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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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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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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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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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We can see that C4and C123are in series

We find for the equivalent capacitance:


1

C1234

1

C123

1

C4C1234

C123C4

C123 C4

= 3.75 nF

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We can see that C5and C1234are in parallel We find for the equivalent capacitance


C12345 C1234 C5 C123C4

C123 C4 C5

(C1 C2 C3)C4

C1 C2 C3C4C5 = 8.75 nF

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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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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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: 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)?

+QQ

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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

chap 26 sol

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