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 State University
Department of Natural Sciences
September 18, 2006
Physics 1112 – Quiz 3
Name _SOLUTION____________________________________
1. A parallel-plate capacitor has square plates of side 10.0 cm separated by 1.00 mm.
a. Calculate the capacitance of the device.
C = 0A/d
C = 8.85 x 10-12 C2/(N-m2)(0.100 m)2/(0.001 m) = 8.85 x 10-11 F = 88.5 pF
b. If the capacitor is charged by 12.0-V battery, how much charge is transferred
from one plate to the other?
C = Q/V
Q = V C = (12.0 V)(88.5 pF) = 1062 pC = 1.06 nC
2. A 2.00 F and 4.00 F capacitors are connected in parallel across an 18.0-V
battery. Find
a. The equivalent capacitance.
Ceq = C1 + C2 = (2.00 F) + (4.00 F) = 6.00 F
b. The charge on the capacitors and the potential difference across each.
V1 =V2 =V = 18.0 V
Q1 = V1 C1
Q1 = (18.0 V)( 2.00 F) = 36.0 C
Q2 = V2 C2
Q2 = (18.0 V)( 4.00 F) = 72.0 C
Clayton State University
Department of Natural Sciences
January 30, 2007
Physics 1112 – Quiz 3
Name ___SOLUTION__________________________________
5
1. An electric field of 8.50 10 V m is desired between two parallel plates, each of
2
area 35.0 cm and separated by 2.45 mm of air.
a. What potential difference should exist between the plates?
V = E d = (8.50 x 105 V/m)(0.00245 m) = 2083 V
b. What is the capacitance of the capacitor?
C = 0 A/d = (8.85 x 10-12 C2/N-m2)(35.0 x 10-4 m2)/(0.00245 m) = 12.6 x 10-12 F = 12.6
pF
c. What charge must be on each plate?
C = Q/V
Q = C V = (12.6 pF)(2083 V) = 26246 pC = 26.2 nC
Clayton State University
Department of Natural Sciences
January 31, 2008
Physics 1112 – Quiz 3
Name ___SOLUTION__________________________________
The plates of a parallel-plate capacitor in vacuum are 5.00 mm apart and 2.00 m2 in
area. A potential difference of 10,000 V is applied across the capacitor.
a. What is the capacitance of the capacitor?
C = 0A/d
C = 8.85 x 10-12 C2/(N-m2)(2.00 m2)/(0.00500 m) = 3.54 x 10-9 F = 3.54 nF
b. What is the charge on each plate?
C = Q/V
Q = V C = (10,000 V)(3.54 x 10-9 F) = 3.54 x 10-5 C = 35.4 C
c. What is the magnitude of the electric field between the plates?
V = E d

V/d = (10,000 V)/(0.00500 m) = 2.00 x 106 V/m
Clayton State University
Department of Natural Sciences
June 12, 2006
Physics 1112 – Quiz 3
Name ____SOLUTION_________________________________
1. The two plates of a capacitor hold 2500 C and 2500 C of charge, respectively,
when the potential difference is 850 V. What is the capacitance?
C = Q/V
C = (2500 x 10-6 C)/(850 V) = 2.94 x 10-6 F
2.
8
A 9500-pF capacitor holds plus and minus charges of 16.5 10 C. What is the
voltage across the capacitor?
C = Q/V
V = Q / C
V = (16.5 x 10-8 C)/( 9500 x 10-12 C) = 17.4 V
3.
How much charge flows from each terminal of a 12.0-V battery when it is connected
to a 7.00-F capacitor?
C = Q/V
Q = C V
Q = (7.00 x 10-6 F)(12.0 V) = 84.0 x 10-6 C
4.
A 0.20-F capacitor is desired. What area must the plates have if they are to be
separated by a 2.2-mm air gap?
C = 0 A/d
A = d C/ 0 = (2.20 x 10-3 m) (0.200 F) / (.00 x 10-6 F) =
Clayton State University
Department of Natural Sciences
June 10, 2008
Physics 1112 – Quiz 4
Name ___SOLUTION__________________________________
One of several 100 F capacitors in defibrillator has a potential difference of 220 V
between its plates.
a. What is the magnitude of charge on each plate of the capacitor?
Q = C V = (100 x 10-6 F)(220 V) = 2.20 x 10-2 C = 0.0220 C
b. What is the energy stored in the capacitor?
Est = ½ C (V)2 = ½ (100 x 10-6 F) (220 V)2 = 2.42 J
c. If this energy were fully utilized to lift a 100 g doughnut with no change in its
kinetic energy, how high could the doughnut be raised?
Est = m g y
y = Est /(m g)
y = (4.84 J)/[(0.100 kg)(9.81 m/s2 g)] = 2.47 m
Clayton State University
Department of Natural Sciences
September 15, 2010
Physics 1112 – Quiz 4
Name __SOLUTION___________________________________
1. How much charge is on each plate of a 4.00 – F capacitor when it is connected
to a 12.0 – V battery?
C
Q
V
Q  CV  (4.00 F )(12.0 V )  48.0 C
2. Two conductors having net charges of +10.0 C and -10.0 C have a potential
difference of 10.0 V between them.
a. Determine the capacitance of the system.
C
Q 10.0 C

 1.00 F
V
10.0 V
b. What is the potential difference between the two conductors if the charges on
each are increased to +100 C and -100 C?
C
Q
V
V 
Q 100 F

 100 V
C 1.00 F