TIME OF COMPLETION____SOLUTION___________    NAME_____________________________

 

 

                                           DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1112, Exam 1                                                                                                             Section 1

Version 1                                                                                                              September 24, 2003

Total Weight: 100 points

 

1.         Check your examination for completeness prior to starting.  There are a total of nine (9) 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 five (5) 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

 

                     1-5

 

                      25

 

 

 

                       6

 

                      25

 

 

 

7

 

                      20

 

 

 

                       8

 

                      15

 

 

 

                       9

 

                      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. Which one of the following electric charges does NOT exist in nature?

 

    1. 27.0 e.

 

    1. –3.00 x 10 125 e.

(5)

    1. 245.5 e.

 

    1. – 180/3 e.

 

 

 

  1. A combination of two electrons and three protons would have a net charge of

 

    1. +1.

 

    1. –1.

(5)

    1. +1.60 x 10–19 C.

 

    1. -1.60 x 10–19 C.

 

 

 

  1. When a battery is placed into a complete circuit, the voltage across its terminals is its

 

    1. Emf

.

    1. Terminal voltage.

(5)

    1. Power.

 

    1. All of these.

 

 

  1. For an ohmic resistor, current and resistance

 

    1. Do not vary with temperature.

 

    1. Are directly proportional to each other.

(5)

    1. Are independent of voltage.

 

    1. None of these.

 

 

 

 

  1. For charged parallel plates, where is the point of lowest electric potential:

 

    1. Near the positive plate.

 

    1. Near the negative plate.

(5)

    1. Midway between the plates.

 

    1. Potential is the same everywhere between the plates.

 

 

 

  1. You have a point charge of magnitude 3.00 x 10–3 C that is attracted to Milly Coulomb yet repelled by Billy Coulomb as shown below.

    1. What is a sign of your charge?

 

(25)         Negative.

 

    1. Find the electric field of Milly and Billy at the position of your charge. Specify both magnitude and direction.

 

EM = ke qM/rM2 = (8.99 x 109 Nm2/C2)(1.00 x 10–3 C)/(1.00 m2) = 8.99 x 106 N/C

EB  = ke qB/rB2 = (8.99 x 109 Nm2/C2)(2.00 x 10–3 C)/(4.00 m2) = 4.50 x 106 N/C

 

EMX = EM cos(qM) = (8.99 x 106 N/C) cos(90.0o) = 0 N/C

 

EMY = EM sin(qM) = (8.99 x 106 N/C) sin(90.0o) = 8.99 x 106 N/C

 

EBX = EB cos(qB) = (4.50 x 106 N/C) cos(0o) =  4.50 x 106 N/C

 

EBY = EB sin(qB) = (4.50 x 106 N/C) sin(0o) = 0 N/C

 

ETOT X = EMX + EBX = 4.50 x 106 N/C

 

ETOT Y = EMY + EBY = 8.99 x 106 N/C

 

ETOT = sqrt(ETOT X2 + ETOT Y2)

 

ETOT = 1.01 x 107 N/C

 

qTOT = tan-1 (ETOT Y /ETOT X ) = 63.4o

 

 

    1.   Find the magnitude and direction of the total electrical force of your charge.

 

 

F = q E

 

F = (3.00 x 10–3 C)(1.01 x 107 N/C) = 3.03 x 104 N

 

q = 63.4o+ 180o = 243o  (Opposite to the electric field.)

 

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

 

           

a.       Find the equivalent capacitance between terminals A and B.

 

C34 = C3 + C4 = 8.00 mF

 

1/C234 = 1/C2 + 1/C34 = 1/(8.00 mF) + 1/(2.00 mF) = 5/(8.00 mF)

 

C2345 = C2 + C345 = (8.00/5 mF) + (2.00 mF) = 18/5 mF

 

1/C12345 = 1/C1 + 1/C2345 = 1/(8.00 mF) + 5/(18 mF)

 

C12345 = 2.48 mF

 

 

(20)

 

b.      You connect the network to a battery and discover that the total charge stored in the network is 0.360 C. What potential difference does the battery provide?

 

C = Q/(DV)

 

DV = Q/C = 145 V

 

 

  1. At 20o Celsius, a silicon rod is connected to a battery with a terminal voltage of 6.00 V. A 0.500-A current results. Given that a = -7.00 x 10-2 (Co)-1,

  

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

 

R = DV/I

 

R = (6.00 V) / (0.500 A) = 12.0 W

 

 

(15)

    1. If the temperature of the rod is increased to 25o Celsius, what is the resistance of the rod?

 

R = RO [ 1 + a (T - TO)]

 

R = (12.0 W) [ 1 + (-7.00 x 10-2 (Co)-1) ( 25o - 20o)] = 7.80 W

 

    1. How much current does the rod then carry?

 

I =  DV/R

 

I = (6.00 V)/(7.80 W) = 0.769 A

 

    1. How much energy is dissipated in this rod in one second at 25o Celsius?

 

P = I2 R

 

P = (0.769 A)2 (7.80 W) = 4.61 W

 

E = 4.61 J

 

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)

 

+e2 - R5 I1 – R1 I1 +e1 – R4 I2 = 0

+e3 - R2 I3 – R4 I2 - e4 – R3 I3 = 0

I1 + I3 = I2