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

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)

• 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

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.

Circuits

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

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

dAC o /

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)

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)

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

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.

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

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?

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

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

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