This exam is based on quizzes given in Spring 98

Solutions to Real Test 2
1. (20 points)

A 1.0-kg piece of metal is taken from a furnace at a temperature of 9000C. It is dropped into 4.0 kg of water at 10 0C. Assume no heat is lost to the surroundings. The specific heat of the metal is 450 J/kg0C. The specific heat of water is 4186 J/kg0C.
 

What is the total change in entropy of the system?

1. Solution Outline: See the solution to problem #4 of real test 1 . Same problem. Different numbers. 
2. (20 points) BALANCING GRAVITY AND ELECTRIC FIELDS. 4B SP12 REFERENCE: QUIZ 1, #82, 33, 35; QUIZ 2, #30-UNIFORM ELECTRIC FIELD PERPENDICULAR TO INFINITE PLANE; 

In the figure below, a small, nonconducting ball of mass m = 1.0x10-3 g and positive charge q = 2.0x10-8 C hangs from an insulating thread that makes an angle = 300 with a vertical, uniformly charged nonconducting sheet. Assume the sheet is infinite. It extends far in all directions within its plane. What is the surface charge density (in C/m2) of the sheet?
2. Solution Outline: See example 23.4
horizontal tension=electric force:
Tsin30 = qE  
vertical tension = vertical weight:
Tcos30 = mg

Divide these two equations to get:
tan 30 = qE/mg. Solve for E = mgtan30/q. Now, we know that  

Solve for the density.

3. ( 20 points) 4B SP12 REFERENCE: SOLID INSULATING SPHERE QUIZ 2, #23, #54

A solid sphere of radius R carries a volume charge density given by a function of r:

for


for


What is the magnitude of the electric field as a function of r for:
(a) (8 points)

(b) (8 points)

(c) (4 points) What is the total charge of the solid sphere?

3.Solution: This is a challenging problem. Before you attempt it, you should understand completely how to do example 24.5 in the textbook. I could put a problem like that example on the real test next Wednesday. Example 24.5 is for a constant charge density. The sample test problem is when the charge density is not constant. I will probably not put a problem with non-constant charge density on  the test. Still, I will outline the solution for your information:

For r < R 

For r > R, the system behaves like  a point charge, so E will be proportional to r-2

 
4. ( 12 points) EXTRA CREDIT 4B SP12 REFERENCE: POTENTIAL OF TWO CONCENTRIC SHELLS: QUIZ 3, #47

Two thin, conducting, spherical shells are shown below. The inner shell has a radius a = 15 cm and a charge 10 nC. The outer shell has a radius b = 40 cm and a charge of -5 nC. Note: 1 nC = 1x10-9C

Find the potential as a function of r in the regions

(a)( 4 points) b < r

(b)( 4 points) a < r < b

(c)( 4 points) r < a
Solution: Please review page 774 of the textbook. This may help you on the real test and on test #3 !
4.(a) Outside for b < r,   the system looks like a point charge. The charge enclosed by a Gaussian sphere of radius r > b is 

10nC - 5nC = 5nc = 5x10-9C  = Qout

Thus,  E = kQout/r2  and so V =  kQout/r

(b) You must integrate the electric field in part (a) from infinity to r = 40 cm = 0.40 m.  This yields kQout/0.40. Then you must continue the integration of E from r = 
40 cm = 0.40 m to a value of r where 
a < r < b. Note that E for a < r < b has a different value of charge than the E in part (a). Now Q enclosed is Qin = 10 nC = 10x10-9 C. Thus the integral is
kQin(1/r - 1/0.40). For this integral, see page 774 of the textbook! Thus, the total potential is
kQout/0.40 + kQin(1/r - 1/0.40)
(c) E = 0 for r < a. Thus, V = constant = value on the surface where r = 0.15 m. Thus,
kQout/0.40 + kQin(1/0.15 - 1/0.40)for r < a.