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TIME OF COMPLETION_______________ NAME_____SOLUTION________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1112, Exam 1 Version 1 Total Weight: 100 points Section 1 October 4, 2004 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work five (5) multiple choice problems and all calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 6:00 p.m. 7:15 p.m PROBLEM POINTS 1-6 30 7 20 8 20 9 15 10 15 TOTAL 100 PERCENTAGE CREDIT CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. Which one of the equations is the correct Kirchhoff’s junction equation for the situation pictured below? a. I1 = I2 + I3. b. I2 = I1 + I3. (6) c. I3 = I1 + I2. d. I1 + I2 + I3 = 0. 2. The unit of electric current, the ampere, is dimensionally equivalent to which of the following? a. Volt x Ohm. b. Volt/ Ohm. (6) c. Ohm x meter. d. Volt/second. 3. When an electric current exists within a conducting wire, which of the following statements is correct? a. Electric field inside the wire is zero. (6) b. Electric field inside the wire is parallel to the current flow. c. Electric field inside the wire is anti-parallel to the current flow. d. Electric field inside the wire is perpendicular to the current flow. 4. If two parallel, conducting plates have equal positive charge, the electric field lines will a. Leave one plate and go straight to the other plate. b. Leave both plates and go to infinity. (6) c. Enter both plates from infinity. d. None of the above. 5. If three 4.00 mF capacitors are connected in parallel, what is the combined capacitance? a. 12.0 mF. b. 0.750 mF. (6) c. 8.00 mF. d. 0.46 mF. 6. A repelling force must occur between two charged objects under which condition? a. Charges are of unlike signs. b. Charges are of like signs. (6) c. Charges are of equal magnitude. d. Charges are of unequal magnitude. 7. Two charges, Q1 = 6.00 C and Q2 = - 6.00 C, are located as shown below. a. Find the total electric field at the origin. Specify both magnitude and direction. E1= ke Q1/r12= (8.99 x 109 Nm2/C2)(6.00 x 10–6C)/(0.150 m)2 = 2.40 x 106 N/C E2 = ke Q2/r22= (8.99 x 109 Nm2/C2)( 6.00 x 10–6C)/(0.235 m)2 = 0.978 x 106 N/C E1x = (2.40 x 106 N/C) cos(90o) = 0 E1y= (2.40 x 106 N/C) sin(90o) = 2.40 x 106 N/C E2x = (0.978 x 106 N/C) cos (0o) = 0.978 x 106 N/C E2y = (0.978 x 106 N/C) sin (0o) = 0 Etotx = 0 N/C + 0.978 x 106 N/C = 0.978 x 106 N/C Etoty = 2.40 x 106 N/C + 0 N/C = 2.40 x 106 N/C Etot =((0.978 x 106 N/C)2 + (2.40 x 106 N/C)2) 1/2 = 2.59 x 106 N/C tan-1Etoty /Etoty) = 67.8o b. What is the magnitude of the total electric force exerted by charges Q1 and Q2 on 2.00 C charge placed at the origin? What is the direction of the force? F = |q| E = (2.00 x 10-6 C)( 2.59 x 106 N/C) = 5.18 N, directed the same as E. 8. You are given a network of capacitors pictured below. C1 = 3.00 F, C2 = 5.00 F, and C3 = 6.00 F. The applied potential is Vab = 24.0 V. a. Find the equivalent capacitance between points a and b. C12 = C1 + C2 = 3.00 F + 5.00 F = 8.00 F 1/C123 = 1/C3 + 1/C12 = 1/(6.00 F) + 1/(8.00 F) = 7/(24.0 F) C123 = 3.43 F b. Calculate charge on the capacitor C3. Q3 = C123 V= (3.43 F) (24.0 V) = 82.3 C c. How much energy is stored in the capacitor C3? EST = Q32/(2C3) EST = (82.3 C)2/(2 (6.00 F)) = 564 J d. Find the potential difference between points a and d. V3 = Q3 /C3 = (82.3 C)/(6.00 F) = 13.7 V Vad = 24.0 V – 13.7 V = 10.3 V 9. A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00oC, the resistance of the carbon resistor is 217.3 What is the temperature on a spring day when the resistance is 215.8 (Hint: Take the reference temperature T0 to be 4.00oC). The temperature coefficient of resistivity of carbon is = -0.500 x 10-3(oC)-1. R = R0 [1 + (T – T0)] R / R0 = 1 + (T – T0) R / R0 - 1 = (T – T0) [R / R0 – 1]/ (T – T0) [R / R0 – 1]/ T0 = T [(215.8 / (217.3 ) – 1]/ (-0.500 x 10-3(oC)-1 13.8C = T T = 17.8 oC 10. Given the network, write equations that would allow you to solve for the currents in each resistor if the values of the emfs and resistances were known. Label and indicate your choices for current directions. DO NOT SOLVE THE EQUATIONS. I4 = I2 + I3 I6 = I5 + I4 I3 = I1 + I6 2+ I2 R6 – I5 R1 + I4 R3= 0 3 + I3 R5 + I6 R4 + I4 R3 = 0 1 + I1 R2 – I2 R6 + I3 R5 = 0 TIME OF COMPLETION_______________ NAME_____________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1112, Exam 1 Version 1 Total Weight: 100 points Section 1 October 2, 2006 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 50 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work five (5) multiple choice and three (3) calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 10:30 a.m. 11:20 a.m. PROBLEM POINTS 1-6 25 7 25 8 25 9 25 10 25 TOTAL 100 PERCENTAGE CREDIT CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. Two charges are separated by a distance d and exert mutual attractive forces of F on each other. If the charges are separated by a distance of d/3, what are the new mutual forces? a. F/9. b. F/3. (5) c. 3F. d. 9F. 2. A surface on which all points are at the same potential is referred to as a. A constant electric force surface. b. A constant electric field surface. (5) c. An equipotential surface. d. An equivoltage surface. 3. A parallel-plate capacitor has a capacitance of C. If the area of the plates is doubled and the distance between the plates is halved, what is the new capacitance? a. C/4. b. C/2. (5) c. 2C. d. 4C. 4. The resistivity of most common metals a. Remains constant over wide temperature ranges. b. Increases as the temperature increases. (5) c. Decreases as the temperature increases. d. Varies randomly as the temperature increases. 5. When two or more resistors are connected in series to a battery a. The total voltage across the combination is the algebraic sum of the voltages across the individual resistors. b. The same current flows through each resistor. (5) c. The equivalent resistance of the combination is equal to the sum of the resistances of each resistor. d. All of the given answers. 