of 32
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When a potential difference of 150 V is applied to the plates of a
parallel-plate capacitor, the plates carry a surface charge density of
30.0 nC/cm2. What is the spacing between the plates?
0 AQ Vd
12 2 2
0
9 2 4 2 2
8.85 10 C N m 150 V
4.42 m
30.0 10 C cm 1.00 10 cm m
Vd
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Four capacitors are connected as shown in
Figure
(a) Find the equivalent capacitance
between points a and b.
(b) Calculate the charge on each capacitor
if Vab = 15.0 V.
1
2.50 F
2.50 6.00 8.50 F
1 15.96 F
8.50 F 20.0 F
s
p
eq
C
C
C
1 1 1
15.0 3.00sC
5.96 F 15.0 V 89.5 CQ C V
89.5 C4.47 V
20.0 F
15.0 4.47 10.53 V
6.00 F 10.53 V 63.2 C on 6.00 F
QV
C
Q C V
89.5 63.2 26.3 C
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Find the equivalent capacitance
between points a and b for the group of
capacitors connected as shown in Figure
P26.27. Take C1= 5.00 F, C2 = 10.0 F,
and C3 = 2.00 F.
1
1
2
1
1 13.33 F
5.00 10.0
2 3.33 2.00 8.66 F
2 10.0 20.0 F
1 16.04 F
8.66 20.0
s
p
p
eq
C
C
C
C
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A parallel-plate capacitor is charged and then disconnected from abattery. By what fraction does the stored energy change (increase
or decrease) when the plate separation is doubled?
2 12d d 2 1
1
2
C C
stored energy doubles. Therefore, the
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Determine (a) the capacitance and (b) the maximum
potential difference that can be applied to a Teflon-filled
parallel-plate capacitor having a plate area of 1.75 cm2
and plate separation of 0.040 0 mm.
12 4 2
110
5
2.10 8.85 10 F m 1.75 10 m 8.13 10 F 81.3 pF4.00 10 m
AC
d
6 5m ax m ax 60.0 10 V m 4.00 10 m 2.40 kVV E d
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A parallel-plate capacitor is constructed using a dielectric material
whose dielectric constant is 3.00 and whose dielectric strength is
2.00 108 V/m. The desired capacitance is 0.250 F, and thecapacitor must withstand a maximum potential difference of 4 000
V. Find the minimum area of the capacitor plates.
8 m axm ax 2.00 10 V m VE
d
600.250 10 F
AC
d
6
2m ax
12 80 0 m ax
0.250 10 4000
0.188 m
3.00 8.85 10 2.00 10
C VCdA
E
3.00
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- A parallel plate capacitor made from 2 squares of metal, 2mm thick and 20cm on a side
separated by 1mm with 1000V between them Find:
a) capacitance b)charge per plate c) charge density d)electric field
e) energy stored f) energy density
pFC
FFd
AC
54.3
1054.3104.35101
1020201085.8
1211
3
4
12
0
nCCVb 54.31054.3Q) 3
27
4
9
/10885.0102020
1054.3Q) mC
Ac
JCVUe
61223
2 1077.12
1054.310
2
1)
33
34
6
/1025.44
101102020
1077.1
volume
Uu) mJf
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Consider the circuit as shown, where C1= 6.00F and C2= 3.00
F and V =20.0V. Capacitor C1
is first charged by closing of
switch S1. Switch S1is then opened and the charged capacitor is
connected to the uncharged capacitor by the closing of S2.
Calculate the initial charge acquired by C1and the final charge
on each.
S1close, S2 openC = Q/V Q = 120 C
After S1open, S2close
Q1 + Q2 = 120 C
Same potential
Q1 /C1 = Q2 / C2
(120-Q2)/C1= Q2/C2(120 - Q2)/6 = Q2/ 3 Q2 = 40 C
Q 1= 80 C
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March 2, 2014 University Physics, Chapter 24 11
An isolated conducting sphere whose radiusRis 6.85 cm has a
charge of q=1.25 nC. How much potential energy is stored in the electric
field of the charged conductor?
