Electrostatic Electrostatic fields fields

Sandra Cruz-Pol, Ph. Sandra Cruz-Pol, Ph. D.D.INEL 4151 ch 4INEL 4151 ch 4ECE UPRMECE UPRMMayagüez, PRMayagüez, PR

Some applicationsSome applications Power transmission, X rays, lightning protectionPower transmission, X rays, lightning protection Solid-state Electronics: resistors, capacitors, FETSolid-state Electronics: resistors, capacitors, FET Computer peripherals: touch pads, LCD, CRTComputer peripherals: touch pads, LCD, CRT Medicine: electrocardiograms, electroencephalograms, Medicine: electrocardiograms, electroencephalograms,

monitoring eye activitymonitoring eye activity Agriculture: seed sorting, moisture content monitoring, Agriculture: seed sorting, moisture content monitoring,

spinning cotton, …spinning cotton, … Art: spray paintingArt: spray painting ……

We will study Electric We will study Electric charges:charges:

Coulomb's Law-Coulomb's Law- Force between e chargesForce between e charges

GaussGauss’’s Law s Law – – Use when Use when charge distributioncharge distribution is is

symmetricalsymmetrical Electric PotentialElectric Potential (uses scalar, (uses scalar,

not vectors)not vectors) Use when Use when potential Vpotential V is known is known

CoulombCoulomb’’s Law (1785)s Law (1785) Force one charge exerts on another Force one charge exerts on another

where where kk= 9 = 9 xx 10 1099

or or kk = 1/4 = 1/4oo

221

R

QkQF

+ +R

Point charges

*Superposition applies

Force with directionForce with direction

12221

12 ˆ4

aR

QQF

o

ExampleExample

3998.1,285.1,004.12.156

4,6,8

81.82

15,15,59

4

13

2

223

1

11

F

rr

rrQQ

rr

rrQQF

x

xx

x

xx

o

Example: Point charges 5nC and -2nC are located at r1=(2,0,4) and r2=(-3,0,5), respectively.

a) Find the force on a 1nC point charge, Qx, located at (1,-3,7)

Apply superposition:

12221

12 ˆ4

aR

QQF

o

Electric Field intensityElectric Field intensity

Is the force per unit Is the force per unit charge when placed in charge when placed in the E fieldthe E field

Example: Point charges 5nC and -2nC are located at (2,0,4) and (-3,0,5), respectively.

b) Find the E field at rx=(1,-3,7).

3

2

223

1

11

4

1

rr

rrQ

rr

rrQE

x

x

x

x

o

[V/m]

If we have many chargesIf we have many charges

dvQv

v

Line charge density, Line charge density,

LL

C/mC/m

Surface charge densitySurface charge density

SS

C/mC/m22

Volume charge densityVolume charge density

vv

C/mC/m33

dSQS

S dlQL

L

The total E-field intensity The total E-field intensity isis

Ro

v

Ro

S

Ro

L

aR

dvE

aR

dSE

aR

dlE

ˆ4

ˆ4

ˆ4

2

2

2

More Charge More Charge distributionsdistributions

Find E fromFind E from Point charge (we just saw this one)Point charge (we just saw this one) Line chargeLine charge Surface chargeSurface charge Volume chargeVolume charge

Find E from LINE Find E from LINE chargecharge

Line charge Line charge w/uniform charge w/uniform charge density, density, LL

dlQB

A L

x

z

B

A

R

dl

dE

Ro

L aR

dzE ˆ

4

'2

0

T

tan' OTz

zRRR

R

asinacos

sec

'dzdl

(0,0,z’)

(x,y,z)

zR R

Rasinacosa

Using

Coulomb’s Law

LINE chargeLINE charge Substituting in: Substituting in:

Ro

L aR

dzE ˆ

4

'2

x

z

B

A

R

dl

dE

0

T

ddz

OTz

]sec0['

tan'2

secR

(0,0,z’)

