PHYS227N S13 COOK Test 2

PHYS227N S13 COOK Test 2 - OLD DOMINION UNIVERSITY PHYSICS...

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Unformatted text preview: OLD DOMINION UNIVERSITY PHYSICS 227N/232N, Universifl Physics Spring 2013 Second Examination 10:00 am — 10:50 am Friday 15 February 2013 Room 142-144, OCNPS PROFESSOR: Dr. D. C. Cook Answer 2 question: All answers should be on the examination paper. If more space is required, attach sheets to paper. Text book may not be used. This examination will be 5% of the total co rse sses ment. TABLE # V Um! Honor Pledge Signatur Candidate's Name CONSTANTS AND E UATIONS so = 8.8510'12 c2 N" m-2 lnC =10'9 c 111C =10'6 C k = l/41r80 = 9.0 109 N m2 C'2 F. = (k qlqz/rz) 1* F. = E q E = (k q1 /r2) r UE = k qlqz /r UE = Vq E = V/d V = k q2 /r E = (k q1 /r2) 1‘ Area of Circle A= m2 @E = I E.dS = q/so Surface area of a sphere AS = 4an Volume of a sphere V= 4/3 (at3 ) lei 1. Electric Flux (Total 50 Marks) la. A flat sheet is in the shape of a rectangle with sides of length 0.400 m and 0.600 m. The sheet is immersed in a uniform electric field of magnitude 75.0 N/C that is directed at 200 from the plane of the sheet as shown in Figure 1. Find the magnitude of the electric flux passing through the sheet. (10 Marks) 9’ awry“ o-‘OOOW‘ “ 01“" IE: ’46 .QN/C go 3% 1,107 'L / —.— 2‘5’. Laglec, d; aiw‘SUS-WIWG = (a , \Lo Elle:- - c... ........ m. ANSWER: Lo. Ko W“ /c [ lb. A wedge shaped object is completely enclosed inside a uniform electric field as shown in Figure 2. The electric field has magnitude of E i 2.50 NC and travels at right angles through the left side square side of the wedge having area A1 = 9.00 m2. The electric field travels parallel to the rectangular base of the wedge having area A3 = 12.00 m2 and also parallel to the two vertical triangular sides of the wedge. The electric field exits the sloped rectangular face of the wedge of area A2. V 2.8” .. 7; SnB g 0058 "' 3;: (i) Calculate the electric flux through the square side of the wedge of area A1. V, Answer: Electric Flux = “’2’ Z'KNN’L /C I l (l: 95‘ (‘1.oow""-(Z.€o We} = 223 “if w W C (ii)Calculate the electric flux through the rectangular base of the wedge of area A3' I L/ N :1 l U K I " - Answer: Electric Flux= m/C. 0. (iii) Calculate the electric flux through the sloped rectangular face of the _,._« wedge of area A2. 5 _ M ’7. '- Answer: Electric Flux F Ll (o . M /C ‘2‘ / J 2 1" o (40 Marks) h C036 = 6 a — J h 7 7 r C. )3 ‘6 d5 1’ n g 7- / "~\ 0 @- “g: 975% = s 03% U9“ U6 End of Problem 1 (0' #4 2. Gauss’ Law (Total 50 Marks) A small solid conducting sphere with radius a is located inside and concentric with a larger conducting spherical shell with inner radius b and outer radius c as shown in Figure 3. The inner solid W’s‘phere has charge of Q1 = -3Q, and the outer conducting spherical shell has a c arge of Q2 = +2Q. Solid Sphere Spherical I 5" a Figure 3. Complete the four parts (i): (iv) below. (i) Explairfivhere and what the/charge is located on the solid sphere. ANSWER \J'” QcQCJ become i-‘r \S 0,. conducive . I W no N 2—) \LL’ Qt L U (4 Marks) (ii) Explain where and what the charge is located on the spherical shell. ANSWER QC beam/JBC O gongém’m" (iii) Use Gauss's Law to find the electric field at any distance r, between 1' = 0 and r = infinity, from the common center of the two spheres. The calculations should be for the following regions, 0<r<a, a<r<b, b<r<c, c<r<oo. After com letin our workin on Pa e 6, place your answers in the appropriate spaces below. Explain answers with working or wrifien explanation. - I" u 'i E: U ANSWER(O<r<a) C/ {V W W/flhu‘cf‘ihz MgrlZs)Vz ANSWER (a<r<b) *'/%% ' w“ 0 0 (8Marks) ANSWER (b <r<c) O 0/ Lt (8Marks) ANSWER (c < r < oo) ( C3} (2‘ (8 Marks) (iv) Sketch the graph of Electric field on the axes below for 0<r<oo. (6 marks) EU) Space for Working Problem 2. End of Problem 2 END EXAM 2 ...
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