Test 4 Aut. 2001, Monday Lecture-Click here for the Tues-lec  test.

1. (36 points) A voltage of exactly V = 4000 Volts is aplied across the cylindrical plates of a capacitor of length L = 0.0200 m = 2.00 cm. Note: Inner radius =a = 0.00060 m = 0.0600 cm and outer radius= b = 0.00120 m = 0.12 cm. Note that the scale used in the diagram may not reflect these dimensions.

(a) (20 points) What is the charge per unit length (in C/m) on the inner cylindrical shell of radius a ?
(b) (10 points) What is the magnitude E of the electric field at a distance r = 0.00100 m from the central axis of the capacitor?

(c) (6 points) Suppose the energy in the capacitor is used to heat up a very small piece of metal of specific heat Cx = 128.0 J/kg 0C and mass
6.00 x10-10 kg. If the initial temperature of the metal is exactly 200 C, then what is the final temperature ? Assume heating only.

Solutions:
(a) Q/4000  =   L/[2k ln(b/a)]
solve for Q/L = 3.21x10-7  C/m
(b) E =  (Q/L)/(2kr)  =  5770000 N/C
(c) CV2/2   =   mC(Tf  -   Ti) so that Tf = 187  0C
This prob. is tight !

 

 2. (30 points) A cylindrical resistor of outer diameter d = 0.000100 m = 0.0100 cm has a hole that runs along the axis. The cross-sectional shape of the hole is a square circumscribed by the outer circular border of the cylinder. The resistor has a length L = 0.300 m. See the diagram below. The resistivity of the material is 1.7x10-8 ohm-m. Suppose a battery of exactly V= 110 Volts is connected to the ends of the resistor. Suppose that the resistor is then submerged in a vat of water of mass m = 1.00 kg. How long does it take the resistor to heat up the water from exactly 10 oC to exactly 90 oC ? Cw = 4186 J/oC kg

 

  

Need this: AREA= CIRCLE AREA - SQUARE AREA =

R= (1.7x10-8) L/AREA  ohms
V2/R = mC (80)/time  so that time = 49.5 seconds !  

3. Extra Credit. (8 points) The battery voltage = 10 Volts. Assume:

How long ( in seconds) does it take resistor R2 to generate 1000 Joules of heat ?

Solutions:   1/Rp = 1/2  +  1/1  +  1/4  =   7/4   
so that Rp =  4/7   ohms. Thus, Rf = 1 + 1 + 4/7 = 2.57 ohms so that i = 10/2.57  =  3.89  A.
i2   =   iRp/R2   =   (3.89)(0.57)/2   = 1.11 A   so that
 P2 = (1.11)2 R2 = (1.11)2 2   = 1000/time   so that
time = 405  seconds. Word is bond, only ONE student got this right. What's up with that ?