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

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%

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

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

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

AqqEF

o

71.43.01085.82

105

2

12

26

2

The electric field = σ/2εo why?

dAC o /

q

-q

q'

q

q

-q'

-q

-q

V

V

V

V'

In 1837 Michael Faraday investigated what happens to thecapacitance 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 knownas 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 constantIn 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 batteryThis is shown in fig.b

q2.

airC C

(25 - 15)

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 airq κq q C κ κCV 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

V VV

Fig.b : Capacitor charge q remains constant

citance / air

q q qC CV V V

(25 - 16)

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

dAC o

Effect of a dielectric : C κC

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

L

CdxdC

LxA

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)

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

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

In general a capacitor systemmay consist of smaller capacitorgroups that can be identified asconnected "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 cC 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)

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.

• 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

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

Circuits

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

summary

• Capacitance• Parallel plates, coaxial

cables, Earth• Series and parallel

combinations• Energy in a capacitor• Dielectrics• Dielectric strength

CVQ

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)/ln(/2/

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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