6. Three identical resistors are connected in parallel to a 12-V battery. What is the voltage of any one of the resistors? a. 36 V. b. 12 V. (5) c. 4 V. d. Zero. 7. Two charges, Q1 = 12.00 C and Q2 = - 20.0 C, are located as shown below. a. Find the total electric field at the origin. Specify both magnitude and direction. E1 = k |q1|/r12 = (8.99 x 109 N-m2/C2) (12.0 x 10-6 C) / (0.450 m)2 = 5.33 x 105 N/C @ -90o E2 = k |q2|/r22 = (8.99 x 109 N-m2/C2) (20.0 x 10-6 C) / (0.500 m)2 = 7.19 x 105 N/C @ 0o E1x = E1 cos(- 90o) = 0 E1y = E1 sin(- 90o) = - 5.33 x 105 N/C E2x = E2 cos(0o) = 7.19 x 105 N/C E2y = E2 sin(0o) = 0 Etot x = E1x + E2x = 7.19 x 105 N/C Etot y = E1y + E2y = - 5.33 x 105 N/C Etot = 8.95 x 105 N/C = -36.5o b. What is the magnitude of the total electric force exerted by charges Q1 and Q2 on - 1.00 C charge placed at the origin? What is the direction of the force? F = |q| E = (1.00 x 10-6 C)(8.95 x 105 N/C) = 0.895 N = -36.5o + 180o = 143.5o 8. In the circuit below, suppose C1 C2 C3 16.0 F. a. What is the equivalent capacitance of the circuit? 1/C23 = 1/C2 + 1/C3 = 1/(16.0 F) + 1/(16.0 F) = 2/(16.0 F) C23 = 8.00 F C123 = C1 + C23 = 16.0 F + 8.00 F = 24.0 F b. How much charge is stored on each capacitor when V 45.0 V? Q1 = C1 V= (16.0 F) (45.0 V) = 720 C Q2 = Q3 = C23 V= (8.00 F) (45.0 V) = 360 C c. What is the voltage across each capacitor? V1 = 45.0 V V2 = Q2 / C2 = 22.5 V V3 = Q3 / C3 = 22.5 V 9. A homemade capacitor is assembled by placing two 20-cm pie pans 5 cm apart and connecting them to the opposite terminals of a 9-V battery. Estimate a. The capacitance, C = 0 A/d = 0 ( R2)/d = (8.85 x 10-12 C2/(N-m2))( (0.100 m)2)/(0.0500 m) = 5.56 x 10-12 F = 5.56 pF b. The charge on each plate, Q = C V = (5.56 x 10-12 F)(9.00 V) = 50.0 x 10-12 C = 50.0 pC c. The work done by the battery to charge the plates. Est = ½ C (V)2 = ½ (5.56 x 10-12 F) (9.00 V)2 = 225 pJ 10. A rubber tube 1.00 m long with an inside diameter of 4.00 mm is filled with a salt solution that has a resistivity of 1.00 x 10-3 -m. a. What is the resistance of the filled tube? R = L/A R = L/( R2) = (1.00 x 10-3 -m) (1.00 m)/( (0.002 m)2) = 0.796 b. Find the current that flows through the tube if the potential difference of 24.0 V is applied across its ends. V = I R I = V / R I = (24.0 V)/(0.796 ) = 30.2 A TIME OF COMPLETION_______________ NAME___SOLUTION__________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1112, Exam 1 Version 1 Total Weight: 100 points Section 1 September 28, 2007 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 50 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work five (5) multiple choice problems and three (3) calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 11:30 a.m. 12:20 p.m. PROBLEM POINTS 1-6 25 7 25 8 25 9 25 10 25 TOTAL 100 PERCENTAGE CREDIT CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. Capacitance of a parallel plate capacitor depends on (A) Area of the plates. (B) Separation distance between the plates. (5) (C) Permittivity of free space. (D) All of the above. (E) None of the above choices is correct. 2. A negative charge is moved from point A to point B along an equipotential surface. (A) The negative charge performs work in moving from point A to point B. (B) Work is required to move the negative charge from point A to point B. (5) (C) Work is both required and performed in moving the negative charge from point A to point B. (D) No work is required to move the negative charge from point A to point B. (E) None of the above choices is correct. 3. Two parallel-plate capacitors are identical in every respect except that one has twice the plate area of the other. If the smaller capacitor has capacitance C, the larger one has capacitance (A) C/4. (B) C/2. (5) (C) C. (D) 2C. (E) 4C. 4. When two or more capacitors are connected in series to a battery, (A) The total voltage across the combination is the algebraic sum of the voltages across the individual capacitors. (B) Each capacitor carries the same amount of charge. (5) (C) The equivalent capacitance of the combination is less than the capacitance of any of the capacitors. (D) All of the given answers are correct. (E) None of the above answers is correct. 5. Kirchhoff’s loop rule is a statement of (A) The low of conservation of momentum. (B) The low of conservation of charge. (5) (C) The law of conservation of energy. (D) The low of conservation of angular momentum. (E) Newton’s second law. 6. A negative charge, if free, tries to move (A) From high potential to low potential. (B) From low potential to high potential. (5) (C) Toward Infinity. (D) Away from infinity. (E) In the direction of the electric field. 7. Four identical 10.0 F capacitors are connected as shown below. a. Find equivalent capacitance of the network. 1/C23 = 1/C2 + 1/C3 = 1/(10.0 F) + 1/(10.0 F) = 2/(10.0 F) C23 = 5.00 F C123 = C1 + C23 = 10.0 F + 5.00 F = 15.0 F 1/C1234 = 1/C123 + 1/C4 = 1/(15.0 F) + 1/(10.0 F) = 5/(30.0 F) C1234 = 6.00 F b. If a potential difference of 24.0 V is applied between points a and b, how much charge is stored on Capacitor 4? Q4 = Q1234 = C1234 V= (6.0 F) (24.0 V) = 144 C c. What is the potential difference across the Capacitor 1? V4 = Q4 / C4 = 14.4 V V1 =24.0 V -V4 = 9.60 V 8. Determine the electric field E at the origin 0 due to the three charges pictured below. Specify magnitude and direction. Let Q = 7.50 C and l = 0.350 m. E1 = k |q1|/r12 = (8.99 x 109 N-m2/C2) (7.50 x 10-6 C) / (0.350 m)2 = 5.50 x 105 N/C @ -90o E2 = k |q2|/r22 = (8.99 x 109 N-m2/C2) (7.50 x 10-6 C) / (0.175 m)2 = 22.0 x 105 N/C @ -90o E3 = k |q3|/r32 = (8.99 x 109 N-m2/C2) (7.50 x 10-6 C) / (0.350 m)2 = 5.50 x 105 N/C @ 0o E1x = E1 cos(- 90o) = 0 E1y = E1 sin(- 90o) = - 5.50 x 105 N/C E2x = E2 cos(- 90o) = 0 E2y = E2 sin(- 90o) = - 22.0 x 105 N/C E3x = E3 cos(0o) = 5.50 x 105 N/C E3y = E3 sin(0o) = 0 Etot x = E1x + E2x + E3x = 5.50 x 105 N/C Etot y = E1y + E2y+ E3y = - 27.5 x 105 N/C Etot = 2.81 x 106 N/C = -78.7o 9. Determine the electric potential V at the origin 0 due to the three charges pictured below. Let Q = 7.50 C and l = 0.350 m. 5 V1 = k q1/r1 = (8.99 x 109 N-m2/C2) (7.50 x 10-6 C) / (0.350 m) = 1.93 x 10 V V2 = k q2/r2 = (8.99 x 109 N-m2/C2) (7.50 x 10-6 C) / (0.175 m) = 3.85 x 105 V V3 = k q3/r3 = (8.99 x 109 N-m2/C2) (-7.50 x 10-6 C) / (0.350 m) = - 1.93 x 105 V Vtot = V1 + V2 + V3 = 3.85 x 105 V How much work needs to be done to move another charge of –7.50 C from infinity to the origin 0? W = U = qVtot = ((-7.50 x 10-6 C)( 3.85 x 105 V) = -2.89 J 10. A potential difference of 12.0 V is found to produce a current of 0.400 A in a 3.20-m length of wire with a uniform radius of 0.400 cm. a. What is the resistance of the wire? V = I R R = V / I = (12.0 V)/(0.400 A) = 30.0 b. What is the resistivity of the wire? R = L/A = R A /L = ( R2) = (30.0 ) ( (0.004 m)2)/(3.20 m) = 4.71 x 10-4 -m c. How will the current in the wire change if the length of the wire is doubled while keeping the potential difference the same? As length doubles, resistance doubles, current decreases to a half of the initial value. TIME OF COMPLETION______NAME____SOLUTION_________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1112, Exam 1 Version 1 Total Weight: 100 points Section 1 February 12, 2004 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 1:30 p.m. 2:45 a.m PROBLEM POINTS 1-6 30 7 20 8 20 9 15 10 15 CREDIT TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 5. At a particular point in space, a charge Q experiences no net force. It follows that a. There are no charges nearby. b. If charges are nearby, they have the opposite sign of Q. (5) c. If charges are nearby, the total positive charge must equal the total negative charge. d. None of the above. 6. Two initially uncharged capacitors of capacitance C0 and 2C0, respectively, are connected in series across a battery. Which of the following is true? a. The capacitor 2C0 carries twice the charge of the other capacitor. b. The voltage across each capacitor is the same. (5) c. The energy stored by each capacitor is the same. d. None of the above. 7. Two wires of the same material with the same length have different diameters. Wire A has twice the diameter of wire B. If the resistance of wire B is R, then what is the resistance of wire A? a. R. (5) b. 2 R. c. R/2. d. R/4. 8. Two resistors are connected in parallel across a potential difference. The resistance of resistor A is twice that of resistor B. If the current is carried by resistor A is I, then what is the current carried by B? a. I. b. 2 I. (5) c. I / 2. d. 4 I. 10. Kirchoff’s loop rule follows from a. Conservation of charge. b. Conservation of energy. (5) c. Conservation of mass. d. Conservation of momentum. 11. The direction of the electric field at any given point is the same as a. The direction of the electric force that acts on a tiny positive test charge. b. The direction opposite to the direction of the electric force that acts on a tiny positive test charge. (5) c. The direction of the electric force that acts on a tiny negative test charge. d. The direction perpendicular to the direction of the electric force that acts on a tiny positive test charge. 12. Two charges, Q1 = 1.50 C and Q2 = - 2.00 C, are located as shown below. A tiny positive charge q = 0.100 C is placed at the origin. a. Find the total electric force exerted on the charge at the origin. Specify both magnitude and direction. F1= keq Q1/r12= (8.99 x 109 Nm2/C2)(0.100 x 10–6C)(1.50 x 10–6C)/(1.50 m)2 = 0.599 x 10 -3N F2 = ke qQ2/r22= (8.99 x 109 Nm2/C2)(0.100 x 10–6C)( 2.00 x 10–6C)/(0.750 m)2 = 3.20 x 10-3 N F1x = (0.599 x 10-3 N) cos(180o) = -0.599 x 10-3 N F1y= (0.599 x 10-3 N) sin(180o) = 0 N F2x = (3.20 x 10-3 N) cos (-90.0o) = 0 N F2y = (3.20 x 10-3 N) sin (-90.0o) = -3.20 x 10-3 N Ftotx = -0.599 x 10-3 N + 0 N = -0.599 x 10-3 N Ftoty = 0 N -3.20 x 10-3 N = -3.20 x 10-3 N Ftot =((-0.599 x 10-3 N)2 + (-3.20 x 10-3 N)2) 1/2 = 3.26 x 10-3 N tan-1Ftoty /Ftoty) = 79.4o + 180o = 259o b. What is the magnitude of the total electric field produced by charges Q1 and Q2 at the origin? (Hint: use the definition of electric field, not the superposition principle to answer this question.) F = |q| E = (3.00 x 10–3 C) (3.00 x 106N/C) = 9.00 x 103 N E = F / |q| = (3.26 x 10–3 N) / (0.100 x 10 –6 C) =32.6 x 10 3 N/C = 3.26 x 10 4 N/C 13. You are given a network of capacitors pictured below. e. Find the equivalent capacitance between terminals A and B. C12 = C1 + C2 = 4.00 mF + 4.00 mF = 8.00 mF 1/C34 = 1/C3 + 1/C4 = 1/(8.00 mF) + 1/(2.00 mF) = 5/(8.00 mF) C34 = 1.60 mF C1234 = C12 + C34 = 8.00 mF + 1.60 mF = 9.60 mF f. You connect the network to a 12-V battery. What is the charge accumulated on the plates of capacitor C3? Q34 = C34 V34= (1.60 mF) (12.0 V) = 19.2 mC Q3 = Q4 = Q34 = 19.2 mC g. How much energy is stored in the capacitor C3? EST = Q2/(2C) EST = (19.2 mC)2/(2 (8.00 mF)) = 23.0 x 10-3 J = 2.30 x 10-2 J 14. A block of carbon is 3.00 cm long and has a square cross-sectional area with sides of 0.500 cm. A potential difference of 8.40 V is maintained across its length. (Resistivity of carbon at 20o Celsius = 3500 x 10-8 m, its temperature coefficient of resistivity = 0.500 x 10-3 (oC)-1). a. What is the resistance of the block at 20o Celsius? R = L /A R = (3500 x 10-8 m) (0.0300 m) / (0.00500 m)2 = 4200 x 10-5 = 0.0420 b. If the temperature of the rod is increased to 45o Celsius, what is the resistance of the block? R = R0 [1 + (T – T0)] R = (0.0420 ) [ 1 + (-0.500 x 10-3 (oC)-1)( 45oC - 20o C)] = 0.0415 c. How much power is dissipated in the block at 20o Celsius? P = (V)2/R P = (8.40 V)2 / (0.0420 ) = 1680 W d. Is the power dissipated in the block at 45o Celsius greater or less than at 20o Celsius. EXPLAIN. GREATER. Power is inversely proportional to the resistance, hence as the resistance decreases, the power dissipated increases. 