TIME OF COMPLETION_______________            NAME_____SOLUTION________________________

 

 

                                           DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1112, Exam 1                                                                                                             Section 1

Version 1                                                                                                              September 25, 2002

Total Weight: 100 points

 

1.         Check your examination for completeness prior to starting.  There are a total of ten (10) problems on nine (9) pages.

 

2.         Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet  (provided by your instructor).

 

3.         You will have 80 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:                10:30 a.m.

Stop:                11:50 a.m

 

 

 

               PROBLEM

 

                 POINTS

 

                 CREDIT

 

                     1-6

 

                      30

 

 

 

                       7

 

                      20

 

 

 

8

 

                      15

 

 

 

                       9

 

                      20

 

 

 

                      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. A repelling force must occur between two charged objects under which conditions?

 

    1. Charges are of unlike signs.

 

    1. Charges are of like signs.

 

    1. Charges are of equal magnitude.

 

    1. Charges are of unlike magnitude.

 

 

  1. At what point is the charge-per-unit area greatest on the surface of irregularly shaped conducting solid in equilibrium?

 

    1. Where surface curves inward.

 

    1. Where surface is flat.

 

    1. Where surface has a sharp point.

 

    1. Where surface curves outward.

 

 

  1. Increasing the separation of the two charged parallel plates of a capacitor which are disconnected from a battery will produce what effect on the capacitor?

 

    1. Increase charge.

 

    1. Decrease charge.

 

    1. Increase capacitance.

 

    1. Decrease capacitance.

 

 

 

 

 

 

 

  1. If two parallel, conducting plates have equal positive charge, the electric field lines will:

 

    1. Leave one plate and go straight to the other plate.

 

    1. Leave both plates and go to infinity.

 

    1. Enter both plates from infinity.

 

    1. None of the above.

 

 

  1. When a superconductor’s temperature drops below the critical temperature, its resistance:

 

    1. Equals that of a semiconductor of equal dimensions.

 

    1.  Increases by two.

 

    1. Drops to zero.

 

    1. Reduces to one half.

 

 

 

  1. Three resistors, each with resistance R1 , are in series in a circuit. They are replaced by one equivalent resistor R. Comparing this resistor to the first resistor of the original circuit:

 

    1. The current through R equals the current through R1.

 

    1. The voltage across R equals the voltage across R1.

 

    1. The power given off by R equals the power given off by R1.

 

    1. R is smaller than R1.

 

 

 

 

 

 

 

  1. Three point charges Q1, Q2, and  Q3 are located as shown below:

 

 

If the charges are Q1 = +6.00 mC, Q2 = -2.00 mC, and Q3 = -4.00 mC , find the electric field at the origin. Provide both the magnitude and direction of the electric field.

 

       E1 = ke Q1 /r12 = (8.99 x 109 N m2 /C2) (6.00 x 10-6 C)/ (0.200 m )2 = 1349 x 103 N/C

 

       E2 = ke Q2 /r22 = (8.99 x 109 N m2 /C2) (2.00 x 10-6 C)/ (0.150 m )2 = 799 x 103 N/C

 

       E3 = ke Q3 /r32 = (8.99 x 109 N m2 /C2) (4.00 x 10-6 C)/ (0.250 m )2 = 575 x 103 N/C

 

      E1x= 1349 x 103 N/C

 

      E1y= 0 N/C

 

      E2x= 799 x 103 N/C

 

      E2y= 0 N/C

 

      E3x= 0 N/C

 

      E3y= - 575 x 103 N/C

 

      Etotx= 2148 x 103 N/C                                                Etot = 2224 x 103 N/C                                               

 

      Etoty= - 575 N/C                                                         q = tan-1 (Etoty / Etotx) = -15.0 degrees

 

 

 

  1. Consider the network of capacitors shown below. Each of the capacitors is 2.00 mF.

    1. Calculate the equivalent capacitance of the network

 

 

(15)        

                     Capacitors C2 and C3 are in series:

 

1/C23 = 1/C2 + 1/C3 = 1/(2.00 mF) + 1/(2.00 mF) = 1/(1.00 mF)

C23 = 1.00 mF

 

Capacitors C4 and C23 are in parallel:

 

C234 = C4 + C23 = 1.00 mF + 2.00 mF = 3.00 mF

 

Capacitors C1, C234 , and  C5  are in series:

 

1/C12345 = 1/C1 + 1/C234  + 1/C5 = 1/(2.00 mF) + 1/(3.00 mF) + 1/(2.00 mF)

              = 8/(6.00 mF)

 

C12345 = 0.750 mF

 

 

    1. The battery provides 12.0 V of the potential difference. What is the charge stored in the network?

 

 

C = Q/(V)             Q = C V        Q = (0.750 x 10-6 F) (12.0 V) = 9.00 x 10-6 C

 

 

    1. How many electrons were moved by the battery in order to build that charge in the network?

 

# electrons = Q/e

 

                  # electrons =  9.00 x 10-6 C/ 1.60 x 10-19 C =  5.625 x 1013

 

 

 

 

 

  1. Four resistors are connected as shown below:

 

 

              

                       The battery provides e = 24.0 V.

 

    1. Calculate the equivalent resistance for the circuit.

 

1/R23 = 1/R2 + 1/R3 = 1/(5.00 W) + 1/(3.00 W) = 8/(15.0W)

R23 = 15/8 W

 

R12345 = R1 + R23  + R4 = 2.00 W + 4.00 W + 15/8 W  

              = 8/(6.00 mF)

 

R12345 = 7.88 W

 

 

(20)

 

 

 

 

 

    1. Calculate the current passing through the 5.00 W resistor.

 

Itot = V/R12345                           Itot = (24.0 V)/(7.88 W) = 3.05 A

 

I2 = V2 /I2 = (5.70 V)/ (5.00 W) = 1.14 A   (see c)

 

 

    1. Calculate potential difference across the 5.00 W  resistor.

 

V1 = R1 Itot = (2.00 W) (3.05 A) = 6.10 V

 

V4 = R4 Itot = (4.00 W) (3.05 A) = 12.2 V

 

V2 = V3 = e - V1 - V4 = 5.70 V

 

 

    1. Calculate the power loss in the 5.00 W resistor.

 

 

P = I2 V2 = (1.14 A) (5.70 V)  = 6.50 W

 

 

 

 

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.

 

 

+e1 - R3 I1 + R4 I2 +e2 – R2 I1 = 0

-e2 - R4 I2 – R5 I3 - e3 – R1 I3 = 0

I1 + I2 = I3