Chapter 25 Solutions 25.1 a) = −̂ b) Δ= ∙ = −̂∙̂ = c) Capacitance is found using = Δ = = THINK: You could stop here but that is a bad idea….this result is pretty meaningless. You know = for a plate. This will simplify the result. = = d) As decreases, capacitance increases! e) Increase area to increase capacitance. f) ≈56pF g) In parallel plates we know electric field magnitude is given by = Δ Rearranging and plugging values gives = Δ ≈ 9V 3×10 V m ≈3μm h) We know = = = = ≈834nC i) Use = ℎ × ℎ ℎ = ℎ = ≈5.2×10 electrons
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Transcript
Chapter 25 Solutions
25.1
a) ��� = ��� �−�
b) Δ� = � ��� ∙ ������������ = � �
�� �−� ∙ ���� = ��
��
c) Capacitance is found using
� = �Δ�
� = � �!�
� = � !� �
THINK: You could stop here but that is a bad idea….this result is pretty meaningless.
You know = "# for a plate. This will simplify the result.
� = � !�$�%&�
� = !�%�
d) As � decreases, capacitance increases!
e) Increase area to increase capacitance.
f) � ≈ 56pF
g) In parallel plates we know electric field magnitude is given by
� = ��
Rearranging and plugging values gives
� = Δ�� ≈ 9V
3 × 102 Vm≈ 3μm
h) We know
� = !�
= !��
�% = !��
� = !�%�
� ≈ 834nC
i) Use
��9� = :;<=>?@ABCℎE@F?� G × : =EF;HI<�?ABA;?CℎE@F?G
;<=>?@ABCℎE@F?� = �
=EF;HI<�?ABA;?CℎE@F?
J = �?
J ≈ 5.2 × 10MNelectrons
25.2
a) ��� = − V"WX @
b) Potential difference is
Δ� = −Y ���Z
�∙ ���
In this case we move radially outwards as we go from one conductor to the other.
This implies ��� = �@@.
Δ� = −Y :−[�@N @G
Z
�∙ ��@@�
Δ� = Y [�@N �@
Z
��@ ∙ @�
Δ� = [�Y �@@N
Z
�
Δ� = [� \−1@]�
Z
Δ� = [� \+1@]Z
�
Δ� = [� \1E −1>]
_` = ab�c − d�cd
c) Capacitance is found using
� = �Δ�
� = �[��> − E�
>E
� = >E[�> − E�
e = fghicd�c − d�
d) When c → ∞
Δ� = [� \1E −1>] →
[�E
elmnodpqrmstquq = fghid
25.3
a) The total charge on the cylinder is � = vw.
The volume charge density is x = 0 (all charge reside on the surface of a conductor in static equilibrium).
The surface charge density on the inner cylinder is = "Nyz{ =
|Nyz.
b) The electric field between the conductors is found from Gauss’s Law due to the symmetry of the situation.
Recall the outer conductor is not inside the Gaussian surface (if the surface lies between the conductors).
As such, the electric field between the conductors arises entirely from the inner conductor.
The electric field between the conductors is thus
��� = 2[�−v�@ @
c) One can obtain potential difference using
Δ� = −Y ��� ∙ ����
�
Now simply turn the crank:
Δ� = −Y 2[�−v�@ @ ∙ ��@@�
Z
�
Minus signs cancel! Also, use @ ∙ @ = 1. Factor out constants.
Δ� = 2[vY �@@
Z
�
Δ� = 2[v ln }>E} d) An interesting alternative style is shown on the next page.
Use the definition of capacitance:
� = �Δ�
� = �2[v ln ~>E~
Don’t forget…by definition we know v = "{!!!
� = �2[ $�w& ln ~
>E~
� = w2[ ln ~>E~
e) Notice we are to find capacitance per unit length!
Divide each side by w and simplify. Notice, in this case we can get a numerical result!
Think: upon plugging in numbers we expect units of �� (Farads per meter).