9. Given the network, write equations that would allow you to solve for the currents in each resistor if the values of the emfs and resistances were known. Label and indicate your choices for current directions. DO NOT SOLVE THE EQUATIONS. (15) I1 = I2 + I3 2- I1R5 + 1 -I2R1 – I1R4+ 4 = 0 3 + I3R2 – I2R1 + I3R3 = 0 TIME OF COMPLETION_______________ NAME___SOLUTION__________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1112, Exam 1 Version 1 Total Weight: 100 points Section 1 February 16, 2006 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 10:30 a.m. 11:45 a.m. PROBLEM POINTS 1-6 30 7 15 8 20 9 20 10 15 TOTAL 100 PERCENTAGE CREDIT CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 9. When resistors are connected in series, a. The same power is dissipated in each one. b. The potential difference across each is the same. (5) c. The current flowing in each is the same. d. More than one of the given answers is true. 10. Three identical capacitors are connected in series to a battery. If a total charge of Q flows from the battery, how much charge does each capacitor carry? a. 3Q. b. Q. (5) c. Q/3. d. Q/9. 11. The total amount of charge that passes through a wire's full cross section at any point per unit of time is referred to as . a. Wattage. (5) b. Current. c. Electric potential. d. Voltage. 12. A kilowatt-hour is equivalent to a. 1000 W. b. 3600 s. (5) c. 3,600,000 J/s. d. 3,600,000 J. 15. Two parallel-plate capacitors are identical in every respect except that one has twice the plate area of the other. If the smaller capacitor has capacitance C, the larger one has capacitance a. C/2. b. C. (5) c. 2C. d. 4C. 16. The direction of the electric field at any given point is the same as a. The direction of the electric force that acts on a tiny positive test charge. b. The direction opposite to the direction of the electric force that acts on a tiny positive test charge. (5) c. The direction of the electric force that acts on a tiny negative test charge. d. The direction perpendicular to the direction of the electric force that acts on a tiny positive test charge. 17. A rectangular solid block made of carbon has sides of lengths 1.0 cm, 2.0 cm, and 4.0 cm, lying along the x, y, and z axes, respectively (Fig. 18–35). Assume the resistivity is 3.0 10 5 m. Determine the resistance for current that passes through the block in a. The x direction, R = L/A Rx = a/(bc) = (3.00 x 10-5 -m) (0.01 m) / ((0.02 m)(0.04 m)) = 0.375 x 10-3 b. The y direction, Ry = b/(ac) = (3.00 x 10-5 -m) (0.02 m) / ((0.01 m)(0.04 m)) = 1.50 x 10-3 c. The z direction. Rz = c/(ab) = (3.00 x 10-5 -m) (0.04 m) / ((0.01 m)(0.02 m)) = 6.00 x 10-3 8. If C1 C2 2C3 22.6 F, a. What is the equivalent capacitance of the network? 1/C23 = 1/C2 + 1/C3 = 1/(22.6 F) + 1/(11.3 F) = 3/(22.6 F) C23 = 7.53 F C123 = C1 + C23 = 22.6 F + 7.53 F = 30.1 F b. How much charge is stored on each capacitor when V 45.0 V? Q1 = C1 V= (22.6 F) (45.0 V) = 1017 C Q2 = Q3 = C23 V= (7.53 F) (45.0 V) = 339 C c. What is the potential difference across each of the capacitors? V1 = 45.0 V V2 = Q2 / C2 = 15.0 V V3 = Q3 / C3 = 30.0 V 9. Calculate the electric field at one corner of a square 1.00 m on a side if the other three 6 corners are occupied by 2.25 10 C charges. E1 = k q1/r12 = (8.99 x 109 N-m2/C2) (2.25 x 10-6 C) / (1.00 m)2 = 20.2 x 103 N/C @ -90o E2 = k q2/r22 = (8.99 x 109 N-m2/C2) (2.25 x 10-6 C) / (1.41 m)2 = 10.1 x 103 N/C @ 225o E3 = k q3/r32 = (8.99 x 109 N-m2/C2) (2.25 x 10-6 C) / (1.00 m)2 = 20.2 x 103 N/C @ 180o E1x = E1 cos(- 90o) = 0 E1y = E1 sin(- 90o) = - 20.2 x 103 N/C E2x = E2 cos(225o) = -7.14 x 103 N/C E2y = E2 sin(225o) = - 7.14 x 103 N/C E3x = E3 cos(180o) = - 20.2 x 103 N/C E3y = E3 sin(180o) = 0 Etot x = E1x + E2x + E3x = - 27.3 x 103 N/C Etot y = E1y + E2y + E3y = - 27.3 x 103 N/C Etot = 38.7 x 103 N/C = 225o 10. Given the network, write equations that would allow you to solve for the currents in each resistor if the values of the emfs and resistances were known. Label and indicate your choices for current directions. DO NOT SOLVE THE EQUATIONS. I leave this one up to you to practice on your own. TIME OF COMPLETION_______________ NAME__SOLUTION___________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1112, Exam 1 Version 1 Total Weight: 100 points Section 1 February 15, 2007 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 10:30 a.m. 11:45 a.m. PROBLEM POINTS 1-6 20 7 20 8 20 9 20 10 20 TOTAL 100 PERCENTAGE CREDIT CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. Three identical resistors are connected in parallel to a battery. If the current of 12 A flows from the battery, how much current flows through any one of the resistors? a. 12 A. b. 4 A. (4) c. 36 A. d. Zero. 2. As more and more capacitors are connected in series, the equivalent capacitance of the combination increases. a. Always true. b. Sometimes true; it depends on the voltage of the battery to which the combination is connected. (4) c. Sometimes true; it goes up only if the next capacitor is larger than the average of the existing combination. d. Never true. 3. A surface on which all points are at the same potential is referred to as a. A constant electric force surface. b. A constant electric field surface. (4) c. An equipotential surface. d. An equivoltage surface. 4. It takes 50 J of energy to move 10 C of charge from point A to point B. What is the potential difference between points A and B? a. 500 V. b. 50 V. (4) c. 5.0 V. d. 0.50 V. 5. Car batteries are rated in "amp-hours." This is a measure of their a. Charge. b. Current. (4) c. Emf. d. Power. 