Answer:
Key Idea: An isolated sphere has a capacitance of C=40R
The energy U stored in a capacitor depends on the charge and the capacitance
according to
and substituting C=40Rgives
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An isolated conducting sphere whose radiusRis 6.85 cmhas a charge of q= 1.25 nC.
Question 2:What is the field energy density at the surface of the sphere?
Answer:
Key Idea: The energy density udepends on the magnitude of the electricfield Eaccording to
so we must first find the Efield at the surface of the sphere. Recall:
2
012
u E
E 1
40
q
R2
u 1
20E
2
q2
3220R4
2.54 105 J/m3 25.4 J/m3
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mVc
Cb
Fd
ACa
/102105
10V/dE)
104.35104.3510CVQ)
104.35105
21085.8)
6
3
4
6104
10
3
12
0
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A parallel-plate capacitor is charged and then disconnected from a
battery. By what fraction does the stored energy change (increase
or decrease) when the plate separation is doubled?
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5- Determine (a) the capacitance and (b) the maximum potential
difference that can be applied to a Teflon-filled parallel-plate
capacitor having a plate area of 1.75 cm2and plate separation of
0.040 0 mm.
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March 2, 2014 University Physics, Chapter 24 16
An air-filled parallel plate capacitor has a capacitance of 1.3 pF.The separation of the plates is doubled, and wax is insertedbetween them. The new capacitance is 2.6pF.
Question:
Find the dielectric constant of the wax.
Answer: Key Ideas:The original capacitance is given by
Then the new capacitance is
Thus
rearrange the equation:
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March 2, 2014 University Physics, Chapter 24 17
Question 1:
Consider a parallel plate capacitor with capacitance C=
2.00 Fconnected to a battery with voltage V= 12.0 V asshown. What is the charge stored in the capacitor?
6 52.50 2.0 10 F 12.0V 6.0 10 Cq CV
q CV 2.00 106 F 12.0 V 2.40 105 C
Question 2:Now insert a dielectric with dielectric constant =2.5 between the plates of the capacitor. What is thecharge on the capacitor?
The capacitance of the capacitor is increased:airC C
The additional charge is provided by the battery.
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Given a 7.4 pFair-filled capacitor. You are asked to
convert it to a capacitor that can store up to 7.4 J witha maximum voltage of 652 V. What dielectric constant
should the material have that you insert to achieve
these requirements?
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One common kind of computer keyboard is based on the idea of capacitance. Each -
key is mounted on one end of a plunger, the other end being attached to a movable
metal plate. The movable plate and the fixed plateform a capacitor. When the key is
pressed, the capacitance increases. The change in capacitance is detected, therebyrecognizing the key which has been pressed.
The separation between the plates is 5.00 mm, but is reduced to 0.150 mm when a key
is pressed. Theplate area is 9.50x10-5m2and the capacitor is filled witha material
whose dielectric constant is 3.50.
Determine the change in capacitance detected by the computer.
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March 2, 2014 University Physics, Chapter 24 20
If each capacitor has a capacitance of 5 nF, what is the capacitance ofthis system of capacitors?
Find the equivalent capacitance
We can see that C1and C2are in parallel,
and that C3is also in parallel with C1and C2
We find C123= C1+ C2+ C3= 15 nF
and make a new drawing
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March 2, 2014 University Physics, Chapter 24 21
We can see that C4and C123are in series
We find for the equivalent capacitance:
and make a new drawing
1
C1234
1
C123
1
C4C1234
C123C4
C123 C4
= 3.75 nF
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March 2, 2014 University Physics, Chapter 24 22
We can see that C5and C1234are in parallel We find for the equivalent capacitance
and make a new drawing
C12345 C1234 C5 C123C4
C123 C4 C5
(C1 C2 C3)C4
C1 C2 C3C4C5 = 8.75 nF
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March 2, 2014 University Physics, Chapter 24 23
We have a parallel plate capacitor
constructed of two parallel plates, each
with area 625 cm2separated by a
distance of 1.00 mm.