�w = 1
2[ ln ~>E~
Alternative solution to problem 25.3
Consider the cross-sectional view shown at right.
Imagine building the total capacitance by considering each of the colored slices as a
single capacitor.
All one need do is add up the small differential capacitances in series.
Said another way
1��� =
1�M +
1�N +⋯ 1
��
��� = 1 1�M +
1�N +⋯ 1
��
where J is the number of slices.
Using calculus to do the sum gives
��� = 1� 1
��
Now we need only determine �� for a single slice.
Consider the green slice at right.
If we slice it thinly, it is well approximately by a parallel plate capacitor.
�� = !�%� = !��2�@w�
�@
Therefore
��� = 1� 1
���zNz
��� = 1� �@
2�!�@wZ�
��� = 2�!�w� �@
@Z�
��� = w2[ ln >E
@
25.4 The result of 23.15 was
����9��� = 2[v : 1� + � +
1� + � − �G
From there, use our standard procedure:
• Determine Δ� from Δ� = −���� ∙ ���. • Take the absolute value and shove into � = "
|��|.
∆� = −Y ��� ∙ ����
�
∆���i�9��� = −Y 2[v : 1� + � +
1� + � − �G ∙ ��
���
���
Since we are going from a plus charge to a minus charge we expect
the result should be ∆���i�9��� � 0.
Use an online integration program to check this result.
If you care, ask me and I’ll write it up more thoroughly.
�
+v −v
� �
� � − �
25.5 On this page I work all the way down to get the equivalent capacitance.
On the next page I work my way back up to get b, _`, and � for each capacitor.
+ 1.5 V -
5 µF 15 µF
8.25 µF
24 µF
+ 1.5 V -
3.75 µµµµF
8.25 µF
24 µF
Get Ceq
for the
two in
series on
the top
+ 1.5 V -
12 µµµµF 24 µF
Get Ceq
for the
two in
parallel
+ 1.5 V -
8 µµµµF
Get Ceq
for the
two in
series
��� = �M�N�M + �N
��� = �5μF��15μF�5μF ^ 15μF
��� � 3.75μF
��� � �M ^ �N
��� � 3.75μF ^ 8.25μF
��� � 12μF
��� ��M�N
�M ^ �N
��� ��12μF��24μF�
12μF ^ 24μF
��� � 8μF
We now have a single Cap in a circuit with the battery.
Notice the term in red is approximately ¦X«¸ after taking the square root. We can tell the red term is larger than the
blue term which implies we should use the positive root. Continues next page…
Δ�Z����W� � ¦
�M � �
�N � �
�¨ � �
�¨ �?
Now we know
� � µ¦N�4� − 1¶ ^ º¦N�4� µ¦N�4� − 12¶
Recall we used � � «·« . If we multiply both sides by � we have solved for �¨.
This was the answer that I asked for.
How could we check this result?
The simplest way might be to use an online simulation.
Pick values for the caps (remember we used �M � �N � �).
Use a known battery and compute the energy �¨ � �.
Then use the values of �, ¦, and � with our formula to compute the predicted value of �¨.
Verify our formula produces the value of �¨ you used in your simulation.
Repeat for several oddball combinations of numbers to increase your confidence in your result.
Is there another, more annoying, method?
Try a simple case with numbers and verify our formula gives the same results as numerical work.
For instance, assume we used ¦ � 1.00 V with � � 1.00 F. We know the energy stored in �M is given by
�M � 12 �Δ��N�M � 0.500 J For the branch with 2 & 3 in series, we expect �N¨ should be smaller than either �N � � or �¨.
This means this branch must have less capacitance for the same potential difference.
We expect it has less stored energy.
This means we should pick a value of � � 0.500 J or our problem won’t make sense!