6. Consider two copper wires. One has twice the length of the other. How do the resistivities of these two wires compare? a. Both wires have the same resistivity. b. The longer wire has twice the resistivity of the shorter wire. (4) c. The longer wire has four times the resistivity of the shorter wire. d. None of the given answers 7. A 100-W lightbulb has a resistance of about 12 when cold (20°C) and 140 when on (hot). Estimate the temperature of the filament when hot assuming an average 1 temperature coefficient of resistivity 0.0060 (Cº ) . R0 = 12.0 at T0 = 20°C R = R0 [1 + (T – T0)] R/R0 = 1 + (T – T0) R/R0 - 1 = (T – T0) (R/R0 – 1)/(T – T0) T = (R/R0 – 1)/ T0 = [(140 )/(12.0 ) -1]/( 0.0060 (o C) -1) + 20.0°C = 1798°C 8. Determine the electric field E at the origin 0 due to the two charges at A and B. Specify magnitude and direction. Let Q = 2.50 C and l = 0.200 m. (20) EA = k |QA|/rA2 = (8.99 x 109 N-m2/C2) (2.50 x 10-6 C) / (0.200 m)2 = 562 x 103 N/C @ -90o EB = k |QB|/rB2 = (8.99 x 109 N-m2/C2) (2.50 x 10-6 C) / (0.200 m)2 = 562 x 103 N/C @ 210o EAx = EA cos(- 90o) = 0 EAy = EA sin(- 90o) = - 562 x 103 N/C EBx = EB cos(210o) = - 487 x 103 N/C EBy = EB sin(210o) = -281 x 103 N/C Etot x = EAx + EBx = - 487 x 103 N/C Etot y = EAy + EBy = - 843 x 103 N/C Etot = 974 x 103 N/C = 9.74 x 105 N/C = 60.0o + 180o = 240o or -120o 9. Determine (20) a. The equivalent resistance of the circuit and R1 = 820 R2 = 680 R3 = 470 R1 and R2 are in parallel: 1/R12 = 1/ R1 + 1/ R2 = 1/(820 ) + 1/(680 ) R12 =372 R12 and R3 are in series: R123 = R12 + R3 R123 = (372 ) + (470 ) = 842 b. The voltage across each resistor. I = V/R123 I = (12 V)/(842 ) = 0.0143 A V3 = I R3= (0.0143 A)( 470 ) = 6.70 V V1 =V2 =V - V3 = 12.0 V - 6.70 V = 5.30 V 10. Given the network, write equations that would allow you to solve for the currents in each resistor. Label and indicate your choices for current directions. DO NOT SOLVE THE EQUATIONS. Assuming that I1 flows in the left branch upward, I2 flows in the central branch downward, and I3 flows in the right branch upward: (-6.00 V) – I2 (6.00 )– I1 (8.00 )– I1 (12.0 ) = 0 (-3.00 V) + I3 (10.0 ) + I2 (6.00 ) + I3 (2.00 ) = 0 I1+ I3 = I2 TIME OF COMPLETION_______________ NAME___SOLUTION__________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1112, Exam 1 Version 1 Total Weight: 100 points Section 1 February 14, 2008 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 50 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work five (5) multiple choice problems and four (4) calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 10:30 a.m. 11:45 a.m. PROBLEM POINTS 1-6 20 7 20 8 20 9 20 10 20 TOTAL 100 PERCENTAGE CREDIT CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. The charge of the proton is the same as (A) The charge of the neutron. (B) The charge of the neutron, but of the opposite sign. (4) (C) The charge of the atom. (D) The charge of the electron. (E) The charge of electron, but with opposite sign. 2. The main difference between conductors and insulators is due to (A) Protons. (B) Neutrons. (4) (C) Valence (outer) electrons. (D) Atoms. (E) None of the above choices is correct. 3. Two parallel-plate capacitors are identical in every respect except that one has twice the separation distance between the plates compared to the other. If the thinner capacitor has capacitance C, the thicker one has capacitance (A) C/4. (B) C/2. (4) (C) C. (D) 2C. (E) 4C. 4. Two parallel metal plates carry positive and negative charge of equal strength as shown below. What is the correct direction of the equipotential lines for this charge configuration? (A) Vertical, perpendicular to the plates. (B) Horizontal, parallel to the plates. (4) (C) Both vertical and horizontal. (D) Tilted at a 45o angle to the plates. (E) Circular, around the plates. 5. When two or more resistors are connected in parallel to a battery, (A) the voltage across each resistor is the same. (B) the total current flowing from the battery equals the sum of the currents flowing through each resistor. (4) C) the equivalent resistance of the combination is less than the resistance of any one of the resistors. (D) all of the given answers are correct. (E) none of the above answers is correct. 6. Four unequal resistors are connected in series. Which one of the following statements is correct about this circuit? (A) The total resistance is equal to any one of the resistors. (B) The total resistance is equal to the average of the resistors. (4) (C) The total resistance is less than the smallest resistor. (D) The total resistance is less than the largest resistor. (E) The total resistance is greater than the largest resistor. 8. Five identical 5.00 F capacitors are connected as shown below. a. Find equivalent capacitance of the network. 1/C34 = 1/C3 + 1/C4 = 1/(5.00 F) + 1/(5.00 F) = 2/(5.00 F) C34 = 2.50 F C2345 = C2 + C34 + C5 = 5.00 F + 2.50 F + 5.00 F = 12.5 F 1/C12345 = 1/C1 + 1/C2345 = 1/(5.00 F) + 1/(12.5 F) = 0.280 F-1 C12345 = 3.57 F b. If a potential difference of 12.0 V is applied between points a and b, how much charge is stored on Capacitor 1? Q1 = Q12345 = C12345V = (3.57 F)(12.0 V) = 42.9 C c. What is the potential difference across the Capacitor 5? V1 = Q1 /C1 = (42.9 C)/(5.00 F) = 8.57 V V5 =V -V1 = 112.0 V – 8.57 V = 3.43 V 8. Determine the electric field E at the origin 0 due to the three charges pictured below. Specify magnitude and direction. Let Q = 1.20 C and l = 0.250 m. E1 = k |Q1|/r12 = (8.99 x 109 N-m2/C2) (1.20 x 10-6 C) / (0.250 m)2 = 1.73 x 105 N/C @ -90o E2 = k |Q2|/r22 = (8.99 x 109 N-m2/C2) (1.20 x 10-6 C) / (0.250 m)2 = 1.73 x 105 N/C @ 180o E3 = k |Q3|/r32 = (8.99 x 109 N-m2/C2) (1.20 x 10-6 C) / (0.125 m)2 = 6.90 x 105 N/C @ 90o E1x = E1 cos(- 90o) = 0 E1y = E1 sin(- 90o) = - 1.73 x 105 N/C E2x = E2 cos(180o) = -1.73 x 105 N/C E2y = E2 sin(180o) = 0 E3x = E3 cos(90o) = 0 E3y = E3 sin(90o) = 6.90 x 105 N/C Etot x = E1x + E2x + E3x = - 1.73 x 105 N/C Etot y = E1y + E2y + E3y = 5.17 x 105 N/C Etot = 5.45 x 105 N/C = -71.5o + 180o = 110o 9. Find the equivalent resistance between points a and b. 1/R23 = 1/R2 + 1/R3 = 1/(12.0 4.00 3.00 R23 = 3.00 R123 = R1 + R23 = 8.00 + 3.00 = 11.0 If the potential drop of 12.0 V is applied to the combination, find the current in each resistor. I123 = I1 = I23 =V/R123 = (12.0 V)/( 11.0 ) = 1.09 A V41= I1 R1 = (1.09 A)(8.00 ) = 8.73 V V2=V3=V -V1= 3.27 V I2 =V2/R2 = (3.27 V)/(12.0) = 0.273 A I3 =V3/R3 = (3.27 V)/(4.00) = 0.818 A 10. An electric space heater has a Nichrome heating element with a resistance of 8.00 at 20.0o C. When 120 V are applied, the electric current heats the Nichrome wire to 1000o C. (Temperature coefficient of resistivity of Nichrome is = 0.4 x 10-3 (o C)-1.) d. What is the initial current in the cold heating element? I =V/R = (120 V)/(8.00) = 15.0 A e. What is the resistance of the heating element at 1000o C? R = R0 [1 + (T – T0)] R = (8.00 ) [1 + 0.4 x 10-3 (o C)-1(1000o – 20.0o)] = 11.1 c. What is the power dissipated in the heating element at 1000o C? P = (V)2 R = (120 V)2 (11.1 ) = 1,297 W TIME OF COMPLETION_______________ NAME_____________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1112, Exam 1 Version 1 Total Weight: 100 points Section 1 June 19, 2006 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 2:30 p.m. 3:50 p.m. PROBLEM POINTS 1-6 20 7 20 8 20 9 20 10 20 TOTAL 100 PERCENTAGE CREDIT CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. The resistivity of a wire depends on a. Its length. b. Its cross-sectional area. (4) c. The material out of which it is composed. d. All of the given answers 2. Negative temperature coefficients of resistivity a. Do not exist. b. Exist in conductors. (4) c. Exist in semiconductors. d. Exist in superconductors. 3. One joule per coulomb is a a. Newton. b. Volt. (4) c. Electron-volt. d. Farad. 4. For a proton moving in the direction of the electric field a. Its potential energy increases and its electric potential decreases. b. Its potential energy decreases and its electric potential increases. (4) c. Its potential energy increases and its electric potential increases. d. Its potential energy decreases and its electric potential decreases. 5. At twice the distance from a point charge, the strength of the electric field a. Is four times its original value. b. Is twice its original value. (4) c. Is one-half its original value. d. Is one-fourth its original value. 6. Two charged objects are separated by a distance d. The first charge is larger in magnitude than the second charge. a. The first charge exerts a larger force on the second charge. b. The second charge exerts a larger force on the first charge. (4) c. The charges exert forces on each other equal in magnitude and opposite in direction. d. The charges exert forces on each other equal in magnitude and pointing in the same direction. 8. Two charges, Q1 = - 2.00 C and Q2 = + 4.00 C, are located as shown below. a. Find the total electric field at the origin. Specify both magnitude and direction. E1 = k |Q1|/r12 = (8.99 x 109 N-m2/C2) (2.00 x 10-6 C) / (0.150 m)2 = 1598 x 103 N/C @ -90o E2 = k |Q2|/r22 = (8.99 x 109 N-m2/C2) (4.00 x 10-6 C) / (0.235 m)2 = 651 x 103 N/C @ 180o E1x = E1 cos(- 90o) = 0 E1y = E1 sin(- 90o) = - 1598 x 103 N/C E2x = E2 cos(180o) = -651 x 103 N/C E2y = E2 sin(180o) = 0 Etot x = E1x + E2x = - 651 x 103 N/C Etot y = E1y + E2y = - 1598 x 103 N/C Etot = 1726 x 103 N/C = 1.726 x 106 N/C = 67.8o + 180o = 248o b. What is the magnitude of the total electric force exerted by charges Q1 and Q2 on - 3.00 C charge placed at the origin? What is the direction of the force? F = |Q| E = (3.00 x 10-6 C) (1.726 x 106 N/C) = 5.18 N @ 67.8o 8. Find the equivalent capacitance of the combination shown below. (20) 1/C12 = 1/C1 + 1/C2 = 1/(3.00 mF) + 1/(9.00 mF) = 4/(9.00 mF) C12 = 2.25 mF 1/C45 = 1/C4 + 1/C5 = 1/(12.0 mF) + 1/(6.00 mF) = 3/(12.0 mF) C45 = 4.00 mF C12345 = C12 + C3 + C45 = 2.25 mF + 11.0 mF + 4.00 mF = 17.25 mF If the potential difference between points a and b is 12.0 V, what is the voltage across the 12.0mF capacitor? Q45 = C45V45 = (4.00 mF)(12.0 V) = 48.0 mC Q4 = Q45 = 48.0 mC V4 = Q4 /C4 = (48.0 mC)/(12.0 mF) = 4.00 V 9. A length of aluminum wire is connected to a precision 10.00-V power supply, and a current of 0.4212 A is precisely measured at 20.0°C. The wire is placed in a new environment of unknown temperature where the measured current is 0.3618 A. What is the unknown temperature? V = I0 R0 R0 = V I0 R0 = (10.0 V) / (0.4212 A) = 23.7 R = V I R = (10.0 V) / (0.3618 A) = 27.6 R = R0 [1 + (T – T0)] R/R0 = 1 + (T – T0) R/R0 - 1 = (T – T0) (R/R0 – 1)/(T – T0) T = (R/R0 – 1)/ T0 = [(27.6)/(23.7 ) -1]/( 0.0039 (o C) -1) + 20.0°C = 62.2°C 10. Given that all resistors in the picture below are 20.0 each, and the battery provides the potential difference of 24.0 V, a. Find the equivalent resistance. R23 = R2 + R3 = 20.0 20.0 40.0 1/R123 = 1/R1 + 1/R23 = 1/(20.0 ) + 1/(40.0 ) = 3/(40.0 ) R12345 = R123 + R4 + R5 = 13.3 + 20.0 20.0 = 53.5 R123 = 13.3 b. Find the current that flows through the resistor 4. I12345 = I4 = V/R12345 = (24.0 V)/( 53.5 ) = 0.449 A c. Find the potential difference across the resistor 4. V4 = I4 R4 = (0.449 A)(20.0 ) = 8.97 V TIME OF COMPLETION_______________ NAME___SOLUTION__________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1112, Exam 1 Version 1 Total Weight: 100 points Section 1 June 19, 2007 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on eight (8) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 1:30 p.m. 2:45 p.m. PROBLEM POINTS 1-6 20 CREDIT 7 20 8 20 9 20 10 20 TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. Note that each of the following multiple choice problems is worth four (4) points. 1. Electric field lines near a positive point charge (A) circle clockwise. (B) circle counter-clockwise. (C) radiate inward. (D) radiate outward. (E) do not exist. 2. The electric field shown below (A) increases to the right. (B) increases down. (C) decreases to the right. (D) decreases down. (E) is uniform. 3. Two parallel metal plates carry positive and negative charge of equal strength as shown below. What is the correct direction of the electric field lines for this charge configuration? (A) Vertical, perpendicular to the plates. (B) Horizontal, parallel to the plates. (C) Both vertical and horizontal. (D) Tilted at a 45o angle to the plates. (E) Circular, around the plates. 