TIME OF COMPLETION______NAME____SOLUTION_________________________

 

 

                                           DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1112, Exam 1                                                                                                             Section 1

Version 1                                                                                                                 February 12, 2004

Total Weight: 100 points

 

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:                1:30 p.m.

Stop:                2:45 a.m

 

 

 

               PROBLEM

 

                 POINTS

 

                 CREDIT

 

                     1-6

 

                      30

 

 

 

                       7

 

                      20

 

 

 

8

 

                      20

 

 

 

                       9

 

                      15

 

 

 

                      10

 

                      15

 

 

 

                 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.

 

 

 

 

 

  1. At a particular point in space, a charge Q experiences no net force. It follows that

 

    1. There are no charges nearby.

 

    1. If charges are nearby, they have the opposite sign of Q.

(5)

    1. If charges are nearby, the total positive charge must equal the total negative charge.

 

    1. None of the above.

 

 

 

  1. Two initially uncharged capacitors of capacitance C0 and 2C0, respectively, are connected in series across a battery. Which of the following is true?

 

    1. The capacitor 2C0 carries twice the charge of the other capacitor.

 

    1. The voltage across each capacitor is the same.

(5)

    1. The energy stored by each capacitor is the same.

 

    1. None of the above.

 

 

 

  1. 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?

 

    1. R.

(5)

    1. 2 R.

 

    1. R/2.

 

    1. R/4.

 

 

 

  1. 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?

 

    1. I.

 

    1. 2 I.

(5)

    1. I / 2.

 

    1. 4 I.

 

 

 

  1. Kirchoff’s loop rule follows from

 

    1. Conservation of charge.

 

    1. Conservation of energy.

(5)

    1. Conservation of mass.

 

    1. Conservation of momentum.

 

 

 

 

 

  1. The direction of the electric field at any given point is the same as

 

    1. The direction of the electric force that acts on a tiny positive test charge.

 

    1. The direction opposite to the direction of the electric force that acts on a tiny positive test charge.

(5)

    1. The direction of the electric force that acts on a tiny negative test charge.

 

    1. The direction perpendicular to the direction of the electric force that acts on a tiny positive test charge.  

 

 

 

 

 

 

  1. Two charges, Q1 = 1.50 mC and Q2 = - 2.00 mC, are located as shown below. A tiny positive charge q = 0.100 mC is placed at the origin.

 

    1. 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

q = tan-1(Ftoty /Ftoty)

 

q = 79.4o + 180o = 259o

 

 

 

    1.   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

 

  1. You are given a network of capacitors pictured below.

 

           

a.       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

 

b.      You connect the network to a 12-V battery. What is the charge accumulated on the plates of capacitor C3?

 

Q34 = C34 DV34= (1.60 mF) (12.0 V) = 19.2 mC

 

Q3 = Q4 = Q34 = 19.2 mC

 

 

c.       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

 

 

  1. 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 r = 3500 x 10-8 Wm, its temperature coefficient of resistivity a = -0.500 x 10-3  (oC)-1).

  

    1. What is the resistance of the block at 20o Celsius?

 

R = r L /A

 

R = (3500 x 10-8 Wm) (0.0300 m) / (0.00500 m)2 = 4200 x 10-5 W =  0.0420 W

 

 

    1. If the temperature of the rod is increased to 45o Celsius, what is the resistance of the block?

 

                   R = R0 [1 + a (T – T0)]

 

                    R = (0.0420 W ) [ 1 + (-0.500 x 10-3  (oC)-1)( 45oC - 20o C)] = 0.0415 W

 

 

    1. How much power is dissipated in the block at 20o Celsius?

 

P = (DV)2/R

 P = (8.40 V)2 / (0.0420 W) = 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

e2- I1R5 + e1 -I2R1 – I1R4+ e4 = 0

e3 + I3R2 – I2R1  + I3R3 = 0