Lecture 10

2. The use of dielectrics

1. Capacitance and E-field energy

Learning Objectives

•To calculate the energy stored in a capacitor

•To derive an expression for the energy

per unit volume

•To obtain an expression for the force

between the two plates of a parallel plate capacitor

•To introduce the use of dielectric materials

Learning Objectives

•To calculate the energy stored in a capacitor

•To derive an expression for the energy per unit volume

•To obtain an expression for the force between the two plates of a parallel plate capacitor

•To introduce the use of dielectric materials

Energy Stored in a Capacitor

Transfer dq between the two conductors whilst at a potential difference of Vq

qdqVdW == Done Work

∫ ∫==Q

qdqVdW0

work WTotal

qCVq = qCdVdq =∴

UCVdVCVWV

qq ===∫ 2

0 21

V

QC =

C

QQVCVU

22

2

1

2

1

2

1 ===

Stored in the electric field between the “plates” of a capacitor

Consider a parallel plate capacitor

Edand Vd

AC == 0ε

2202

2

1

2

1dE

d

ACVU

ε==

( ) 20

20 region field of volume

2

1

2

1EdAEU εε ==

d

€

ρ E( ) =1

2ε0E 2

Energy Density (energy stored per unit volume)

Note this is generally valid

€

U = ρ E( )dV∫

What is the Force Experienced by a Capacitor Plate?

Cannot use Coulomb’s law directly(integration very complicated.)

Easiest approach is to use potential energy

dx

dUForce −=F Use (STMR)

( ) 202

1AxExU ε=

€

∴F = −dU

dx= -

1

2ε0AE 2

€

Using E =σ

ε0

=Q

Aε0

,

QEF2

1−=

€

F = −1

2ε0AE 2 = −

1

2ε0AE

Q

Aε0

= −1

2QE

QEF2

1−=

Why is it negative?An attractive force between the plates

€

F = −dU

dx

€

dU

dx=

1

2ε0AE 2

€

F = −1

2ε0AE 2 = −

1

2ε0A

V

d2

2

€

F = −1

2ε0AE 2 = −

1

2ε0A(

Q2

A2ε02 ) = −

1

2(

Q2

Aε0

)

Summary on electrostatics

Point charge: E ,V, U

Electric dipole: E and V

Infinite plane of charge: E

Charged cylinder, Charged sphere (solid as well as shell)

Capacitors in planar geometry, cylindrical geometry,and spherical geometry

The second half of the lecture module: magnetism

24-5Dielectrics

Review and Summary•Electrostatic potential energy U of a charged capacitor, is given by

C

QQVCVU

22

2

1

2

1

2

1 ===

•U represents the work required to charge the capacitor

•Stored energy density with an electric field

( ) 202

1 EE ερ =

Dielectrics

E-Field in a DielectricInduced Surface ChargePolarisation inside a Dielectric

Note that this material in non-examinable, but it will be needed for your second year course

What is a dielectric?

A dielectric is an insulator – e.g. air, glass, paper, teflon

Where is it used?It is placed between conducting surfaces

What are the functions of a dielectric?

•To keep the surfaces physically separated

•To raise the C of a capacitor

•To reduce the chance of electric breakdown. Larger V possible (more charge and energy stored)

+ + + + + + +

_ _ _ _ _ _ _

Consider a parallel plate capacitor

Perfect voltmeter

V0

C0

+ + + + + + +

_ _ _ _ _ _ _

Perfect voltmeter

V0

C0

Consider a parallel plate capacitor

+ + + + + + +

_ _ _ _ _ _ _

Perfect voltmeter

Vd < V0

Cd

Charge is the same, ∴ Cd > C0

Define the dielectric constant as

KC

Cd =0

Consider a parallel plate capacitor

The electric permittivity ε is defined by the equation

ε = Kε0

K is often referred to as the relative permittivity

Dielectrics are chosen to have the following properties:

K independent of V or E - linearity

K independent of the orientation of the dielectric - isotropy

K has the same value for every part of the dielectric - homogeneity

Linear, isotropic, homogeneous (LIH) dielectrics

What is the effect of K on electrostatic formulae?

For a LIH dielectric replace ε0 with ε, i.e.

00 εε K→

24-6Molecular View of a Dielectric

Molecular Explanation of Dielectric Behaviour

Schematic representation of a molecule

Molecular Explanation of Dielectric Behaviour

The molecule is polarised

Eind

Polarisation of a dielectric gives rise to thin layers of bound charges on the surfaces.

E0

E = E0/K

iσσ −=Charge Surface Net

dielectric noFor 0εσ

=oE

0

0 dielectric a εσσ i

KE

EWith−

==

00

εσσ

εσ i

K−

=∴

⎟⎠

⎞⎜⎝

⎛ −=K

11i σσ

(Gauss’s law in a dielectric)

( )0

surfaceGaussian inside charges .

εK

freeAdE

surface

∑=∫

Note: conductors have free charge carriers - they are completely polarised in that inside they make E = zero

Review and Summary

•Electrostatic potential energy U of a charged capacitor, is given by

C

QQVCVU

22

2

1

2

1

2

1 ===

•U represents the work required to charge the capacitor

•Stored energy density with an electric field

( ) 202

1 EE ερ =

Review and Summary

Capacitance with a Dielectric

•Capacitance increases by K (dielectric constant)

Gauss’s Law with a Dielectric

0

.εK

QAdE freeencl

surface

−=∫

Next

An introduction to Magnetic Fields and their Properties

Cosmic rays bombard the Earth from outer space and experiments show that they are mostly positively charged protons arriving uniformly over the surface at a rate of 1 proton m-2 s-1 with an average energy of 10000 MeV. By treating the Earth as a conducting sphere of radius 6371 km calculate the rate of change of potential as a result of this bombardment.

At this rate how long would it take for the Earth to charge sufficiently to repel all the cosmic rays?

In view of the facts that the age of the Earth is approximately 4.5 109 years and that cosmic rays are arriving today, comment on your result.

RR

q

r

drqEdrV

RR

Sphere 00

20

4C 44

πεπεπε

=∴=−=−= ∫∫∞∞

Number of protons hitting the Earth per second = 1R4 2 ×πeR

dt

dQ 24π=

1-

00

2

0 s V 1150

4

4 4C .===∴===

επεππε eR

ReR

dtdV

dtdV

CdtdQ

CV QR

Therefore time to reach 1010 volts necessary to deflect protons =

years 2765 years 101431150

107

10

=×× ..

.