when a potential difference of 150 v is applied to the plates of a parallel-plate capacitor, the...

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

09 2 4 2 2

8.85 10 C N m 150 V4.42 m

30.0 10 C cm 1.00 10 cm m

Vd

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 ΔVab = 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 1 115.0 3.00sC

5.96 F 15.0 V 89.5 CQ C V

89.5 C4.47 V

20.0 F15.0 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

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 μF, C2 = 10.0 μF, and C3 = 2.00 μF.

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

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?

2 12d d 2 112

C C

stored energy doubles. Therefore, the

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 2110

5

2.10 8.85 10 F m 1.75 10 m8.13 10 F 81.3 pF

4.00 10 m

AC

d

6 5max max 60.0 10 V m 4.00 10 m 2.40 kVV E d

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 μF, and the capacitor must withstand a maximum potential difference of 4 000 V. Find the minimum area of the capacitor plates.

8 maxmax 2.00 10 V m

VE

d

60 0.250 10 FA

Cd

62max

12 80 0 max

0.250 10 40000.188 m

3.00 8.85 10 2.00 10

C VCdA

E

3.00

- 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

412

0

nCCVb 54.31054.3Q ) 3

274

9

/10885.0102020

1054.3Q ) mC

Ac

JCVUe 6

12232 1077.1

2

1054.310

2

1 )

3334

6

/1025.44101102020

1077.1

volume

Uu ) mJf

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 S1 is then opened and the charged capacitor is connected to the uncharged capacitor by the closing of S2. Calculate the initial charge acquired by C1 and the final charge on each. S1 close, S2 open

C = Q/V Q = 120 CAfter S1 open, S2 close Q1 + Q2 = 120 CSame potential Q1 /C1 = Q2 / C2 (120-Q2)/C1= Q2/C2

(120 - Q2)/6 = Q2/ 3 Q2 = 40 C

Q 1= 80 C

April 18, 2023 University Physics, Chapter 24 11

• An isolated conducting sphere whose radius R is 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=40R gives

• An isolated conducting sphere whose radius R is 6.85 cm has 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 u depends on the magnitude of the electric

field E according to

so we must first find the E field at the surface of the sphere. Recall:

20

1

2u E 2

0

1

2u E

E 1

40

q

R2

u 1

20E2

q2

32 20R4 2.54 10 5 J/m3 25.4 J/m3

mVc

Cb

Fd

ACa

/102105

10V/dE )

104.35104.3510CVQ )

104.35105

21085.8 )

63

4

6104

103

120

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?

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 cm2 and plate separation of 0.040 0 mm.


• An air-filled parallel plate capacitor has a capacitance of 1.3 pF. The separation of the plates is doubled, and wax is inserted between 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:


Question 1:Consider a parallel plate capacitor with capacitance C = 2.00 F connected to a battery with voltage V = 12.0 V as shown. 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 10 6 F 12.0 V 2.40 10 5 C

Question 2:Now insert a dielectric with dielectric constant = 2.5 between the plates of the capacitor. What is the charge on the capacitor?

The capacitance of the capacitor is increased:airC C

The additional charge is provided by the battery.

Given a 7.4 pF air-filled capacitor. You are asked to convert it to a capacitor that can store up to 7.4 J with a maximum voltage of 652 V. What dielectric constant should the material have that you insert to achieve these requirements?

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 plate form a capacitor. When the key is pressed, the capacitance increases. The change in capacitance is detected, thereby recognizing 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. The plate area is 9.50x10-5m2 and the capacitor is filled with a material whose dielectric constant is 3.50.Determine the change in capacitance detected by the computer.


If each capacitor has a capacitance of 5 nF, what is the capacitance of this system of capacitors?

Find the equivalent capacitance

• We can see that C1 and C2 are in parallel, • and that C3 is also in parallel with C1 and C2

• We find C123 = C1 + C2 + C3 = 15 nF• … and make a new drawing


• We can see that C4 and C123 are in series• We find for the equivalent capacitance:

• … and make a new drawing

1

C1234

1

C123

1

C4

C1234 C123C4

C123 C4

= 3.75 nF


• We can see that C5 and C1234 are 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 C3 C4

C5 = 8.75 nF


• We have a parallel plate capacitor constructed of two parallel plates, each with area 625 cm2 separated by a distance of 1.00 mm.

Question: What is the capacitance of this parallel plate capacitor?

Answer: A parallel plate capacitor constructed out of square conducting plates 25 cm x 25 cm separated by 1 mm has a capacitance of about 0.5 nF.

-12 20

10

8.85 10 F/m 0.0625 m

0.001 m

5.53 10 F

= 0.553nF

AC

d


• 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 km x 10.6 km (6 miles x 6 miles) separated by 1 mm are required to produce a capacitor with a capacitance of 1 F.

