Clayton College & State University
Department of Natural Sciences
September 2, 2004
Physics 1112 – Quiz 1
Name ______SOLUTION_______________________________
1. Three non-conducting balls supported by insulating threads hang from a support. We
know that ball X is positively charged. When ball X is brought near balls Y and Z
without touching them, it attracts Y and repels Z. We can conclude that
a. Y has a negative charge.
b. Z has a negative charge.
c. Y is neutral.
d. Z is neutral.
2. Doug rubs a piece of fur on a hard rubber rod, giving the rod a negative charge. What
happens?
a. Protons are removed from the rod.
b. Electrons are added to the rod.
c. The fur is also charged negatively.
d. The fur is left neutral.
3. A metal sphere is grounded and a positively charged balloon is brought near it. The
switch is opened and the balloon is taken away. The sphere is now
a. Neutral.
b. Negatively charged.
c. Positively charged.
d. Charged, but we cannot know its polarity.
Clayton College & State University
Department of Natural Sciences
September 15, 2004
Physics 1112 – Quiz 3
Name _______SOLUTION______________________________
A capacitor is made of two flat parallel plates 1.50 mm apart. The magnitude of charge
on each of the plates is 0.0180 C with the potential difference of 200 V.
a. What is the capacitance?
C = Q / V
C = (0.0180 x 10 -6 C) / (200 V) = 0.900 x 10 -10 F
b. What is the area of each plate?
C = o A / d
A = C d / o
A = (0.900 x 10 -10 F) (1.50 x 10 -3 m) / ( 8.854 x 10 -12 C/N m2) = 0.0152 m
c. What is the total energy stored ?
EST = ½ Q V
EST = ½ (0.0180 x 10 -6 C) (200 V) = 1.80 x 10 -6 J
Clayton College & State University
Department of Natural Sciences
September 22, 2004
Physics 1112 – Quiz 4
Name _____SOLUTION________________________________
You apply a potential difference of 4.50 V between the ends of a wire that is 2.50 m in
length and 0.654 mm in radius. The resulting current through the wire is 17.6 A.
a. What is the resistance of the wire?
V = I R
R = V  I
R = (4.50 V) / (17.6 A) = 0.256 
b. What is the resistivity of the wire’s material?
R  L /A
 = R AL = R ( r2) / L
 = (0.256 ) ( (0.654 x 10-3 m)2) / (2.50 m) = 0.138 x 10-6 -m
The temperature increases from 20o C to 0.0039 (o C) -1Given  = 0.0039 (o C) -1, what is
the new resistance of the wire? R = RO [ 1 + (T - TO)]
R = R0 [1 + (T – T0)]
R = (0.256 ) [1 + 0.0039 (o C) -1(30.0o C - 20.0o C)] = 0.266
Clayton College & State University
Department of Natural Sciences
October 13, 2004
Physics 1112 – Quiz 5
Name _____SOLUTION________________________________
1. Find the direction of the magnetic force acting on the positively charged particle
for given orientations of particle’s velocity and constant magnetic field.
2. A proton is released such that its initial velocity is from right to left across this
page. The proton’s path, however, is deflected in a direction toward the bottom
edge of the page due to the presence of a uniform magnetic field. What is the
direction of this field?
a. Out of the page.
b. Into the page.
c. From bottom edge to top edge of the page.
d. From right to left across the page.
Clayton College & State University
Department of Natural Sciences
October 25, 2004
Physics 1112 – Quiz 6
Name ________SOLUTION___________________________
1. In the Figure below, let B = 0.600 T, V = 8.00 m/s, L = 15.0 cm, and R = 25.0 ,
and assume that the resistance of the rods and rails is negligible. Find
a. The induced emf in the circuit.

 = BVL = (0.600 T)(8.00 m/s)(0.150 m) = 0.720 V
b. The current in the circuit.
I = / R = (0.720 V) / (25.0  0.0288 A
c. The direction of the current in the circuit.
Clockwise.
d. The force needed to move the rod with constant velocity.
F = I B L = (0.0288 A)(0.600 T)(0.150 m) = 0.00259 N
Clayton College & State University
Department of Natural Sciences
November 10, 2004
Physics 1112 – Quiz 7
Name _____SOLUTION______________________________
A 500-Hz ac generator with a rms voltage output of 100 V is connected in series to a
24.0- resistor, a 10.0-F capacitor, and a 50.0-mH inductor.
a. What is the impedance of the circuit?
XL = 2f L = 2(500.0 Hz) (50.0 x 10-3 H) = 157 
XC = 1/(2f C) = 1/(2(500.0 Hz) (10 x 10-6 F)) = 31.8 
Z = (R2 + (XL – XC)2)1/2 = ((24.0 )2 + (157  – 31.8 )2)1/2 = 127 
b. Find the rms current flowing through the resistor.
IRMS = VRMS/Z
IRMS = (100 V) / (127 ) = 0.787 A
c. Find the rms voltage across the inductor.
VR, RMS= IRMS R
VR, RMS= (0.787 A) (24.0 ) = 18.9 V
d. Does the voltage lead the current or the current lead the voltage?
Explain.
 tan-1 ((XL – XC)/R)
 tan-1 ((157  – 31.8 )/ (24.0 )) = 51.4o
 >0, hence voltage leads current.
Clayton College & State University
Department of Natural Sciences
November 17, 2004
Physics 1112 – Quiz 8
Name ______SOLUTION_____________________________
1. A ray of light is incident on a flat surface of a block of crown glass that is
surrounded by water. The angle of refraction is 19.6o. Find the angle of reflection.
The index of refraction of water is 1.333, the index of refraction of crown glass is
1.52.
n1 sin(1) = n2 sin(2)
sin(1) = n2 sin(2)/ n1
sin(1) = (1.52) sin(19.6o)/(1.333) = 0.383
1 = 22.5o
2. Find the speed of light in water.
n = c/v
v = c/n = (3.00 x 108 m/s )/ 1.333 = 2.25 x 108 m/s
Clayton College & State University
Department of Natural Sciences
November 22, 2004
Physics 1112 – Quiz 9
Name _______SOLUTION____________________________
A 15.0-cm tall candle stands 120 cm from the vertex of a concave mirror having a radius
of curvature of 60.0 cm.
a. How far is the image from the mirror?
p = 120 cm
f = R/2 = 30.0 cm
1/p + 1/q = 1/f
1/q = 1/f – 1/p = 1/(30.0 cm) – 1/(120 cm) = 3/(120 cm)
q = 40.0 cm
b. What is the height of the candle’s image?
M = -q/p = - (40.0 cm)/(120 cm) = -1/3
M = h’/h
h’ = M h = (-1/3)(15.0 cm) = -5.00 cm
(Continued on the next page.)
c. Draw a ray diagram to check your answers.
d. Is the image real or virtual, upright or inverted, enlarged or reduced?
Real, inverted, and reduced
Clayton College & State University
Department of Natural Sciences
December 1, 2004
Physics 1112 – Quiz 10
Name ____SOLUTION_______________________________
A lens forms an image of an object. The object is 16.0 cm from the lens. The image is
12.0 cm from the lens on the same side as the object.
a. What is the focal length of the lens?
p = 16.0 cm
q = -12.0 cm
1/f = 1/p + 1/q
f = -48.0 cm
b. Is the lens converging or diverging?
Diverging (f < 0)
c. If the object is 8.50 mm tall, how tall is the image?
M = -q/p = -(-12.0 cm)/(16.0 cm) = 0.750
h’ = M h = 0.750 (8.50 mm) = 6.375 mm
d. Is the image upright or inverted?
Upright (M > 0)
e. Draw a ray diagram.