cm each. Find the totalelectric force, rFE, acting on q2. If you are clever, you don’t have to work sohard.By Symmetry, rrrrrr rrFF r cm mFFFF FFkq qrkqrCmNNFNiNENmC12 3212 32 42 424224222992244244010001008 988 10 15 0 1001000 0002022 2 022 102 022 10 2 022 10 180 022+= = ==++====´´==´=- ´ = ´ °×----;. [email protected] 1150 / Exam 1 Fall 2006 Page 3(b) If instead of +15.0 nC, what charge would q4 have to have if the total electric field, rE, acting on q1 is zero?EkqrCmNCEkqrCmNCEEE NC NCNCEkqrqErkNC mNmCNmCNmC222992332992423442442298 988 10 15 0 100050053 9308 988 10 15 0 10010013 48053 930 13 48067 41067 410 01508988 101222222==´´===´´==- + =- +=-===-´=-×-×-×...,/...,/,/ ,/,/,/...chchbgchchbgbgb gbgbgch688 10 16887´=--CnC.You can put the negative sign in by hand – since q2 and q3 areboth positive charges, q4 has to be negative for Etotal = 0.Physics 1150 / Exam 1 Fall 2006 Page 4A single charge q5 = -72.5 × 10 -9 C and mass m = 0.0135 kg, sits in a constantelectric field rENCi=+1190 /$. Find (c) the electric force vector, rFE, and (d)the acceleration in the x-direction, ax .(c) rrFqEFqE C NCNNEE===- ´=- =- ´--72 5 10 11900 00008628 8 628 1095./..chbg(d) FmaaFmNkgmsxxxx===-´=--8 628 10001350 00639152.../(e) Find the number of electrons added or subtracted which would be needed to neutralize q5.qNeNqeCC=±==-´´=---72 5 101602 10452 600 000 000919..,,,Since the charge is negative, we have to SUBTRACT or REMOVE452,600,000,000 electrons to neutralize q3Physics 1150 / Exam 1 Fall 2006 Page 5A Capital Capacitance Experience (50,000 points) If you don’t get an answer to (a), use C = 621 pF.2.) A parallel plate capacitor consists of two metal plates, each 0.0700 m wide by 0.124 m long, with a gapd = 1.15 mm. (a) Calculate the capacitance, C, of this capacitor. e0 = 8.85×10-12 F/mAmm mCAdFmmmFpF====´=´ =--0 0700 0124 0 008680885 100 0086800 001156 680 10 66 802012211.. ../....bgbgchchbge (b) Find the charge ±Q on the plates with DV = 555 volts.CQVQ CV F voltsC=== ´=´--6 680 10 5553707 10118..chbg(c) Find the strength of the E-field between the plates with DV = 555 volts.VEdEVdvoltsmVm=== =5550 00115482 600.,/ OR sse==´==´==´´=----QACmCm CmECmFmVm3707 100 0086800 000004271 4 271 104271 10885 10482 5008226206212..././././,/(d) Sketch the E-field for this capacitor, both in between the plates and include a couple of field lines “at theedges”.Physics 1150 / Exam 1 Fall 2006 Page 6 (e) Four of our capacitors are connected together as shown in two configurations, A and B. Find the equivalentcapacitance, Ceq , of each arrangement. Are they the same? If not, which is bigger, A or B ?A BA CCCCCCCCCCC pFeqeqeq=+==+====21121222166 80. B 111222266 80CCCCCCCCCCpFeqeqeq=+===+==.A and B are the SAME(but only because all the capacitors are the