Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction...

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Transcript of Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction...

Page 1: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.
Page 2: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

Summary

• More circuits• Forces between

plates of a capacitor• Dielectrics• Energy and the

distinction about constant Q or V.

Page 3: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

C1 = 5μF, C2 = 10 μF, C3 = 2 μFCalculate the equi. Capacitance.

a) 6 μFb) 2 μFc) 4 μFd) 8 μF

Page 4: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

a. Find the equivalent capacitance between points a and b. b. What fraction of the total energy is stored in the 7 µF capacitor , if the circuit has been charged by a 12 volt battery?

a. 12.92 µFb. 46%

Page 5: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

C123 2.4 µF, q = 28.8 µC

C2 C24 = 12 µF

C1234 = 3 µF q =36 µC

Δq = 7.2 µC

C1 = C3 = 8.00 μF, C2 = C4 = 6.00 μF, V = 12VWhen the switch S is closed, how much charge flows through point P

Page 6: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.
Page 7: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.
Page 8: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

Question

• A parallel-plate capacitor has a plate area of 0.3m2 and a plate separation of 0.1mm. If the charge on each plate has a magnitude of 5x10-6 C then the force exerted by one plate on the other has a magnitude of about:

• A. 0 B. 5N C. 9N D. 1 x104 N E. 9 x 105 N

Page 9: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

Question• A parallel-plate capacitor

has a plate area of 0.3m2 and a plate separation of 0.1mm. If the charge on each plate has a magnitude of 5x10-6 C then the force exerted by one plate on the other has a magnitude of about:

A. 0B. 5NC. 0. 9ND. 1 x104 NE. 9 x 105 N

N

A

qqEF

o

71.43.01085.82

105

2

12

26

2

The electric field = σ/2εo why?

Page 10: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

dAC o /

Page 11: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

q

-q

q'

q

q

-q'

-q

-q

V

V

V

V'

In 1837 Michael Faraday investigated what happens to the

capacitance of a capacitor when the gap between the plates

is completely filled with an insulator (a.k.a. dielectri

C

Capacitor with a dielectric

c)

Faraday discovered that the new capacitance is given by :

Here is the capacitance before the insertion

of the dielectric between the plates. The factor is known

as the dielectric co

airairC CC

nstant of the material.

Faraday's experiment can be carried out in two ways:

With the voltage across the plates remaining constant

In this case a battery remains connected to the plates .

This is

V1.

shown in fig.a

With the charge of the plates remaining constant.

In this case the plates are isolated from the battery

This is shown in fig.b

q2.

airC C

(25 - 15)

Page 12: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

q

-q

q'

q

q

-q'

-q

-q

V

V

V

V'

airC C

This is bacause the battery remains connected to the plates

After the dielectric is inserted between the capacitor plates

the plate charge changes from tq

Fig.a : Capacitor voltage V remains constant

o

The new capacitance air

q κq q C κ κC

V V V

q q

This is bacause the plates are isolated

After the dielectric is inserted between the capacitor plates

the plate voltage changes from to

The ne

w capa

VV

V

Fig.b : Capacitor charge q remains constant

citance

/ air

q q qC C

V V V

(25 - 16)

Page 13: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

If the areas are A1 and A-A1.

2112 (

AAd

C o

C123 2.4 µF, q = 28.8 µC

C2 C24 = 12 µF

C1234 = 3 µF q =36 µC

212

d

AC o

Effect of a dielectric : C κC

Page 14: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

The force on a filling dielectric as it is inserted between the parallel plates of a capacitor.

L

Cdx

dC

L

xA

dCCC o

1

1121

x

L

With the battery connected, U1 = ½CV2

With the battery disconnected, U2 = Q2/2C

L

Udx

dUF

11

1

L

Udx

dUF

12

2

With the battery connected, since x is increasing downwards, a negative force is upwards, pushing the dielectric away.With the battery disconnected, the force is positive and pointed downwards, pulling in the dielectric.The force is proportional to (κ-1) and inversely to L.To make the argument simple, we have omitted the fact that the field at the end includes the fringe field which depends on distance between the plates. The result including fringe fields is quantitatively different. See S. Margulies, American Journal of Physics, V. 52, p 515 (1984)

Page 15: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

A questionWhat is the equivalent capacitance between the points A and B?

A. 1 μFB. 2 μFC. 4 μFD. 10μFE. None of these

A B

What would a 10V battery do, i.e. how much charge will it provide, when it is connected across A and B? 40 μC (20 μC) on 2 μF

Page 16: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

HITT

A parallel-plate capacitor has a plate area of 0.2m2 and a plate separation of 0.1 mm. To obtain an electric field of 2.0 x 106 V/m between the plates, the magnitude of the charge on each plate should be:A. 8.9 x 10-7 C B. 1.8 x 10-6 C C. 3.5 x 10-6 C D. 7.1 x 10-6 C E. 1.4 x 10-5 C

Page 17: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.
Page 18: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

In general a capacitor system

may consist of smaller capacitor

groups that can be identified as

connected "in parallel" or

"in series"

More complex capacitor systems

12 1 2

1 2

12 3

In the example of the figure and in fig.a are connected .

They can be substituted by the equivalent capacitor as

shown in fig.b. Cap

in parallel

acitors and in fig.b are c

C C C

C C

C C

123

123123 12 3

onnected .

They can be substituted by a single capacitor as shown in fig.c

is given by the equation:

in series

1 1

1

C

CC C C

(25 - 12)

Page 19: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

All C’s are 8.00 nF. The battery is 12 V.What is the equivalent capacitance?

C12 = 4 nFC123 = 12 nFQ123 = C123 x V = 144 nCQ3 = C3 x V = 96 nCQ12 = C12 x V = 48 nC

U123 = ½ C123V2 = ½ x 12x10-9 x122 = 864 nJU1 = ½ C1V1

2 = ½ x 8x10-9 x62 = 144 nJ = U2

U3 = ½ x 8x10-9 x122 = 576 nF

C3 stores most energy, also the highest electric field and most charge, the most stressed part of the circuit.

Page 20: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

• Series combinations reduce the capacitance. Equal C reduce by the number involved.

• In parallel the capacitance increases.A basket of 4 capacitors, each of C = 6 nF. How

can you arrange them to get a) 1.5 nFb) 2 nF g) 2.4 nFc) 3 nF h) 3.6 nFd) 4 nF i) 4.5 nF j) 6 nFe) 12 nF k) 18 nFf) 24 nF

Page 21: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

clicker

All C’s are 8.00 nF. The battery is 12 V.What is the equivalent capacitance?

a. 4 nFb. 6 nFc. 8 nFd. 10 nFe. 12 nF

Page 22: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.
Page 23: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.
Page 24: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.
Page 25: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.
Page 26: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.
Page 27: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.
Page 28: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

Circuits

• All capacitors being the same, rank the equivalent capacitances of the four circuits.

Page 29: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

summary

• Capacitance• Parallel plates, coaxial

cables, Earth• Series and parallel

combinations• Energy in a capacitor• Dielectrics• Dielectric strength

CVQ

abaab

abC

abL

dAC

oo

o

o

44

)/ln(/2

/

.....111

2121

CCCCCC

C

QCVU

22

1 22

CC

Page 30: Summary More circuits Forces between plates of a capacitor Dielectrics Energy and the distinction about constant Q or V.

A question

• Each of the four capacitors shown is 500 μF. The voltmeter reads 1000V. The magnitude of the charge, in coulombs, on each capacitor plate is:

A. 0.2 B. 0.5 C. 20 D. 50 E. none of these