electrostatics. outline electric force, electric fields electric flux and gau law electric...

ELECTROSTATICS

Outline

• Electric Force, Electric fields

• Electric Flux and Gaub law

• Electric potential

• Capacitors and dielectric (Electric storage)

The physics of charged objects

• Study of electricity aims to understand the interaction between different charged objects.

+ -

The physics of charged objects

• Study of electricity aims to understand the interaction between different charged objects.

+ +

- -

Structure of Matter

• Fundamental building blocks of the matter are atoms.

+++

++

+ + -

-

-

-

-

--

Structure of Matter

• Neutral atom – electron = Positive ion

+++

++

+ + -

-

-

-

-

- -

C101.602charge electron -191

Structure of Matter

• Fundamental building blocks of the matter are atoms.

+++

++

+ + -

-

-

-

-

--

Structure of Matter

• Neutral atom + electron = negative ion.

+++

++

+ + -

-

-

-

-

--

-

ELECTRICALLY CHARGING OBJECTS

+ - +

+- - +

+ - +

-

-

-

-

+

+

-

+

ELECTRICALLY CHARGING OBJECTS

+ - +

+- - +

+ - +

-

-

-

-

+

+

-

+-

• In metals outer atomic electrons are not bound to any atoms (electron see).

Charging by Induction

++

+ +

+ +

+

++

+


• In metals outer atomic electrons are not bound to any atoms (electron see).

++

+ +

+ +

+

++

+

-

• Same atoms have weakly bound electrons.

Electric Polarization

+++

++

+ +

-

-- -

-

-

-


• Same atoms have weakly bound electrons.

+++

++

+ +

-

-

-

-

-

-

- +

The Electric Force

Coulomb’s Law

• Quantifies the electric force between two charges.

baba QQF

Coulomb’s Law


2

1

baba r

F

Coulomb’s Law


baba

baba

ba

baba r

rQQ

krr

QQF

ˆˆ4

122

0

229

212

/10988.8

/10854.8

CNmk

and

NmC

where

0

Electric Force Field

• Gravitational force field:


+Q


+Q+q


• Definition of Electric field:

q

FE qQ


• Definition of Electric field:

qQrrkQ

EqQ

ˆ2


1r 5r

2r

+

+

+

+

+

3r

4r


• The electric field due to a number of source charges is given by the expression

N

ii

i

i

N

iii

rrq

k

rEE

12

1

ˆ

)(

Electric Force Field(Linear distribution of charge)

dL

dLdq

density charge Linear

?dEr

2rdq

kdE


dL

dLdq


?dE

2rdL

kdE


dL

dLdq


?dE

2r

dLkE

Electric Force Field(Surface distribution of charge)

dadq

ondistributi charge Surface

r?dE

da 2rdq

kdE


dadq


r?dE

da2rda

kdE


dadq


r?dE

da

surface r

dakE

2

Electric Dipole

• Electric dipole consists of a pair of point charges with equal size but opposite sign separated by a distance d.

d+ -

dqp

Electric Dipole


d+ -

Electric Dipole


d+ -

p

Electric Dipole


d+ -

dqp

Electric Dipole

• Water molecules are electric dipoles

+ +

-waterp

Exercise 1 A point charge q = -8.0 nC is located at the

origin. Find the electric field vector at the point x = 1.2 m, y = -1.6 m

m 2.1

-

m 6.1mr 0.2

m 2.1

-

m 6.1mr 0.2

jEiEE yxˆˆ

m 2.1

-

m 6.1mr 0.2

)ˆsinˆ(cos jiEE

m 2.1

-

m 6.1mr 0.2

jCN iCN E ˆ/14ˆ/11

Exercise 2

An electric dipole consists of a positive charge q and negative charge –q separated by a distance 2a, as shown in the figure below. Find the electric field due to these charges along the axis at the point P, which is the distance y from the origin. Assume that y>>a.

q q

r

aa

r

Vector Flux

Vector Flux

• Definition of flux:

Av

Electric Flux

Electric Flux

AdEE

Gaub’s Law

Surface Enclosed

EnclosedQAdE

0

Gaub’s Law

dA EAdEd E

+r

Gaub’s Law

dA EE

+r

Gaub’s Law

dArkQ

E 2

+r

Gaub’s Law

0

22

0

44

Q

rr

QE

+r

Exercise 3dE

Solution

Coulomb’s Law

22 RdAk

Rkdq

dE

Solution

Infinitesimal area of disk

rdrdA 2

Solution

Infinitesimal area of disk

2

2

Rrdrk

dE

Solution

Y-component of E-field element

cos2

cos2Rrdrk

dEdEy

Solution

RL

cos

Solution


RL

Rrdrk

dEy 2

2

Solution


04

1

k

Solution


2/3220 )(2 Lr

rdrLdEy

Solution


0

02/322

02/322

0

2

1

)(

)(2

y

y

E

LLrrdr

Lrrdr

LE

identity the Using

Two Oppositely charged Parallel Plates (Capacitor)


?E

Exercise 4

00

LQAdE

Gauss

0

0

2

2

rE

LrLE

Area rL2cylinder a of

Electric Potential

+Q+q

Electric Potential

+Q

Electric Potential

test

test

QQ on

oodneighboreh its in source due Ppoint aat potential

)( PWork

Electric

Electric Potential

PSource

P

P

test

rdrr

kQ

rdE

rdFQ

Electric

ˆ

1

2

oodneighboreh its in source due Ppoint aat potential

Electric Potential

rQ

rkQ SourceSource

04V(r)

1J/C V 1:Unit

Electric Potential

N

i i

i

rQ

104

1PV

Electric potential at position P due to a system of N source charges is given by:

Electric Potential

• Potential difference:

Electric Potential

b

a

r

r

absourceab

rdE

rrkQrVrVV

11)()(


constantE

V


Ed

rrE

drE

rdEV

ba

r

r

r

r

b

a

b

a

)(

Capacitors and di-electrics

• Capacitors store electric potential energy

Battery

VV

VQ

C

plates across Voltage

plate each on stored ChargeeCapacitanc

)(1/11 FfaradVCC


E

E

AQ

EEE

0

0

resultant



• We can therefore express the voltage across the capacitor plates as follows:

dA

QEdV

0

• Hence

dA

AQd

QVQ

C 0

0

Exercise 3

A parallel-plate capacitor has an area of A = 2 cm2 and a plate separation of d =1mm.

(a) Find its capacitance. (answer = 1.77pF)

(b) If the plate separation of this capacitor is increased to 3 mm, find the capacitance. (answer = 0.59 pF).


• Capacitors in Parallel


• Di-electric material inside a parallel Electric field.


• Di-electric material between parallel capacitor plates.



vacuumelectricdi KCC



vacuumvacuumelectricdielectricdi VCVCQ

constant remains Charge



vacuumKVelectric-diV



vacuumKE

Ed

electric-diE

V Since

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