2/17/2014
HW 4
Due: 11:00pm on Sunday, February 16, 2014
HW 4
^
You will receive no credit for items you complete after the assignment is due. Grading Policy
Introduction to Capacitance
Learning Goal:
To understand the meaning of capacitance and ways of calculating capacitance
When a positive charge q is placed on a conductor that is insulated from ground, an electric field emanates from the
conductor to ground, and the conductor will have a nonzero potential V relative to ground. If more charge is placed on the
conductor, this voltage will increase proportionately. The ratio of charge to voltage is called the capacitance C of this
conductor: C = q/V.
Capacitance is one of the central concepts in electrostatics, and specially constructed devices called capacitors are
essential elements of electronic circuits. In a capacitor, a second conducting surface is placed near the first (they are often
called electrodes), and the relevant voltage is the voltage between these two electrodes.
This tutorial is designed to help you understand capacitance by assisting you in calculating the capacitance of a parallel-plate
capacitor, which consists of two plates each of area A separated by a small distance d with air or vacuum in between. In
figuring out the capacitance of this configuration of conductors, it is important to keep in mind that the voltage difference is the
line integral of the electric field between the plates.
Part A
What property of objects is best measured by their capacitance?
ANSWER:
O ability to conduct electric current
O ability to distort an external electrostatic field
Xability to store charge
Part B
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E
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HW4
Assume that charge -q is placed on the top plate, and +g is placed on the bottom plate. What is the magnitude of the
electric field E between the plates?
Express E in terms of q and other quantities given in the introduction, in addition to Eŋ and any other
constants needed.
You did not open hints for this part.
ANSWER:
E
AE
Part C
What is the voltage V between the plates of the capacitor?
Express V in terms of the quantities given in the introduction and any required physical constants.
You did not open hints for this part.
ANSWER:
V =
&d
AE
Part D
Now find the capacitance C of the parallel-plate capacitor.
Express C in terms of quantities given in the introduction and constants like €0.
ANSWER:
C=
E. A
ठ
Part E
Consider an air-filled charged capacitor. How can its capacitance be increased?
You did not open hints for this part.
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ANSWER:
Increase the charge on the capacitor.
O Decrease the charge on the capacitor.
HW 4
Increase the spacing between the plates of the capacitor.
Decrease the spacing between the plates of the capacitor.
O Increase the length of the wires leading to the capacitor plates.
Part F
Consider a charged parallel-plate capacitor. Which combination of changes would quadruple its capacitance?
ANSWER:
�� Double the charge and double the plate area.
Double the charge and double the plate separation.
O Halve the charge and double the plate separation.
O Halve the charge and double the plate area.
Halve the plate separation; double the plate area.
�� Double the plate separation; halve the plate area.
Capacitors in Parallel
Learning Goal:
To understand how to calculate capacitance, voltage, and charge for a paralle! combination of capacitors.
Frequently, several capacitors are connected together to form a collection of capacitors. We may be interested in determining
the overall capacitance of such a collection. The simplest configuration to analyze involves capacitors connected in series or
in parallel. More complicated setups can often (though not always!) be treated by combining the rules for these two cases.
Consider the example of a parallel combination of capacitors: Three capacitors are connected to each other and to a battery
as shown in the figure. The individual capacitances are C, 2C, and
3C, and the battery's voltage is V.
C
,
20
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༣།། ༣
3C
ད། 6
V
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Part A
HW 4
If the potential of plate 1 is V, then, in equilibrium, what are the potentials of plates 3 and 6? Assume that the negative
terminal of the battery is at zero potential.
You did not open hints for this part.
ANSWER:
OV and V
O 2V and 3V
V and 0
V
��
/ and /
V
3
Part B
If the charge of the first capacitor (the one with capacitance C) is Q, then what are the charges of the second and third
capacitors?
You did not open hints for this part.
ANSWER:
O
2Q and 3Q
Q
0 / and /
O Q and Q
O 0 and 0
Part C
Suppose we consider the system of the three capacitors as a single "equivalent" capacitor. Given the charges of the
three individual capacitors calculated in the previous part, find the total charge Qtot for this equivalent capacitor.
Express your answer in terms of V and C.
ANSWER:
Q tot = 6Q
= 6CV
C
Part D
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HW 4
Using the value of Q tot find the equivalent capacitance Ceq for this combination of capacitors.
Express your answer in terms of C.
You did not open hints for this part.
ANSWER:
Qest
6Q
Ceg
V
6C
V
Capacitors in Series
Learning Goal:
To understand how to calculate capacitance, voltage, and charge for a combination of capacitors connected in series.
