PHYSICS DEPARTMENT

PHY 2054                                                                                  Exam 1                                       1 June 2010

Prof. F.E. Dunnam

 

Exam 1 Solutions

Answer (1) is the correct answer for all questions except  #10 where correct answer is (4).  Most  solutions are sketched: not all are worked out in detail. Please see one of us if you have difficulties.

[NOTE: My html editor refuses to acknowledge the existence of the Greek 'delta' so  ' *V' means 'delta V'.]


  

 

1. Two particles have charges +Q and -Q (equal magnitude and opposite sign) and lie along a straight line. For a net force of zero to be exerted on a third charge it must be placed:

                  (1) at none of the places listed (thereÕs no such location).

                  (2) on the perpendicular bisector of the line joining Q and -Q, but not on that line itself.

                  (3) on the line joining Q and -Q, to the side of Q opposite -Q.

                  (4) on the line joining Q and -Q, to the side of -Q opposite Q.

                  (5) midway between Q and -Q.

                   

Try it....there is no location where you can place a test charge (or any charge!) near this configuration such that the resultant force is zero!

 

2. A particle with charge 2µC charge is placed at the origin. An identical particle, with the same charge, is placed 2 m from the origin on the x axis, and a third identical particle, with the same charge, is placed 2 m from the origin on the y axis. The magnitude of the force on the particle at the origin is:

            (1) 1.3 x 10-2N   (2) 6.4 x 10-3N     (3) 9.0 x 10-3N     (4) 1.8 x 10-2N     (5) 3.6 x 10-2

Sketch the force diagram. The resultant of the two (repulsive) forces on the charge at the origin is = 2 F cos 45 deg.  

Since F = k q2/r2  ==>  net force =  (2 k q /r2) cos 45 deg =  1.27 e^-2 N

 

3. A 200-N/C electric feld is in the positive x direction. The force on an electron in this feld is:

                  (1) 3.2 x 10-17 N, in the negative x direction

                  (2) 200 N in the positive x direction

                  (3) 3.2 x 10-17N, in the positive x direction

                  (4) 200 N in the negative x direction

                  (5) zero

 

. F = eE and the force is in the opposite direction of E since e is negative.

 

 

4. A particle with a charge of 5.5 x 10-8C is fixed at the origin. A particle with a charge of −2.3 x 10-8C is moved from x = 3.5cm on the x axis to y =3.5cm on the y axis. The change in the potential energy of the two-charge system is:

(1) zero     (2) −3.2 x 10-4J     (3) 9.3 x 10-3J     (4) −9.3 x 10-3J     (5) 3.2 x 10-4J

 

The moved charge q remains the same distance from the fixed charge Q.  Since V (Q) is the same at both locations, no work is done and thus no change in PE.          [Remember, the E-field is conservative!]

 

5. During a lightning discharge, 30 C of charge moves through a potential difference of 1.0 x 108V in 2.0 x 10-2s. The energy released by this lightning bolt is:

(1) 3.0 x 109J     (2) 1.5 x 1011J     (3) 6.0 x 107J     (4) 3.3 x 106J     (5) 1500 J

W = Q *V=  (30C x 108 V ) = 3 x 109 J  [remember, V = J/C]

6. A conducting sphere with radius R is charged until the magnitude of the electric field just outside its surface is E. The electric potential of the sphere, relative to the potential far away, is:

(1) ER     (2) E/R     (3) E/R2     (4) zero     (5) ER2

. E = k Q/R2  and V = k Q/R,  hence V = ER

 

7. A metal sphere of radius 20 cm carries a charge of 5 x 10-9C and is at a potential of 400 V, relative to the potential far away. The potential 10 cm from the center of the sphere is:

(1) 400 V     (2) -400V     (3) 2 x 10-6V     (4) 0     (5) none of these

 Metal ==> conducting;   for a conducting sphere, V inside is the same as at the surface,  
So inside at 10 cm, V = 400V, as given.

8. The work required to carry a particle with a charge of 6.0-C from a 5.0-V equipotential surface to a 6.0-V equipotential surface and back again to the 5.0-V surface is:

(1) zero     (2) 1.2 x 10-5J     (3) 3.0 x 10-5J     (4) 6.0 x 10-5J     (5) 6.0 x 10-6J

 

 Moving on a 'round trip' between two equipotential surfaces involves no net change in potential energy so W = 0

 

9. The diagram shows six 6-µF capacitors. The capacitance between points a       
and b is:

(1) 4µF     (2) 6µF     (3) 3µF     (4) 9µF     (5) 1µF

  For  three series capacitors,  1/Ceq  = 1/6 + 1/6 + 1/6 = 3/6, hence Ceq  = 6/3 = 2µF .

The two sets of series capacitors are connected in parallel, so Cab = 2 Ceq=  4µF

 

10. A certain wire has resistance R. Another wire, of the same material, has half the length and half the diameter of the Þrst wire. The resistance of the second wire is:

    (1) R/4     (2) R/2     (3) R     (4) 2R    (5) 4R

R is proportional to the conductors' (length/area).  

