Electrostatic fieldsece.uprm.edu/~pol/pdf/Electrostatics.pdf · R More Charge distributions Find E from n Point charge (we just saw this one) n Line charge n Surface charge nVolume

Dr. S. Cruz-Pol, INEL 4151-Electromagnetics I

Electrical Engineering, UPRM (please print on BOTH sides of paper) 1

Electrostatic fields Sandra Cruz-Pol, Ph. D. INEL 4151 ch 4 ECE UPRM Mayagüez, PR

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

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

spinning cotton, … n  Art: spray painting n  …

We will study Electric charges:

n Coulomb's Law- n Use when charge distribution is known

n Gauss’s Law – n Use when charge distribution is symmetrical

n Electric Potential (uses scalar, not vectors)

n Use when potential V is known

221

RQkQF =

encS

QSdD =⋅=Ψ ∫!!

∫∞

⋅−=r

ldErV!!

)(

Coulomb’s Law (1785)

n  Force one charge exerts on another

where k= 9 x 109 or k = 1/4πεo ε0=8.85 x 10-12

221

RQkQF =

+ + R

Point charges

*Superposition applies

Force with direction

!F12 =

Q1Q24πεoR

2 â12

â12 =!r2 −!r1!r2 −!r1

Force that Q1 exerts on Q2

Note: Observation point goes First!

Example

!Fx =

14πεo

Q1Qx!rx −!r1( )

!rx −!r13 +

QxQ2!rx −!r2( )

!rx −!r23

⎡

⎣⎢⎢

⎤

⎦⎥⎥

!F = 9

5 −1,−3,3( )(1+ 9+ 9)⎡⎣

⎤⎦3 −2 4,−3,2( )156.2

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟= −1.004,−1.285,1.3998( )nN

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:



Electric field intensity

n  Is the force per unit charge when placed in the E field E =

FQ

E = Q4πεoR

2 âRExample: Same 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

41

rrrrQ

rrrrQE

x

x

x

x

oπε

!= −1.0,−1.29,1.4( )V /m

If we have many charges

dvQv

v∫= ρ

Line charge density, ρL

C/m

Surface charge density ρS

C/m2

Volume charge density ρv

C/m3

dSQS

S∫= ρdlQL

L∫= ρ

E = FQ

E = Q4πεoR

2 âR

The total E-field intensity is

E =ρLdl( )4πεoR

2∫ âR

E =ρSdS( )4πεoR

2∫ âR

E =ρvdv( )4πεoR

2∫ âR

Epoint charge

=Q

4πεoR2 âR More Charge distributions

Find E from n  Point charge (we just saw this one) n  Line charge n  Surface charge n  Volume charge

Results Preview

!E = ρS

2εoân

!E = Q

4πεor2 âr

!E= ρL

2πεoρâρLine charge

Sheet charge

Volume Charge

We will derive these 3 cases!!1. Using Coulomb!

2. Using Gauss!

Find E from LINE charge n  Line charge w/uniform

charge density, ρL *use cylindrical coordinates

dlQB

A L∫= ρ

x

z

B

A

R

dl

dE

α

Ro

L aRdzE ˆ

4'2∫= πε

ρ

0

T

αρ tan' −=OTz

zRRRR

âsinâcossec

αα

αρ

ρ +=

=!

'dzdl =

(0,0,z’)

(x,y,z)

zR RR âsinâcosâ αα ρ +==!



Defining angles α1 and α2 α1 =imaginary perpendicular line with the back α2=imaginary perpendicular line with the front

x

z

B

A

dl

dE

0

T α 2

α 1

LINE charge n  Substituting in:

Ro

L aRdzE ˆ

4'2∫= πε

ρ

z

ααρ

αρ

ddzOTz

]sec0['tan'

2−=

−=

αρ sec=R

x

B

A

R

dl

α

0

T

zR âsinâcosâ αα ρ +=

!E = ρL[−ρ sec

2α]dα4πεoρ

2 sec2αα1

α2

∫ [cosα âρ + sinα âz ]

finite Line Charge:!E = ρL

4πεoρ[−(sinα2 − sinα1 )âρ + (cosα2 − cosα1 )âz ]

ρρπερ

α

â2

)90( Charge Line infinite o1,2

o

LE =

= ∓

More Charge distributions

n  Point charge n  Line charge n  Surface charge n  Volume charge

Find E from Surface charge

n  Sheet of charge w/uniform density ρS

dSdQ Sρ=

y zhR â)â( +−= ρρ

RR

R

!=â

ρφρ dddS =

[ ][ ] 2322

zρ

4

ââ

h

hdddE

o

S

+

+−=

ρπε

ρρφρρ

Ro

S aRdSdE ˆ

4 2περ

=

z

Observation point is at z-axis:

Element of area is:

SURFACE charge n  Due to SYMMETRY the ρ component cancels

out. [ ]∫∫

∞

== +=

0 2322

2

04 ρπ

φ ρ

ρρφ

περ

h

dhdEo

Sz

no

SE â2

:Charge Surface infinite

ερ

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+

−∞

=22

24

0hh

o

SzE

ρπ

περ

More Charge distributions

n  Point charge n  Line charge n  Surface charge n  Volume charge



Find E from Volume charge n  Sphere of charge w/

uniform density, ρv dvdQ vρ=

x

Law of cosines:

R2 = z2 + r '2− 2zr 'cosθ 'r '2 = z2 + R2 − 2zRcosα

'''sinzrRdRd =θθ

.cos

symmetry toDue

survivesdEdE

only

z α=

Ro

v aRdvdE ˆ

4 2περ

=

θ’

φ’

(r’,θ’,φ’)

P(0,0,z)

ρv

dE

(Eq. *) α

Differentiating (Eq. *)

''''sin'2 drddrdv φθθ=R

z

Find E from Volume charge n  Substituting…

zo

vz aR

dvdE ˆcos4 2

απερ

=

x

'''sinzrRdRd =θθ

''''sin'2 drddrdv φθθ=

θ’

φ’

(r’,θ’,φ’)

P(0,0,z)

ρv

dE

2

222'

'

2

0'

2

0

12

'''

''4 RzR

rRzdrzrRdRrdE

rz

rzR

a

ro

vz

−+= ∫∫∫

+

−===

π

φ

φπερ

E = 14πεoz

243πa3ρv

⎛

⎝⎜

⎞

⎠⎟ âz =

Q4πεor

2 âr

De donde salen los limites de R?

R r '2 = z2 + R2 − 2zRcosα

P.E. 4.5 n  A square plate at plane z=0 and

carries a charge mC/m2 . Find the total charge on the plate and the electric field intensity at (0,0,10).

2,2 ±≤±≤ yx

dyydxdyydxQyxyx∫∫∫∫=−=−=−=

==2

0

2

2

2

2

2

2

)2(1212

y12

mC1922)2(124

2

0

2y⋅=

!E = ρs

4πεor2 dS∫ âr 3'

'4 rr

rrdS

o

s!!

!!

−

−= ∫ πε

ρ

)10,','()0,','()10,0,0(' yxyxrr −−=−=− !!

dSQS

S∫= ρ

10 --

Cont…

( ) 2/3222

2

2

2 100)10,,(

412

++

−−= ∫ ∫

− −= yxyxdxdyyE

y oπε

!

( ) ( ) ( ) ⎥⎥⎦

⎤

⎢⎢⎣

⎡

+++

++

−+

++

−⋅= ∫ ∫ ∫∫ ∫ ∫

− − −=−= − −=

2

2

2

2

2

22/322

2

2

2

2

2

22/3222/322

6

100

ˆ10

100

ˆ

100

ˆ10108

x

z

x x

yx

yx

adxdyyy

yx

adxdyyy

yxaxdxdyy

sheet of charge

x=2

y=2

z

Due to symmetry only Ez survives:

( )mMVa

yxaydxdyE

z

z

/ˆ5.16100

ˆ102101082

2

2

02/322

6

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++⋅⋅= ∫ ∫

−

!

Chapter Outline

n Coulomb's Law- n Use when charge distribution is known

n Gauss’s Law – n Use when charge distribution is symmetrical

n Electric Potential (uses scalar, not vectors)

n Use when potential V is known

221

RQkQF =

encS

QSdD =⋅=Ψ ∫!!

∫∞

⋅−=r

ldErV!!

)(

Electric Flux Density

!D = εo

!E = ρvdv

4πR2∫ âR [C /m2 ]

Then the electric flux is:

Ψ =!D ⋅d!S∫ [C]

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



Gauss’s Law

∫∫

∫∫

∫

⋅∇=⋅

⋅==

=⋅=Ψ

vS

Svenc

encS

dvDSdD

SdDdvQ

QSdD

!!!

!!

!!

ρ

Dv!

⋅∇=ρThis is the 1st of the Maxwell’s equations derived here."

Therefore:

Gauss’s Law

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

∫∫

∫=⋅==Ψ

==

vv

Senc

Rv

o

dvSdDQ

aRdvED

ρ

πρ

ε!!

!!ˆ

4 2

The key is to choose the Gauss surface to simplify the problem."Follow the symmetry of the particular case. "Pick surface so that D is ⊥

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

Point Charge is at the origin.

n  Choose a spherical dS n  Note where D is perpendicular to

this surface. 24 rDdSDQ r

Sr π== ∫

∫ ⋅=S

SdDQ!!

D

P

r

charge

rarQD ˆ4 2π

=

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

Infinite Line Charge

n  Choose a cylindrical dS n  Note that integral =0 at top and

bottom surfaces of cylinder

∫ ⋅==S

l SdDQdl!!