Question: What is the capacitance of this
parallel plate capacitor?
Answer:A parallel plate capacitor constructed out ofsquare conducting plates 25 cm x 25 cm separated by 1 mmhas a capacitance of about 0.5 nF.
-12 2
0
10
8.85 10 F/m 0.0625 m
0.001 m
5.53 10 F
= 0.553nF
AC
d
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March 2, 2014 University Physics, Chapter 24 24
We have a parallel plate capacitor
constructed of two parallel plates
separated by a distance of 1.00 mm.
Question: What area is required to
produce a capacitance of 1.00 F?
Answer:Square conducting plates with dimensions 10.6 kmx 10.6 km(6 miles x 6 miles) separated by 1 mmarerequired to produce a capacitor with a capacitance of 1 F.
A Cd
0
1 F 0.001 m
8.85 10-12
F/m
1.13
10
8m
2
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March 2, 2014 University Physics, Chapter 24 25
: A storage capacitor on a random access memory (RAM) chiphas a capacitance of 55 nF. If the capacitor is charged to 5.3
V, how many excess electrons are on the negative plate?
Answer:Idea:We can find the number of excess electrons on thenegative plate if we know the total charge qon the plate.
Then, the number of electrons n=q/e, where eis the electroncharge in coulomb.Second idea:The charge qof the plate is related to thevoltage Vto which the capacitor is charged: q=CV.
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SinceQ= CVand the two capacitors are
identical, the one that is connected to
the greater voltagehas more charge,
which is C2in this case.
Capacitor C1is connected across a
battery of 5 V. An identical
capacitor C2is connected across a
battery of 10 V. Which one has
more charge?
1) C1
2) C2
3) both have the same charge
4) it depends on other factors
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Since Q = CV, in order to increase the charge that a
capacitor can hold at constant voltage, one has to
increase its capacitance. Since the capacitance is
given by , that can be done by either
increasingAor decreasing d.
1) increase the area of the plates
2) decrease separation between the plates
3) decrease the area of the plates
4) either (1) or (2)
5) either (2) or (3)
d
AC
0
What must be done to a
capacitor in order to
increase the amount of
charge it can hold (for a
constant voltage)?
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Since the battery stays connected, the voltage
must remain constant! Since ,
when the spacing dis doubled, the capacitance
Cis halved. And since Q= CV, that means the
charge must decrease.
+Q Q
d
AC
0
A parallel-plate capacitor
initially has a voltage of 400 V
and stays connected to thebattery. If the plate spacing is
now doubled,what happens?
1) the voltage decreases
2) the voltage increases
3) the charge decreases
4) the charge increases
5) both voltage and charge change
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Once the battery is disconnected, Qhas to remain
constant,since no charge can flow either to or from
the battery. Since , when the
spacing dis doubled, the capacitance Cis halved.
And since Q= CV, that means the voltage must
double.
A parallel-plate capacitor initially has a
potential difference of 400 Vand is
then disconnected from the chargingbattery. If the plate spacing is now
doubled(without changing Q), what is
the new value of the voltage?
1) 100 V
2) 200 V
3) 400 V
4) 800 V
5) 1600 V
+Q Q
d
AC
0
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The 2 equal capacitors in series add
up as inverses, giving 1/2C. These
are parallel to the first one, which
add up directly. Thus, the total
equivalent capacitance is 3/2C.
o
o
C CCCeq
1) Ceq = 3/2C
2) Ceq = 2/3C
3) Ceq = 3C
4) Ceq = 1/3C
5) Ceq = 1/2C
What is the equivalent
capacitance, Ceq, of the
combination below?
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1) V1 = V2
2) V1 > V2
3) V1
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C1= 1.0 F C3= 1.0 F
C2= 1.0 F
10 V
We already know that the voltage
across C1is 10 V and the voltage
across both C2and C3is 5 V each.
Since Q= CVand Cis the samefor
all the capacitors, we have V1> V2
and thereforeQ1> Q2.
1) Q1 = Q2
2) Q1 > Q2
3) Q1