Consider the combination of capacitors shown in the figure. Three
capacitors are connected to each other in series, and then to the
battery. The values of the capacitances are C, 2C, and 3C, and
the applied voltage is AV. Initially, all of the capacitors are
completely discharged; after the battery is connected, the charge
on plate 1 is Q.
C
20
30
1
2
3 4
5 6
+Q
-Q +Q
+Q
Part A
What are the charges on plates 3 and 6?
You did not open hints for this part.
ANSWER:
����
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O +Q and +Q
O -Q and -Q
+Q and -Q
O -Q and +Q
O 0 and +Q
O 0 and -Q
HW 4
Part B
If the voltage across the first capacitor (the one with capacitance C) is AVI, then what are the voltages across the
second and third capacitors?
You did not open hints for this part.
ANSWER:
O2AV₁ and 3AV₁
AV₁ and AV1
O AV and AVI
O 0 and AV₁
C=² = V =
۵۷۱
= Q
4��₂ = Q₂ = Q
łz
=
الاله
2
0V3
Q3
Q
C3
3C
3
Part C
Find the voltage AV₁ across the first capacitor.
Express your answer in terms of AV.
You did not open hints for this part.
OV = ov₁tov₂ +O��3
=��V₁+ OV! + EVI
6+3+2 OV₁ ==/ OV₁
6
الاه
ANSWER:
AV₁ =
ov
ㅠ
Part D
Find the charge Q on the first capacitor.
Express your answer in terms of C and AV₁.
ANSWER:
Q = CV = C. &V₁
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Q= COV
HW 4
Part E
Using the value of Q just calculated, find the equivalent capacitance Ce for this combination of capacitors in series.
Express your answer in terms of C.
You did not open hints for this part.
ANSWER:
Ceq
=
Q
الان
김
C
60
��
Finding the Capacitance
A parallel-plate capacitor is filled with a dielectric whose dielectric constant is K, increasing its capacitance from C₁ to
KC₁. A second capacitor with capacitance C₂ is then connected in series with the first, reducing the net capacitance back
to C1.
Part A
What is the capacitance C₂ of the second capacitor?
KC,
6₂
Express your answer in terms of K, C₁, and constants.
�����
You did not open hints for this part.
ANSWER:
Cz
=
C
K-1
"
���֧�
W
KC,
51-
K-1
11
KC₁
K
K-1
Force between Capacitor Plates
Consider a parallel-plate capacitor with plates of area A and with separation d.
Part A
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نیا
+
You
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HW 4
Find F(V), the magnitude of the force each plate experiences due to the other plate as a function of V, the potential
drop across the capacitor.
Express your answer in terms of given quantities and to.
You did not open hints for this part.
ANSWER:
E = 25
1
Gauss law =
��
==
260
-
V = (El·d = [El d
F= QE =2€ A/6)²
F(V) =
GAV²
2d2
LESA
Go A V²
262
The Capacitor as an Energy-Storing Device
Learning Goal:
To understand that the charge stored by capacitors represents energy; to be able to calculate the stored energy and its
changes under different circumstances.
An air-filled parallel-plate capacitor has plate area A and plate separation d. The capacitor is connected to a battery that
creates a constant voltage V.
Part A
Find the energy Uo stored in the capacitor.
0
Express your answer in terms of A, d, V, and co. Remember to enter e as epsilon_0.
You did not open hints for this part.
ANSWER:
Uo= { Cv² = _GAV²
c = ८० वं
Part B
The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until the
separation reaches 3d. Find the new energy U1 of the capacitor after this process.
Express your answer in terms of A, d, V, and E.
You did not open hints for this part.
C'=
C² = to
A
3d
Q2 does not change
before pulling
ANSWER:
Q = CV
U₁ =
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=홀.
E. AV²
EAV
2 Go A
श्व
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��
2/17/2014
U1
=
लाल
GAV²
d
HW4
Part C
The capacitor is now reconnected to the battery, and the plate separation is restored to d. A dielectric plate is slowly
moved into the capacitor until the entire space between the plates is filled. Find the energy U2 of the dielectric-filled
capacitor. The capacitor remains connected to the battery. The dielectric constant is K.
Express your answer in terms of A, d, V, K, and to.
ANSWER:
U2
=
KG AV²
zd
C" = KE 1
U₂ = {CV² =
KGAV²
zd
Capacitors with Partial Dielectrics
Consider two parallel-plate capacitors identical in shape, one aligned so that the plates are horizontal, and the other with the
plates vertical.
