MT1--2008--solutions - 02/04/2009 16:18 IFAX a Mike...

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Unformatted text preview: 02/04/2009 16:18 IFAX a Mike HaSinoff 001/006 First Letter of Last Name: l:| The University of British Columbia Midterm 1 term 2- February 13, 2008 Physics 153 ' ' 7 Elements of Physics TIME: 60 minutes CANDIDATE’S NAME: 6 U l J Last Name First Name or Initials SIGNATURE: STUDENT NUMBER: ' THIS EXAMINATION CONSISTS OF 4 PAGES. CHECK TO ENSURE THAT THIS PAPER IS COMPLETE. ' INSTRUCTOR’S NAME: SECTION NUMBER: (a) Each candidate should'be prepared to produce, upon request, his/ her library/AMS card. (b) Read and observe the following additional rules: No candidate shall be permitted to enter the examination room after the expiration of one—half hour, or leave during the first half—hour of the examination. Candidates are not permitted to ask questions of the invigilators, except in cases of supposed errors or ambiguities in examination questions. CAUTION: Candidates guilty of any of the following, or similar, dishonest practices shall be immediately dismissed from the examination and shall be liable to disciplinary action. (a) Making use of any books, memoranda, calculators, cell phones, audio or Visual players or other memory aid devices and electonics, other than authorized by the examiners. (b) Speaking or communicating with other candidates. Each Question is 10 points: 10 10 10 Total Mark out of 30 points. Total: 30 02/04/2009 16:18 IFAX a Mike Hasinoff 002/006 02/04/2009 16:19 IFAX » Mike Hasinoff 003/006 CW} seamsm Q 2. An non—conducting spherical ball of radius a has a total charge Q which is distributed uniformly throughout its volume. This ball is situated at the centre of a hollow metal shell with inner radius b and outer radius c as shown. The metal shell has a net charge 2 +4Q. Name: vacuum Using Gauss’ law (a) (3 points) What is the E field in region-1 where r < a? (b) (1 point) What is the E field in region—2 where a < r < b? c) (1 point) What is the Ill field in region—3 where b < r < c? d) (2 points) What is the electric potential at the point r = b? Take V = 0 at r 2 oo. 6) ( (3 points) How much work do you have to do in order to move a very small test charb +q from b to a? n3 3 9‘1912 W (a) fish: ElwfikfigLL 4%6: 96%;: - 96).— (W ) 3 image.» 60 W Elm) : Aft- Mm EM: WW): 65.9 => 3F 56? L (Jaw) \MC) 12 AWLer (+512 Ohm/Cw. «2 \l (c): lg ( :35?) 5Q Ema Ea? \lblfvm: 7 War“ M 02/04/2009 16: l 9 I F AX 02/04/2009 16:19 IFAX —> Mike Hasinoff 005/006 3. Capacitor problem Cylindrical capacitor C1 with value 20 pF is constructed frOm a wire with diameter 3.68 mm centered in an air-filled metal tube with inner diameter 10 mm. Capacitor C2 with value 5 nF is constructed from a 50 mm x 50 mm square of plastic film that is 10 microns thick, coated with metal foil on both sides. Capacitor C3” with value 30 pF is constructed of parallel 65.7 mm diameter metal disks separated by air. The inner wire of Cl is connected to one side of C2, and the other side is connected to a switch. One end of C3 is connected to the other side of the switch, and the other end of C3 is connected to the outer tube of C1. Initially the switch is open, C3 has charge 1 11C, and C1 and C2 have zero charge. ' ‘ ' 01:20pF 02=5nF _ ‘ C3=30 A. Find the dielectric constant K for the plastic in C2 B. Find the electric field in C3 before the switch is closed C. Find the voltage for C1, C2, and C3 a long time after the switch is closed. D. Find the length of C]. Cd A. For parallel plates, C = K80 £=> K = —. d 80A The capacitanceis C = 5x10'9. The area is A = (0.050)2. The gap is d =10x10‘6. (5x10'9)(10x10'6) s 19*— O (8.85x10-12)(0.050)2 = 2.26. B. We know the charge and capacitance, so we can find V = Q [C . If we knew the gap, we could get the electric field from E = V/d. We can get the gap from the capacitance and the plate area: C = so A/d => 0' = so A/C . Combining the above, = K = 9/0 = 2. d 50 A/C 50A ' Bquivalently, the surface charge density is a = Q [A , and by Gauss’ Law E = 0/80 for parallel plates (the two plates cancel the factor of 2 in the formula for a single plate). 0.0657 ‘ 2 Numerically, the charge is Q = 1x109, and the area is A = m2 = Jr( J = 3.39x10'3,so —9 5:3: “‘10 = 3.33x1041 80A 8.85x10‘” 3.39x10'3 m 02/04/2009 16:20 IFAX —> Mike Hasinoff 006/006 C. Cl and C2 are in series with each other. I l 1 The e uivalent ca acitance is iven b —= —+ —, which oives Cl P g 3’ C12 Cl C2 e Cpl (20x10‘12)(5x10‘9) C1+C2 _ 20x10-12+5x10-9 This is only slightly different from C1 alone. Cl2 = =1992x10‘”. After the switch is closed, C12 is in parallel with C3, giving CM = C12 +C3 = (1992+ 30) x10‘12 = 49.92x10‘12 The initial charge Q = l ><10'9 on the total capacitance gives Vml = Q [CW = 20.03 V. The final voltage on C3 will be exactly this voltage: V3 = 20.03 V. The sum of V1 and V2 will also be this voltage. One way to get the individual voltages is to use the fact that for capacitors in series, the charges are equal. But we still need to find the charge. We can get it from the difference between the initial charge and final charge on C3. The final charge is Q, = C3V3 = (30x10"2)(20.03) = 0.600940”, so the charge on the other capacitors is Q12 = (1.0— 0.6009) x109 = 0.3991x10‘9. Then V, = le/CJ = 19.96 v and V2 = Qn/Cz = 0.079 v. Another way to get the voltages is by the capacitive—divider formula: 1 V1: LV = 0.9960 V llcwl for {of ‘/2 = LEI)! = V'Ui' l/CI +1] 2 1 D. For cylindrical geometry, Gauss” Law gives (DE = 27mLE(r) = Q—= E = Q —. 80 2315801, 1‘ We integrate this from the inner radius to the outer radius to get the potential difference r, d V: Q L: Q ln(r2/r1). ZJFEOL '1 r 215501, 2 L The capacitance is then C = 9= —Q—= mo so V Q “(B/n) ln( 7/2/11) 23reoL ln r r 20 l "‘2 In l0 3.68 L: E (2/1) = X‘Ou#= 0.3596 m. 80 25c 8.85x10" 2.7: ...
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This note was uploaded on 02/03/2010 for the course PHYS PHYS 153 taught by Professor Billmcutcheon during the Spring '09 term at The University of British Columbia.

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MT1--2008--solutions - 02/04/2009 16:18 IFAX a Mike...

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