(x,y,z)zR asinacosa

]asina[cossec4

]sec[22

22

zo

L dE

]a)cos(cosa)sin(sin[4

:Charge Line finite

1212 zo

LE

a2

)90( Charge Line infinite o1,2

o

LE

Using

Coulomb’s Law

More Charge More Charge distributionsdistributions

Point chargePoint charge Line chargeLine charge Surface chargeSurface charge Volume chargeVolume charge

Find E from Surface Find E from Surface charge charge

Sheet of charge Sheet of charge w/uniform density w/uniform density SS

dSdQ S

yzhR a)a(

R

RR

a

dddS

2

322

zρ

4

aa

h

hdddE

o

S

Ro

S aR

dSdE ˆ

4 2

z

Usin

g

Coulomb’s Law

SURFACE chargeSURFACE charge Due to Due to SYMMETRYSYMMETRY

the the component component cancels out.cancels out.

0 2322

2

04

h

dhdE

o

Sz

no

SE a2

:Charge Surface infinite

22

24

0h

ho

SzE

Using

Coulomb’s Law

More Charge More Charge distributionsdistributions

Point chargePoint charge Line chargeLine charge Surface chargeSurface charge Volume chargeVolume charge

Find E from Volume Find E from Volume chargecharge

sphere of charge sphere of charge w/uniform density, w/uniform density, vv

dvdQ v

x

cos2'

'cos'2'

:cosines of Law

222

222

zRRzr

zrrzR

'''sin

zr

RdRd

.

cos

symmetry toDue

survives

dEdE

only

z

Ro

v aR

dvdE ˆ

4 2

’

’

(r’,’’

P(0,0,z)

v

dE

(Eq. *)

Differentiating (Eq. *)

''''sin'2 drddrdv

Using

Coulomb’s Law

Find E from Find E from Volume chargeVolume charge

Substituting… Substituting… zo

vz a

R

dvdE ˆcos

4 2

x

'''sin

zr

RdRd

''''sin'2 drddrdv

’

’

(r’,’’

P(0,0,z)

v

dE

2

222'

'

2

0'

2

0

1

2

''

'''

4 RzR

rRzdr

zr

RdRrdE

rz

rzR

a

ro

vz

ro

zo

v ar

Qa

r

aE ˆ

4ˆ

3 22

3

De donde salen los limites de R?

Using

Coulomb’s Law

Electric Flux DensityElectric Flux Density

][

:isflux electric Then the

]/[ˆ4

22

CSdD

mCaR

dvED R

vo

D is independent of the medium in which the charge is placed.

GaussGauss’’s Laws Law

vS

S

venc

enc

S

dvDSdD

SdDdvQ

QSdD

Dv

GaussGauss’’s Laws Law

The The total electric total electric flux flux ,, through any through any closed surface is closed surface is equal to the equal to the total total charge enclosedcharge enclosed by by that surface.that surface.

v

v

S

enc

Rv

o

dvSdDQ

aR

dvED

ˆ

4 2

Some examples:Some examples: Finding Finding D D at point P from the charges:at point P from the charges:

Point Charge is at the Point Charge is at the origin.origin.

Choose a spherical dSChoose a spherical dS Note where D is Note where D is

perpendicular to this perpendicular to this surface.surface.

24 rDdSDQ r

S

r

S

SdDQ

D

P

r

charge

rar

QD ˆ

4 2

Some examples:Some examples: Finding Finding D D at at point P from the charges:point P from the charges:

Infinite Line ChargeInfinite Line Charge

Choose a cylindrical dSChoose a cylindrical dS Note that integral =0 at Note that integral =0 at

top and bottom surfaces top and bottom surfaces of cylinderof cylinder

S

l SdDQdl

D

P

Line

charge

lDdSDQS

2

aD L ˆ

2

Some examples:Some examples: Find Find DD at at point P from the charges:point P from the charges:

Infinite Sheet of chargeInfinite Sheet of charge

Choose a cylindrical Choose a cylindrical box cutting the sheetbox cutting the sheet

Note that D is parallel to Note that D is parallel to the sides of the box.the sides of the box.