This is like two capacitors
in para holl
(1-A
C₁ = E₂ CHA
C₂ = KE₂ +A
= T
B
Ca₂ = C₁+ Cz
EA ((1-4)+KF)
Part A
d/2
use Gauss's law
modified
A
4
شاه
1Th vacuum (upper half) =
Ej in dielectric lower half) = I
»V = − S,˜ˆÈ · dễ = − (S,' 'E'; de + Si? És de)
-
=-(IE⋅ $+ (E₁1. £)
2
KE
슬픔 (1+k) =
-
2/ (1 + 2) 228
Q
The horizontal capacitor is filled halfway with a material that has dielectric constant K. What fraction ƒ of the area of the
vertical capacitor should be filled (as shown in the figure) with the same dielectric so that the two capacitors have equal
capacitance?
Express your answer in terms of K.
You did not open hints for this part.
C =
› @₂A ((1-4)+kf)
d
ZEA
da+k)
26. A
d(1+*)
=) (1-f+kf)(1 + 1 ) = 2
1+ * + +(K-1)(1 + 2) = 2
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ANSWER:
HW 4
f(k+1-1-k) = 1-*
The
=
K-I
(K+1)(K-1)
(k+1) (K-1) - K+)
f=
1-*
K-1
*-*
k-1
Video Tutor: Discharge Speed for Series and Parallel Capacitors
First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment.
Then, close the video window and answer the question at right. You can watch the video again at any point.
LAUNCH VIDEO
Part A
The capacitors in each circuit are fully charged before the switch is closed. Rank, from longest to shortest, the length of
time the bulbs (resistors) stay lit in each circuit.
You did not open hints for this part.
ANSWER:
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HW 4
Energy of a Capacitor in the Presence of a Dielectric
An dielectric-filled parallel-plate capacitor has plate area A and plate separation d. The capacitor is connected to a battery
that creates a constant voltage V. The dielectric constant is K.
Part A
Find the energy U₁ of the dielectric-filled capacitor. The capacitor remains connected to the battery.
Express your answer in terms of A, d, V, K, and €0.
You did not open hints for this part.
ANSWER:
C = KE A
�ܧ֧�
U₁ = = cv²
=
KEAV²
Id
U₁ =
KE AV
2ď
Part B
The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy
U2 of the capacitor at the moment when the capacitor is half filled with the dielectric.
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Express your answer in terms of A, d, V, K, and €0.
You did not open hints for this part.
ANSWER:
U₂ =
2
6A (KH)V²
41
HW 4
I
=>
C=C+C = KE A+ E. A
= GA (k+1)
U₂ = {CV² = {ACK+I) V��
4 d
after disconnection.
Part C
The capactor is now disconnected from the battery, and the dielectric plate is then slowly removed the rest of the way out
of the capacitor. Find the new energy of the capacitor, U3.
Express your answer in terms of A, d, V, K, and co.
You did not open hints for this part.
before disconnect
Q = CV =
Q doesn't change
ANSWER:
FA
ZICKH). V
after the dielectric is
new capacitance after the dielectric is removed
C' = Eo
removed
U3 =
€��ACK+1)² V²
&d
1/3=
(A²(KH)}~
d
402
6. A
€ACK+U'V
80
Part D
In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is
done by the extemal agent acting on the dielectric?
Express your answer in terms of A, d, V, K, and 0.
You did not open hints for this part.
ANSWER:
w= U₂-U₂ =
E₂A(K+1)V²
4 d
(k+1 -1)
2
Exercise 24.11
A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and
the outer is positively charged; the magnitude of the charge on each is 16.0pC. The inner cylinder has a radius of 0.450mm
1
the outer one has a radius of 7.40mm, and the length of each cylinder is 20.0cm.
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Part A
HW 4
��
What is the capacitance?
Use 8.854��10-12F/m for the permittivity of free space.
Vout-Vin
ANSWER:
C=
16��10-12
4.026
11
3.97 X10
-12
F
E=
-��
zar E
V=-S₁₂ E· de
-S. HE|17|
(180�� betwee Ĕ de)
��
SAVE dr
Part B
What applied potential difference is necessary to produce these charges on the cylinders?
ANSWER:
V= 4,026
V
��
QL
lm
la
����
16��10717/0.2
7.4��103
2��LY 8,854��10
0.45��10-3
V
4.026V
Exercise 24.18
Two capacitors are connected in series. Let 3.00µF be the capacitance of first capacitor, 5.40µF the capacitance of the
second capacitor, and Vab = 54.0V the potential difference across the system.
Part A
Calculate the charge on each capacitor.