Notice this is a good check on our result! Our result gives a negative number under the square root if � � 0.500 J! Let us assume then that �¨ � � � 0.200 J. Our formula predicts
� � µ 1.00N4�0.200��1.00� − 1¶ ^ º �1.00�N
4�0.200��1.00� µ �1.00�N4�0.200��1.00� − 12¶
� � �0.25� ^ ½1.25�0.75�
� � 1.1875 � 1916
�¨ � 1.1875�
Now you have a number for e³ you could use to check with series parallel rules.
Use ¦ � 1.00 V, �M � �N � � � 1.00 F & �¨ � 1.1875 F. Verify the stored energy on 3 is �¨ � 0.200 J.
25.13
a) To simplify the circuit, imagine grabbing the two nodes and
stretching the circuit. If you connect at nodes X & Y, imagine
grabbing those two nodes and stretching the circuit vertically.
The left branch of the circuit (in red) has two caps in series.
This branch has capacitance
����� � ��� ^ � � �2
Same thing for the right branch (in blue) gives
�W�¾¿� � ��� ^ � � �2
All three branches are in parallel giving eq¡ � e ^ eoqÀp ^ eulÁtp � ´e
b) If we connect at X & Z, we grab those
two nodes and stretch.
The red branch has equivalent
capacitance of �W�� � «««Â« � «N.
That branch is in parallel with the
black branch giving
�W��&Z���V � �2 ^ � � 32 �
The combo of red and black is in series
with the green branch giving
�W��,¾W���&Z���V � � $32 �&� ^ 32 � � 35 �
Add the blue branch in parallel to get ��� � �Z�Å� ^ �W��,¾W���&Z���V
��� � � ^ 35 �
eq¡ � ± e � Æ. ®e
Z
X
Y
Original Circuit
Z
X
Y
Grab XY & Stretch
Z
X
Y
Original Circuit
Z
X
Y
Grab XZ & Stretch Grab XZ & Stretch
Z
X
Y
25.14 First combine 2 & 3 in parallel to give �N¨ � �N ^ �¨ � 5�.
Next combine 1, the 2-3 combo, and 4 in series: 1��� � 1�M ^ 1�N¨ ^ 1�
1��� � 1� ^ 15� ^ 14�
1��� � 2020� ^ 420� ^ 520�
1��� � 2920�
HEY YOU!!! Don’t forget to flip the final answer upside down!
��� � 2029 �
Think: when combining caps in series, equivalent capacitance is always smaller than the smallest cap. Checks out!
The total charge is ��� � ¦��� � N�NÇ ¦�.
Since we combined 1, 2-3 combo and 4 in series we know �M � �N¨ � � � N�NÇ ¦�.
Now use � � �� to find
Δ�N¨ � �N¨�N¨ � 2029 ¦� 5� � 429 ¦
While unnecessary to determine each cap’s charge, I found Δ�M � "È«È � X�Xɦ« « � N�NÇ ¦ and Δ� � "Ê«Ê � X�Xɦ«
« � �NÇ ¦.
Notice the sum of the voltages adds up to the voltage across the ideal battery. This is a check on my work.
Moving on, since we combined 2 & 3 in parallel we know Δ�N � Δ� � NÇ ¦.
Now use � � �� to find
�N � Δ�N�N � 829 ¦�
�¨ � Δ� �¨ � 1229 ¦�
Double check this result. Since 2 & 3 were in parallel we expect �N ^ �¨ � �N¨…checks out!
25.15 This is one of those circuits where it helps to imagine
b) I made a video for this one: https://www.youtube.com/watch?v=WhyCoUhofQg
c) We know �M ^ �N � 4¦� � �Mª ^ �Nª which can be rearranged to give �Nª � 4¦� − �Mª Δ�M′ � Δ�N′ �Mª�M � �Nª�N
�Mª� � �Nª2�
2�Mª � �Nª 2�Mª � 4¦� − �Mª
�Mª � 43 ¦�
From this we can now compute
�Nª � 4¦� − �Mª � 83 ¦�
In the original picture we see ¦� is represented by 3 plus charges. The final picture should show 4 charges
on �M and 8 charges on �N.