S

s SdDQdS

bottomtop

sS dSdSDQA

zS aD ˆ

2

AADA sS

sheet of charge

D

DArea A

P.EP.E. .

4.74.7

A A pointpoint charge of 30nC is located at the origin, charge of 30nC is located at the origin, while while planeplane y=3 carries charge 10nC/m y=3 carries charge 10nC/m22..

Find Find DD at (0, 4, 3) at (0, 4, 3)

ns

rQ aar

QDDD ˆ

2ˆ

4 2

2

3

9

nC/m ˆ057.0ˆ08.5

ˆ5)3,4,0(54

1030

zy

y

aa

anD

P.E. 4.8P.E. 4.8If C/m2 . If C/m2 . FindFind

:: volume charge density at (-1,0,3)volume charge density at (-1,0,3)

Flux thru the cube defined by Flux thru the cube defined by

Total charge enclosed by the cubeTotal charge enclosed by the cube

zyx axaxyazyD ˆˆ4ˆ2 2

CQ 2

10,10,10 zyx

1

0

1

0

1

0

4 dzdydxxdvQv

venc

3C/m44)3,0,1( xDv

ReviewReview

nS aD ˆ

2

rar

QD ˆ

4 2

aD L ˆ2

Point charge or volume

Charge distribution

Line charge distribution

Sheet charge distribution

We will study Electric We will study Electric charges:charges:

Coulomb's Law (general cases)Coulomb's Law (general cases) GaussGauss’’s Law (symmetrical cases)s Law (symmetrical cases) Electric PotentialElectric Potential (uses scalar, (uses scalar,

not vectors)not vectors)

Electric Potential, VElectric Potential, V The work done to move a charge The work done to move a charge QQ from from AA to to B B isis

The (-) means the work is done by an external force.The (-) means the work is done by an external force. The total work= potential energy required in The total work= potential energy required in

moving Q:moving Q:

The energy per unit charge= potential difference The energy per unit charge= potential difference between the 2 points:between the 2 points:

ya dlEQ

dlFdW

B

A

ldEQW

VC

J

B

A

AB ldEQ

WV

V is independent of the path taken.

The The PotentialPotential at any point is the potential at any point is the potential difference between that point and a difference between that point and a chosen reference point at which the chosen reference point at which the potential is zero. (choosing infinity):potential is zero. (choosing infinity):

For many For many Point charges at Point charges at rrkk::

(apply superposition)(apply superposition)

For For Line ChargesLine Charges::

For For Surface chargesSurface charges::

For For Volume chargesVolume charges::

v

v

o rr

dvrrV

'ˆˆ''ˆ

4

1)ˆ(

V ˆˆ

Q

4

1 )(

n

1k

k

ko rr

rV

L

L

o rr

dlrrV

'ˆˆ''ˆ

4

1)ˆ(

V 4

'

1

4 ˆ'ˆ

'4)(

2 r

Q

r

Qadra

r

QldErV

o

r

or

r

ro

r

S

s

o rr

dSrrV

'ˆˆ''ˆ

4

1)ˆ(

P.E. P.E. 4.104.10

A point charge of -4A point charge of -4C is located at (2,-1,3)C is located at (2,-1,3)

A point charge of 5A point charge of 5C is located at (0,4,-2)C is located at (0,4,-2)

A point charge of 3A point charge of 3C is located at the originC is located at the origin

Assume V(∞)=0 and Assume V(∞)=0 and Find the potential at (-1, Find the potential at (-1, 5, 2)5, 2)

Crr

QrV

k ko

k

3

1 4)(

30

3

18

5

46

4

109/1

10)2,5,1(

30)0,0,0()2,5,1(

18)2,4,0()2,5,1(

46)3,1,2()2,5,1(

9

6

3

2

1

V

rr

rr

rr

=10.23 kV

ExampleExampleA line charge of 5nC/m is located on line x=10, y=20A line charge of 5nC/m is located on line x=10, y=20