4. When two or more resistors are connected in series to a battery (A) the total voltage across the combination is the algebraic sum of the voltages across the individual resistors. (B) the same current flows through each resistor. (C) the equivalent resistance of the combination is equal to the sum of the resistances of each resistor. (D) all of the given answers are correct. (E) none of the above answers is correct. 5. A battery charges a parallel-plate capacitor fully and then is removed. The plates are immediately pulled apart. (With the battery disconnected, the amount of charge on the plates remains constant.) What happens to the potential difference between the plates as they are being separated? (A) It increases. (B) It decreases. (C) It remains constant. (D) cannot be determined from the information given 6. A negative charge is moved from point A to point B along an equipotential surface. (A) The negative charge performs work in moving from point A to point B. (B) Work is required to move the negative charge from point A to point B. (C) Work is both required and performed in moving the negative charge from point A to point B. (D) No work is required to move the negative charge from point A to point B. (E) None of the above choices is correct. 7. The network of three capacitors is shown below. a. Find the equivalent capacitance of the network. C12 = C1 + C2 = 2.00 F + 4.00 F = 6.00 F 1/C123 = 1/C3 + 1/C12 = 1/(3.00 F) + 1/(6.00 F) = 3/(6.0 F) C123 = 2.00 F b. Find the charge on the 3.00-F capacitor when the system is connected to a 6.00V battery. Q3 = C123 V= (2.00 F) (6.00 V) = 12.0 C c. What is the potential difference across the 2.00-F capacitor? V3 = Q3 /C3 = (12.0 C)/(3.00 F) = 4.00 V V1 = 6.00 V – 4.0 V = 2.00 V 8. Determine the electric force exerted on the charge located at the origin 0 due to the two charges at A and B. Specify magnitude and direction. Let Q = 2.50 C and l = 0.200 m. FAO = k |QA||QO|/rAO2 = (8.99 x 109 N-m2/C2) (2.50 x 10-6 C) (2.50 x 10-6 C) / (0.200 m)2 = 1.41N @ 90.0o FBO = k |QB||QO|/rBO2 = (8.99 x 109 N-m2/C2) (2.50 x 10-6 C) (2.50 x 10-6 C) / (0.200 m)2 = 1.41 N @ 30.0o FAOx = FAO cos(- 90.0o) = 0 FAOy = FAO sin(- 90.0o) = - 1.41 N FBOx = FBO cos(30.0o) = 1.22 N FBOy = FBO sin(30.0o) = 0.705 N Ftot x = FAOx + FBOx = 1.22 N Ftot y = FAOy + FBOy = - 0.705 N Ftot = 1.41 N = -30.0o 9. A quantity of mercury (30.0 cm3) is poured into a glass tube that has a diameter of 4.00 mm. The resistance of the mercury column is determined to be 0.01138 . a. What is the resistivity of mercury? R = L/A = RA/L A = r2 = (0.00200 m)2 = 1.26 x 10-5 m2 V=AL L = V/A = (30.0 x 10-6 m3)/( 1.26 x 10-5 m2) = 2.38 m = (0.01138 )(1.26 x 10-5 m2)/(2.38 m) = 6.02 x 10-8-m b. What potential difference must be applied across the mercury column to produce a current of 1.00 A in it? V = I R = (1.00 A)(0.01138) = 0.01138 V c. Assuming that the electric field inside the mercury column is uniform, what is its strength? V = E L E = V/L = (0.01138 V)/(2.38 m) = 0.00478 V/m 10. Pictured below is a circuit containing 5 resistors and a battery. (20) a. Find equivalent resistance. Let R1 = 12.0 Let R2 = 8.00 Let R3 = 6.00 Let R4 = 2.00 Let R5 = 10.0 R12 = R1 + R2 = 12.0 8.00 20.0 1/R123 = 1/R3 + 1/R12 = 1/(6.00 ) + 1/(20.0 ) R123 = 4.62 R12345 = R123 + R4 + R5 = 4.62 + 2.00 10.0 = 16.6 b. Find the current flowing in the 10.0- resistor. I12345 = I5 = V/R12345 = (3.00 V)/( 16.6 ) = 0.181 A c. Find the power dissipated in the 10.0 - resistor. P = I2 R = (0.181 A)2 (10.0 ) = 0.327 W TIME OF COMPLETION_______________ NAME___SOLUTION__________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1112, Exam 1 Version 1 Total Weight: 100 points Section 1 June 17, 2008 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work five (5) multiple choice problems and four (4) calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 1:30 p.m. 2:45 p.m. PROBLEM POINTS 1-6 20 7 20 8 20 9 20 10 20 TOTAL 100 CREDIT PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. An equipotential surface must be (A) Parallel to the electric field at each point. (B) Anti-parallel to the electric field at each point. (4) (C) Equal to the electric field at each point. (D) Perpendicular to the electric field at each point. 2. The electron-volt is a unit of (A) Charge. (B) Electric potential. (4) (C) Electric force. (D) Electric field. (E) Energy. 3. The length of a certain wire is halved while its radius is doubled. What is the change in the resistance of the wire? (A) It stays the same. (B) It is reduced by a factor of 4. (4) (C) It is increased by a factor of 4. (D) It is reduced by a factor of 8. (E) It is increased by a factor of 8. 4. When atom A loses an electron to atom B, (A) Atom A becomes more negative than atom B. (B) Atom A acquires more neutrons than atom B. (4) (C) Atom A acquires less neutrons than atom B. (D) Atom A becomes a positive ion and atom B becomes a negative ion. (E) Atom A becomes a negative ion and atom B becomes a positive ion. 5. Four unequal resistors are connected in parallel. Which of the following statements is correct about this combination? (A) The total resistance is less than the smallest resistor. (B) The total resistance is equal to the average of the resistance of all the resistors. (4) (C) The total resistance is more than the largest resistance. (D) None of the above answers is correct. 6. The work done in moving a positive charge against an electric field does not depend on the path chosen to move the charge in that field. Based on the statement, what kind of force is the electrostatic force? (A) Discrete. (B) Quantized. (4) (C) Polarized. (D) Conservative. (E) Resistive. 9. Five identical 10.00 F capacitors are connected as shown below. a. Find equivalent capacitance of the network. 