25.20 This time use �M − �N � 2¦� � �Mª ^ �Nª which can be rearranged to give �Nª � 2¦� − �Mª .
One finds �Mª � N ¦� and �Nª � ¦�. I hope drawing the pictures gives you a feel for how charge is conserved. I
hope you are beginning to see how charge conservation and energy conservation apply to capacitor circuits.
25.21
a) Once the switch is closed there is no way for the 24.0 mC of excess charge to neutralized (assuming no
sparks across the capacitors’ gaps). Charge is conserved. When the switch closes, the excess charge
redistributes between the two capacitors. Note: strictly speaking electrons move from the top plate of �N
towards the top plate of �M. This partially cancels the excess positive charge on the upper plate of �M while
leaving and absence of electrons (an excess positive charge) on the upper plate of �N. �M ^ �N � �Mª ^ �Nª �M ^ 0 � �Mª ^ �Nª �M � �Mª ^ �Nª
Furthermore, energy must be conserved going around the loop. In practice, this tells us Δ�Mª � Δ�Nª This can be rewritten in terms of charge using � � Δ�� �Mª�Mª � �Nª�Nª
In this problem, we haven’t modified the capacitors at all…just the charges on them.
We know �Mª � �M and �Nª � �N. Rewriting the previous line with this in mind gives �Mª�M � �Nª�N
The rest of the original 120.0 mC must reside on the other cap since charge is conserved. bª � ¯i. i mCmCmCmC
Now use � � �� for each cap to verify the size of each potential difference is the same (as predicted by
conservation of energy). I found Δ�Mª � "ÈΫÈÎ � 20.0 V & Δ�Nª � "XΫXÎ � 20.0 V. Looks good!
d) I found �Mª � 400 mJ and �Nª � 800 mJ. This gives ��9���ª � �Mª ^ �Nª � 1200 mJ. Notice the energy change is %Δ� � ¸ÐÑ¸Ò¸Ò × 100% � −7.41%.
Be careful with the way you say things:
• The change is −7.41%.
• The capacitors lost 7.41% of the energy.
• Do not say “the capacitors lost −7.41%” because that is stating the negative sign twice.
25.23
a) When wired in parallel to the battery, each capacitor should have the same potential difference as the
battery. We know each capacitor has � � 12.0 V. Initially we have �M � 24.0 mC and �M � 144 mJ while �N � 48.0 mC and �N � 288 mJ. The total initial stored energy is ��9� � 432 mJ.
b) We know negative charges will move around in the wires. We assume no negatives can jump the gap in
the capacitors. The excess negative charge on the upper plate of �N is larger than he excess positive charge
on the upper plate of �M. Negative charges will flow over to �M and cancel the excess positive.
Furthermore, additional excess negative charges are available and will spread out to be shared between the
two top plates. In the end we expect the top plates of each capacitor will be negative.
c) Since the excess charge on the top plates are opposite signs is positive we know �M − �N � �Mª ^ �Nª
After switch is closed the net potential difference going around the circuit must be zero (energy is
conserved). Said in equation form, we know Δ�Mª � Δ�Nª �Mª�Mª � �Nª�Nª
We know �Mª � �M and �Nª � �N (capacitors have not been modified). �Mª�M � �Nª�N
�M � 64.0 mC �N � 16.0 mC �¨ � 48.0 mC ��� � 64.0 mC Δ�M � 16.0 V Δ�N � 8.00 V Δ� � 8.00 V Δ��� � 24.0 V �M � 512 mJ �N � 64.0 mJ �¨ � 192 mJ �q¡ � °®¯ mJmJmJmJ b) My guess, your guess, any guess goes tonight!
c) In this problem we insert the dielectric. Capacitor �¨ should increase by factor Í.
We know � ª � Í�¨ � 18.00 mF which changes the equivalent to ���ª � M� � 3.333 mF.
d) If the switch is open while the capacitors and messed with, you are supposed to realize ���ª � ��� .
Think: there is no path for charges going to or leaving from the top plate in the network (top plate of �M).