Assume V(0,0,0)=0 and Assume V(0,0,0)=0 and Find the potential at A(3, 0, 5)Find the potential at A(3, 0, 5)

adaldErV

o

L ˆˆ2

)ˆ(

0=|(0,0,0)-(10,20,0)|=22.36 and A=|(3,0,5)-(10,20,0)|= 21.2

8.40

lnln2

ln2

)ˆ(

A

Aoo

LAorigin

o

L

V

VV

CrV

VA=+4.8V

P.E. P.E. 4.114.11QUIZ #2: A point charge of 7nC is QUIZ #2: A point charge of 7nC is

located at the located at the originorigin

V(0,3,-5)=2V and Find CV(0,3,-5)=2V and Find C

Cr

QV

o

4

P.E. P.E. 4.114.11

A point charge of 5nC is located at the A point charge of 5nC is located at the originorigin

V(0,6,-8)=2V and V(0,6,-8)=2V and Find the potential at A(-3, 2, 6)Find the potential at A(-3, 2, 6)

Find the potential at B(1,5,7), the potential difference Find the potential at B(1,5,7), the potential difference VVABAB

Cr

QV

o

4 C

n

r

o

104

52

10)8,6,0()0,0,0(

5.2C

VCn

Vo

A 93.3)0,0,0()6,2,3(4

5

Vn

Vo

B 696.25.2)0,0,0()7,5,1(4

5

VVVV ABAB 233.1

Relation between E and Relation between E and VV

0

ldEVV

VV

BAAB

BAAB

A

B

*Esto aplica sólo a campos estáticos.

Significa que no hay trabajo NETO en mover una carga en un paso cerrado donde haya un campo estático E.

0 SdEldES

V is independent of the path taken.

Static E satisfies:Static E satisfies:

dzEdyEdxE

ldEdV

zyx

0 E

dzz

Vdy

y

Vdx

x

VdV

A

BCondition for Conservative field = independent of path of integration

VE

ExampleExample Given the potential Given the potential

Find D at . Find D at .

cossin10

2rV

VED oo

0,2

,2

2/ˆ1.22 mCaD r

a

V

ra

ra

rE r ˆsin

10ˆcoscos

10ˆcossin

20333

In spherical coordinates:

aaaED roo ˆ0ˆ0ˆ8

20)0,2/,2(

a

V

ra

V

ra

r

VE r ˆ

sin

1ˆ

1ˆ

P.E. P.E. 4.124.12

Given that Given that E=(3xE=(3x22+y)a+y)axx +x a +x ayy kV/m, kV/m, find the work donefind the work done in in moving a -2moving a -2C C charge from (0,5,0) to (2,-1,0)charge from (0,5,0) to (2,-1,0) by by taking the straight-line path.taking the straight-line path.

xdydxyxdlEQ

W 23

1

5

2

0

23 xdydxyxQ

W

)3(353 2 dxxdxxxdlEQ

W

2

0

2 563 dxxxQ

W

)1218)(( QW

610128 Q

W

a) (0,5,0)→(2,5,0) →(2,-1,0)

b) y = 5-3xdxdy 3

mJW 12)2(6

mJW 12

Electric DipoleElectric Dipole

Is formed when 2 point charges of equal Is formed when 2 point charges of equal but opposite sign are separated by a small but opposite sign are separated by a small distance.distance.

21

12

21 4

11

4 rr

rrQ

rr

QV

oo

2

cos

4 r

dQV

o

P

y

r1

r2

r

z

d

Q+

Q-

For far away observation points (r>>d):

Energy Density in Energy Density in Electrostatic fieldsElectrostatic fields

It can be shown It can be shown that the total that the total electric work done electric work done is: is:

v

o

v

E dvEdvEDW 2

22

1