1/C23 = 1/C2 + 1/C3 = 1/(10.0 F) + 1/(10.0 F) = 2/(10.0 F) C12 = 5.00 F C234 = C23 + C4 = 5.00 F + 10.0 F = 15.0 F 1/C12345 = 1/C1 + 1/C234 + 1/C5 = 1/(10.0 F) + 1/(15.0 F) + 1/(10.0 F) = 8/(30.0 F) C12345 = 3.75 F b. If a potential difference of 36.0 V is applied between points a and b, how much charge is stored on Capacitor 5? Q12345 = C12345V = (3.75 F)(36.0 V) = 135 C Q5 = Q12345 = 135 C c. What is the potential difference across the Capacitor 4? V4 = V234 = Q234 /C234) = (135 C)/( 15.0 F) = 9.00 V 8. Determine the electric field E at the origin 0 due to the three charges pictured below. Specify magnitude and direction. Let Q = 3.50 C and l = 0.480 m. E1 = k |q1|/r12 = (8.99 x 109 N-m2/C2) (3.50 x 10-6 C) / (0.480 m)2 = 1.37 x 105 N/C @ 180o E2 = k |q2|/r22 = (8.99 x 109 N-m2/C2) (3.50 x 10-6 C) / (0.240 m)2 = 5.48 x 105 N/C @ 180o E3 = k |q3|/r32 = (8.99 x 109 N-m2/C2) (3.50 x 10-6 C) / (0.240 m)2 = 5.48 x 105 N/C @ 90o E1x = E1 cos(180o) = - 1.37 x 105 N/C E1y = E1 sin(180o) = 0 E2x = E2 cos(180o) = - 5.48 x 105 N/C E2y = E2 sin(180o) = 0 E3x = E3 cos(90o) = 0 E3y = E3 sin(90o) = 5.48 x 105 N/C Etot x = E1x + E2x + E3x = -6.85 x 105 N/C Etot y = E1y + E2y+ E3y = 5.48 x 105 N/C Etot = 8.77 x 105 N/C = -38.6o + 180o = 141o 9. Determine the electric potential V at the origin 0 due to the three charges pictured below. Let Q = 3.50 C and l = 0.480 m. V1 = k q1/r1 = (8.99 x 109 N-m2/C2) (-3.50 x 10-6 C) / (0.480 m) = -6.56 x 104 V V2 = k q2/r2 = (8.99 x 109 N-m2/C2) (3.50 x 10-6 C) / (0.240 m) = 13.1 x 104 V V3 = k q3/r3 = (8.99 x 109 N-m2/C2) (3.50 x 10-6 C) / (0.240 m) = 13.1 x 104 V Vtot = V1 + V2 + V3 = 1.96 x 105 V How much work is required to bring a -2.40 C charge from infinity to the origin without changing the charge’s kinetic energy? W = qVtot = (-2.40 x 10-6 C)(1.96 x 105 V) = -0.470 J 10. A nichrome wire ( = 10-6 -m and = 0.4 x 10-3 (o C)-1 at 20.0o C) has a radius of 0.650 mm. f. What length of wire is needed to obtain a resistance of 2.00 at 20.0o C? R = L/A L = R A/ = (2.00 ) (0.650 x 10-3 m)2/(10-6 -m) = 2.65 m g. How cold must be the wire for its resistance to be half of that? TIME OF COMPLETION_______________ NAME____SOLUTION_________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 1112, Exam 1 Version 1 Total Weight: 100 points Section 1 October 30, 2009 1. Check your examination for completeness prior to starting. There are a total of nine (9) problems on six (6) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 50 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and three (3) calculation problems. Work five (5) multiple choice and three (3) calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 12:00 p.m. 12:50 p.m PROBLEM POINTS 1-6 25 7 25 8 25 9 25 TOTAL 100 PERCENTAGE CREDIT CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 13. When two charged conductors are connected to each other by a conducting wire, the one that gains electrons or, equivalently, loses positive charge is said, in comparison with the other conductor, to have had a. A greater electric potential energy. b. A lower capacitance. (5) c. A lower dielectric constant. d. A higher potential. 14. Which one of the following electric charges does NOT exist in nature? a. 27.0 e. b. –3.00 x 10 125 e. (5) c. 245.5 e. d. – 180/3 e. 15. When an electric current exists within a conducting wire, which of the following statements is correct? a. Electric field inside the wire is zero. b. Electric field inside the wire is parallel to the current flow. (5) c. Electric field inside the wire is anti-parallel to the current flow. d. Electric field inside the wire is perpendicular to the current flow. 4. If three 4.00 mF capacitors are connected in parallel, what is the combined capacitance? a. 12.0 mF. b. 0.750 mF. (5) c. 8.00 mF. d. 0.46 mF. 5. The electrostatic force between a negative electron and a neutral neutron is a. Attractive. b. Repulsive. (5) c. Zero. d. Sometimes attractive and sometimes repulsive. 6. It takes 50.0 J of energy to move 10.0 C of charge from point A to point B. What is the magnitude of the potential difference between points A and B? a. 500 V. b. 50.0 V. (5) c. 5.00 V. d. 0.500 V. 7. A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 5.00o C, the resistance of the carbon resistor is 221.4 What is the temperature on a spring day PHYS 1112 Exam 1, Version 1 Fall 2004 70 when the resistance is 213.2 (Hint: Take the reference temperature T0 to be 5.00o C). The temperature coefficient of resistivity of carbon is = -0.500 x 10-3(oC)-1. R = R0 [1 + (T – T0)] R/ R0 = [1 + (T – T0)] R/ R0 - 1 = (T – T0) (R/ R0 – 1)/ T – T0 T = T0 + (R/ R0 – 1)/ T = 79.1o C (This temperature is, of course, unrealistic) 8. Determine the electric field E at the origin 0 due to the two charges at A and B. Specify magnitude and direction. Let Q = 1.50 C and l = 0.400 m. E1 = k |q1|/r12 = (8.99 x 109 N-m2/C2) (1.50 x 10-6 C) / (0.400 m)2 = 8.43 x 104 N/C @ -90o E2 = k |q2|/r22 = (8.99 x 109 N-m2/C2) (1.50 x 10-6 C) / (0.400 m)2 = 8.43 x 104 N/C @ 30o E1x = E1 cos(- 90o) = 0 PHYS 1112 Exam 1, Version 1 Fall 2004 71 E1y = E1 sin(- 90o) = - 8.43 x 104 N/C E2x = E2 cos(30o) = 7.30 x 104 N/C E2y = E2 sin(30o) = 4.21 x 104 N/C Etot x = E1x + E2x = 7.30 x 104 N/C Etot y = E1y + E2y = - 4.21 x 104 N/C Etot = 8.43 x 104 N/C = -30.0o 9.You are given a network of capacitors pictured below. C1 = 8.00 F, C2 = 5.00 F, C3 = 6.00 F and C4 = 6.00 F. The applied potential is Vab = 24.0 V. h. Find the equivalent capacitance between points a and b. 1/C34 = 1/C3 + 1/C4 = 1/(6.00 F) + 1/(6.00 F) = 2/(6.00 F) C34 = 3.00 F C234 = C2 + C34 = 5.00 F + 3.00 F = 8.00 F 1/C1234 = 1/C1 + 1/C234 = 1/(8.00 F) + 1/(8.00 F) = 2/(8.00 F) i. Calculate the charge on the capacitor C1. Q1 = Q1234 = C1234 V= (4.0 F) (24.0 V) = 96.0 C PHYS 1112 Exam 1, Version 1 Fall 2004 72 C1234 = 4.00 F j. What is the potential difference across the capacitor C1? V1 = Q1 / C1 = 12.0 V k. What is the potential difference across the capacitor C2? V2 = V - V1 = 24.0 V – 12.0 V = 12.0 V PHYS 1112 Exam 1, Version 1 Fall 2004 73 PHYS 1112 Exam 1, Version 1 Fall 2004 74