A Cd

0

1 F 0.001 m

8.8510-12 F/m

1.13108 m2


: A storage capacitor on a random access memory (RAM) chip has 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 the

negative plate if we know the total charge q on the plate. Then, the number of electrons n=q/e, where e is the electron charge in coulomb.

Second idea: The charge q of the plate is related to the voltage V to which the capacitor is charged: q=CV.

Since QQ = = CVCV and the two capacitors are

identical, the one that is connected to

the greater voltagegreater voltage has more chargecharge,

which is CC22 in this case.

Capacitor C1 is connected across a battery of 5 V. An identical capacitor C2 is connected across a battery of 10 V. Which one has more charge?

1) CC11

2) CC22

3) both have the same charge

4) it depends on other factors

Since Q = CVQ = CV, in order to increase the charge that a

capacitor can hold at constant voltage, one has to

increase its capacitanceincrease its capacitance. Since the capacitance is

given by , that can be done by either

increasing increasing AA or decreasing decreasing dd.

1) increase the area of the plates increase the area of the plates

2) decrease separation between the platesdecrease separation between the plates

3) decrease the area of the plates

4) either (1) or (2)

5) either (2) or (3)

dAC 0

What must be done to a What must be done to a capacitor in order to capacitor in order to increase the amount of increase the amount of charge it can hold (for a charge it can hold (for a constant voltage)?constant voltage)?

+Q –Q

Since the battery stays connected, the voltage Since the battery stays connected, the voltage

must remain constant!must remain constant! Since

, when the spacing d is doubled, the

capacitance C is halved. And since QQ = = CVCV, that

means the charge must decreasecharge must decrease.

+Q –Q

dAC 0

A parallel-plate capacitor initially has a voltage of 400 V and stays connected to the battery. If the plate spacing is now doubled, what happens?

1) the voltage decreasesthe voltage decreases

2) the voltage increasesthe voltage increases

3) the charge decreasesthe charge decreases

4) the charge increasesthe charge increases

5) both voltage and charge changeboth voltage and charge change

Once the battery is disconnected, Once the battery is disconnected, QQ has to remain has to remain

constantconstant, since no charge can flow either to or from

the battery. Since , when the

spacing d is doubled, the capacitance C is halved.

And since QQ = = CVCV, that means the voltage must voltage must

doubledouble.

A parallel-plate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If the plate spacing is now doubled (without changing Q), what is the new value of the voltage?

1) 100 V 100 V

2) 200 V200 V

3) 400 V

4) 800 V

5) 1600 V

+Q –Q

dAC 0

The 2 equal capacitors in seriesseries add

up as inversesinverses, giving 1/21/2CC. These

are parallelparallel to the first one, which

add up directlydirectly. Thus, the total

equivalent capacitance is 3/23/2CC.

o

o

C CCCeq

1) 1) CCeq eq = 3/2= 3/2CC

2) 2) CCeq eq = 2/3= 2/3CC

3) 3) CCeq eq = 3= 3CC

4) 4) CCeq eq = 1/3= 1/3CC

5) 5) CCeq eq = 1/2= 1/2CC

What is the equivalent

capacitance, Ceq , of the

combination below?

1) 1) VV11 = = VV22

2) 2) VV11 > > VV22

3) 3) VV11 < < VV22

4) all voltages are zero4) all voltages are zero

CC11 = 1.0 = 1.0 FF CC33 = 1.0 = 1.0 FF

CC22 = 1.0 = 1.0 FF

10 V10 V

The voltage across C1 is 10 V. The

combined capacitors C2 + C3 are

parallel to C1. The voltage across C2

+ C3 is also 10 V. Since C2 and C3 are

in series, their voltages add. Thus the voltage across C2 and C3 each

has to be 5 V, which is less than V1.

How does the voltage V1 across

the first capacitor (C1) compare to

the voltage V2 across the second

capacitor (C2)?

CC11 = 1.0 = 1.0 FF CC33 = 1.0 = 1.0 FF

CC22 = 1.0 = 1.0 FF

10 V10 V

We already know that the voltage

across C1 is 10 V and the voltage

across both C2 and C3 is 5 V each.

Since QQ = = CVCV and C is the samesame for

all the capacitors, we have have VV11 > V > V22

and therefore QQ11 > Q > Q22.

1) 1) QQ11 = = QQ22

2) 2) QQ11 > > QQ22

3) 3) QQ11 < < QQ22

4) all charges are zero4) all charges are zero

How does the charge Q1 on the first

capacitor (C1) compare to the charge

Q2 on the second capacitor (C2)?

when a potential difference of 150 v is applied to the plates of a parallel-plate capacitor, the...

Documents