With the switch open during this monkey business, it is possible for voltages to change!!!
I use this value of ���ª to re-determine every value in the table.
Note: due to this particular configuration of caps, nothing changes for �M!!! That is not always true.
�Mª � 64.0 mC �Nª � 6.40 mC �ª � 57.6 mC ���ª � 64.0 mC Δ�Mª � 16.0 V Δ�Nª � 3.20 V Δ�ª � 3.20 V Δ���ª � 19.2 V �Mª � 512 mJ �Nª � 10.24 mJ �ª � 92.16 mJ �q¡ª � ®Æf. f mJmJmJmJ e) Answer in table above. Notice bringing in the dielectric increases capacitance but lowers total energy.
Recall the switch was open while dielectric was inserted and equivalent charge was held constant.
The dielectric increases capacitance and makes it easier to separate the charge.
Less energy is required to maintain the charge separation.
25.26
a) I found the following answers tabulated at right. I used series and parallel rules to figure out the charge on
and potential difference across each cap (and the equivalent). To check my work, I know the first three
energies should add up to the equivalent. This check always works regardless of the arrangement of caps.
�M � 10.0 mC �N � 4.00 mC �¨ � 6.00 mC ��� � 10.0 mC Δ�M � 10.0 V Δ�N � 2.00 V Δ� � 2.00 V Δ��� � 12.0 V �M � 50.0 mJ �N � 4.00 mJ �¨ � 6.00 mJ �q¡ � ®i. i mJmJmJmJ b) My guess, your guess, any guess goes tonight!
c) This time we remove the dielectric. Capacitor �¨ should decrease by factor Í.
We know � ª � MÓ �¨ � 1.00 mF which changes the equivalent to ���ª � 0.750 mF.
d) If the switch is closed while the capacitors and messed with, you are supposed to realize Δ���ª � Δ��� .
Think: the battery is still available to push charges around the circuit and change the charges on the caps.
I use this value of Δ���ª to re-determine every value in the table.
�Mª � 9.00 mC �Nª � 6.00 mC �ª � 3.00 mC ���ª � 9.00 mC Δ�Mª � 9.00 V Δ�Nª � 3.00 V Δ�ª � 3.00 V Δ���ª � 12.0 V �Mª � 40.5 mJ �Nª � 9.00 mJ �ª � 4.50 mJ �q¡ª � ±f. i mJmJmJmJ e) Answer in table above. Notice removing the dielectric decreases both capacitance & total energy.
Recall the switch was closed while dielectric was removed.
Summary of previous two problems:
Monkey business with closed switch
• Δ���ª � Δ���
• If ��� goes up, ��� goes up.
Monkey business with open switch
• ���ª � ���
• If ��� goes up, ��� goes down!
25.27 c, f, g, i, & k are all true; the rest are false.
25.28 e, g, h, j, & l are all true; the rest are false.
25.29 For part h I meant to say bª � bƪ � ²e. Given that change: c, f, h, k, & l are all true; the rest are false. It
is a bit surprising that h is true but since the capacitors are in series and disconnected from the battery the charge on
either one will not change!
Note: part i is false. Assume capacitor �N is the one changing.
We know �Nª � ∆�Nªe′ (new capacitance) but �Nª ≠ ∆�Nªe (original capacitance).
25.30 b, f, g, k, & l are all true; the rest are false. It would probably be wise to also write down �Mª � ∆�Mª� & �Nª � ∆�Nª�′. Again, here I am assuming that C2 is the capacitor that changed.
25.31 The left capacitor �M acts like two caps in series. 1�M � 1��9� ^ 1�Z9� � 1 !��ÕI2 ^ 1
Í!��ÕI2 � I2!��Õ ^ I2Í!��Õ � �Í ^ 1�I2Í!��Õ
Remember to flip!!!
eÆ � ´ÖÖ ^ Æ ∙ hi×Øp
The right capacitor �N acts like two caps in parallel.