ANSWER:
Q₁ =
104
Part B
ANSWER:
Q₂ =
104
www.w
Q1=Q₂
C₁
C₂
HH
V
�ڧ�
V₁+V₂ = V =
+음
अनु
Q =
V
54
3x1067
5.4��106
Part C
Calculate the potential difference across each capacitor.
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104 X10
=104 �ާ�
-4
= 1.04��10 C
سایا
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104��10-6
C₁
3.0��10-6 =
V
2/17/2014
ANSWER:
V₁ = 3469
34.7
HW 4
V₁ = Q
Part D
ANSWER:
V₂ =
196
19-3
V
Exercise 24.28
A parallel-plate vacuum capacitor has 7.06J of energy stored in it. The separation between the plates is 3.20mm. If the
separation is decreased to 1.75mm,
Part A
what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the
plates was changed?
ANSWER:
Q = constant
Q2
U=0.541x U₂ = 3.86 (J)
W₂ =
1/4
����
J
U = { @2
= = =
d
1.95��103
C² = 6₂
do
A
3.2x10.
d
=0.547
Part B
What is the energy now stored if the capacitor remained connected to the potential source while the separation of the
plates was changed?
ANSWER:
V is constant
U= 1.83 U₁ = 12,9 (J)
W₁ =
U = = C² V²
J
dc
من ان
do
}}
d
= 1.83
Exercise 24.36
A parallel-plate capacitor has a capacitance of Co= 3.80pF when there is air between the plates. The separation between
the plates is 2.60mm.
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HW 4
Part A
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the
plates is not to exceed 3.00x104V/m?
Qmax
ANSWER:
Part B
-10
2.96x10
wwwwww wwww
Vmax
Co
C
Qmax = | Eman | .d.co
3x10" x 2.6x10" x 3.8��10-12
=
A dielectric with a dielectric constant of 3.10 is inserted between the plates of the capacitor, completely filling the volume
between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates
is not to exceed 3.00��104V/m?
Qmax = [ Emax) .d. c' ; c'= K Co
ANSWER:
-10
Q
9.2��10
=
3x104 x²-6x10-3
x 3.1 x 3,8 �� 10-12
C
A
9.2��10"
Exercise 24.45
A parallel-plate capacitor has the volume between its plates filled with plastic with dielectric constant K. The magnitude of
the charge on each plate is Q. Each plate has area A, and the distance between the plates is d.
Part A
Use Gauss's law to calculate the magnitude of the electric field in the dielectric.
Express your answer in terms of the given quantities and appropriate constants.
ANSWER:
Q
E=
KE. A
Part B
A
modified Gauss's law
Ô KỂ LÃ
Qfree-end
E.
KĒI-|A|
=
گاره
IEI=
KE. A
Use the electric field determined in part A to calculate the potential difference between the two plates.
Express your answer in terms of the given quantities and appropriate constants.
ANSWER:
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V= |ov|= Qd
KE. A
HW 4
140
-S, E. de
SV = -
11
-Si REA
Q de
Q d
KEGA
Part C
Use the result of part B to determine the capacitance of the capacitor.
Express your answer in terms of the given quantities and appropriate constants.
ANSWER:
C=
V
KEGA
ठ
Problem 24.53
A 20.0uF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected
to those of an uncharged 9.00µF capacitor.
Part A
Zo��F
800V
=> Q = cv
CV
Compute the original charge of the system.
ANSWER:
20��108 x800
= 1-6��102
Q=1.6��10-2
C
Part B
Compute the final potential difference across capacitor.
ANSWER:
AV=
550 V
Part C
Compute the final energy of the system.
ANSWER:
www.www.
Q
equilibrium:
V₁ = V₂
Q₁
Q2
C₁
V
Q₁+Q₂ = Q
Q₁ + c A Q₁ = Q
�� charges are
shared
-6
Q₁ = Q~
C₁
C₁+Cz
1.6x102
20 x10
{0+9) xc6
1-1��102
C
V₁ = Q₁
1-1��10-2
20��10-6
550 V
16/17
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U =
4.39 (J)
www
HW 4
U = U₁ +Uz
J
= = c₁vi² + { C₂V₁₂
=
{ V = (C₁ + G ) = { x 5507 (20+9) �� 106
Part D
Compute the decrease in energy when the capacitors are connected.
ANSWER:
V; = ½ cv ²
13��20��10 6 �� 800
=
~2
J
=6.4
AU =
1V+- U₁ = 14.39-6.41:
Score Summary:
Your score on this assignment is 0.0%.
You received 0 out of a possible total of 15 points.
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