For the second wire, ½ length ==> R' = R/2. But  ½ diameter reduces area by 4 and increases R by 4, 

      so  net effect is R(4/2) = 2R

 

11. A small plastic sphere has a conductive coating of aluminum paint.           If its mass is 10 grams and it has a charge of −

50 microC, what electric Þeld (in V/m) will exactly balance its weight?

   (1) 1960, downward   (2) 1960, upward   (3) 5.1 x 10-4 upward   (4) 5.1 x 10-4 downward   (5) insufficient data

For balanced forces,  gravitational force = electrostatic force  ==> mg = qE;

Thus E = mg/q = 1960 N/C.  q is negative so for the E-force to be upward against gravity, E must be downward.

 

 

12. An electron with a speed of 2 x 106 m/s moves into a uniform electric field of 500 N/C that is parallel to the electron's motion. How long does it take, in nanoseconds, for the electron to come to rest?

(1) 23     (2) 35     (3) 11.4     (4) 2300     (5) never

'Parallel' = in the direction of E, so the electron is slowing down.  

Impulse = momentum  ==> Ft = mv  ==> t = mv/F = mv/eV = 2.27e^-8 s  = 23 ns

 

13. The electron in the above question initially possessed what kinetic energy, in electron volts?

    (1) 11.4     (2) 35      (3) 23      (4) 2300      (5) 3200

K = ½ mv2 = 3.64 e^-18 J

3.64 e^-18 J /1.6 e^-19 J/eV = 11.37 eV

 

14. Four point charges are placed on the rim of a circle of radius 10 cm. The charges, in microC, are +0.5, +1.5, −1.0, and -0.5. If electrical potential at the center of the circle due to the the +0.5 charge alone is 45 kilovolts, what is the total potntial in kilovolts at this point?

     (1) 45    (2) 18     (3) -45     (4) -18      (5) zero

The net charge on the rim = (+0.5 + 1.5  - 0.5 - 1.0) = +0.5µC,

            so the potential is just that of the +0.5µC charge = 45kV.

 

15. What is the potential energy in Joules of a two-charge system consisting of a 3.4 microC charge and a 6.6 microC charge, 30 cm apart?

(1) 0.67      (2) 0.34      (3) -0.34      (4) -0.75      (5) 3.2

The PE of a system of charges q1 and q2, distance d apart, is just  [V(q1)] x q2

            =(kq1/d) x q2  =  0.673 J

 

16. A parallel-plate capacitor of capacitance C has charge Q placed on it, at potential V.   If the plates are carefully moved twice as far apart,

                  (1) the potential becomes 2V

                  (2) the potential becomes V/2

                  (3) the charge becomes 2Q

                  (4) the charge becomes Q/2

                  (5) the capacitance becomes 2C

                   

C ~ A/d, so if plate separation d is doubled, C' = C/2

But C = Q/*V . The system is isolated,  so Q is constant.

Therefore *V' = Q/C' = Q/(C/2) = 2Q/C =  2 *V

 

 

17. The capacitor in the above question [same initial conditions] has a slab of dielectric (κ = 2) carefully inserted between the plates without touching them. Now

                  (1) the potential becomes V/2

                  (2) the potential becomes 2V

                  (3) the charge becomes 2Q

                  (4) the charge becomes Q/2

                  (5) the capacitance becomes C/2

 

. Insertion of the dielectric results in C' = κC = 2C .

            As in #16, above, *V' = Q/C' = Q/2C = ½ Q/C =*V/2

 

 

18. A resistor of resistance R is connected to a battery (voltage V) and dissipates power P . Replacing the resistor with one of resistance 2R will result in what power dissipated?

(1) P/2     (2) 2P      (3) P/4      (4) 4P      (5) none of these

P = *V2/R ;   P' =*V2/2R = ½ *V2/R = P/2

 

19. An electric car runs off a bank of 12-V batteries with total stored energy of 30 megaJ at maximum charge. If the motor draws 6 kW when propelling the car at a steady 10 m/s, how far can the car travel, in km?

(1) 50      (2) 25      (3) 100      (4) 150     (5) 75

Total energy W available from batteries = t x (power) = tP    ==> t = W/P  

So   30 e^6 J/(6 e^3 J/s) = 5 e^3 s .   Distance = speed x time =  10 m/s x 5 e^3 s = 50 km

 

20. A standard incandescent light bulb is labeled "100 W, 120V". Its resistance in ohms while burning, and its approximate resistance when "cold" are likely to be, respectively,

(1) 144, less      (2) 144, greater      (3) 240, less      (4) 120, greater      (5) 12, less
During operation, P = *V2/R ==> R = *V2/P = (120V) 2 / 100W = 144 ohms . This is at incandescence --- a high temperature.   Resistance when cold will be appreciably lower.