ρ

Dρ

P

Line

charge

lDdSDQS

πρρρ 2== ∫ρπρ

ρ aD L ˆ2

=!

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

n  Infinite Sheet of charge

n  Choose a cylindrical box cutting the sheet

Note that D is parallel to the

sides of the box.

∫∫ ⋅==S

s SdDQdS!!

ρ

⎥⎥⎦

⎤

⎢⎢⎣

⎡+== ∫∫bottomtop

sS dSdSDQAρ

zS aD ˆ2ρ

=!

[ ]AADA sS +=ρ

sheet of charge

D

DArea A

P.E. 4.7

A point charge of 30nC is located at the origin, while plane y=3 carries charge 10nC/m2.

Find D at (0, 4, 3)

ns

rQ aarQDDD ˆ

2ˆ

4 2ρ

πρ+=+=

!!!

( )[ ] ya

nD ˆ210)0,0,0()3,4,0(

344

10303

22

9

+−+

⋅=

−

π

!

( )2

3

9

nC/m ˆ057.0ˆ08.5

ˆ5)3,4,0(54

1030

zy

y

aa

anD

+=

+⋅

=−

π

!



P.E. 4.8 If C/m2 . Find : n  volume charge density at (-1,0,3)

n  Flux thru the cube defined by

n  Total charge enclosed by the cube

( ) zyx axaxyazyD ˆˆ4ˆ2 2 +++=!

CQ 2=Ψ=

10,10,10 ≤≤≤≤≤≤ zyx

∫ ∫ ∫∫ ===Ψ1

0

1

0

1

0

4 dzdydxxdvQv

venc ρ

3C/m44)3,0,1( −==⋅∇=− xDv!

ρ

Review

nS aD ˆ2ρ

=!

rarQD ˆ4 2π

=

ρπρρ aD L ˆ2

=!

Point charge or volume

Charge distribution

Line charge distribution

Sheet charge distribution

We will study Electric charges:

n Coulomb's Law (general cases) n Gauss’s Law (symmetrical cases) n Electric Potential (uses scalar, not vectors)

Electric Potential, V n  The work done to move a charge Q from A to B is

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

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

yâ dlEQ

dlFdW

⋅−=

⋅−=!

!

∫ ⋅−=B

A

ldEQW!!

[ ]VCJ =⎥⎦⎤

⎢⎣

⎡⋅−== ∫B

AAB ldEQ

WV!!

V is independent of the path taken.

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

For many Point charges at rk: (apply superposition)

For Line Charges: For Surface charges:

For Volume charges: ( )

∫ −= vv

o rrdvrrV'ˆˆ''ˆ

41)ˆ( ρπε

[ ]V ˆˆ

Q4

1 )(n

1k

k∑= −

=ko rr

rVπε

( )∫ −= L

L

o rrdlrrV'ˆˆ''ˆ

41)ˆ( ρπε

[ ]V 4

'

14

ˆ'ˆ'4

)( 2 rQ

rQadra

rQldErV

o

r

or

r

ro

r

πεπεπε==⋅−=⋅−=

∞∞∞∫∫

!!

( )∫ −= S

s

o rrdSrrV'ˆˆ''ˆ

41)ˆ( ρπε

P.E. 4.10 A point charge of -4µC is located at (2,-1,3) A point charge of 5µC is located at (0,4,-2) A point charge of 3µC is located at the origin Assume V(∞)=0 and Find the potential at (-1, 5, 2)

Crr

QrVk ko

k +−

=∑=

3

1 4)(

πε

⎥⎦

⎤⎢⎣

⎡++

−

⋅=−

=−−=−

=−−−=−

=−−−=−

−

303

185

464

109/110)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



Example A line charge of 5nC/m is located on line x=10, y=20 Assume V(0,0,0)=0 and Find the potential at A(3, 0, 5)

ρρ ρρπερ adaldErVo

L ˆˆ2

)ˆ( ∫∫ ⋅−=⋅−=!

ρ0=|(0,0,0)-(10,20,0)|=22.36 and ρA=|(3,0,5)-(10,20,0)|= 21.2

[ ]

8.40

lnln2

ln2

)ˆ(

−=−

−−=−

+−=

A

Aoo

LAorigin

o

L

V

VV

CrV

ρρπερ

ρπερ

VA=+4.8V

P.E. 4.11 QUIZ #2: A point charge of 7nC is located at the origin

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

Cr

QVo

+=πε4

P.E. 4.11 A point charge of 5nC is located at the origin V(0,6,-8)=2V and Find the potential at A(-3, 2, 6)

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

Cr

QVo

+=πε4 ( ) C

nr

o

+=

=−−=

10452

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

πε5.2−=∴C

VCnVo

A 93.3)0,0,0()6,2,3(45

=+−−

=πε

VnVo

B 696.25.2)0,0,0()7,5,1(45

=−−

=πε

VVVV ABAB 233.1−=−=

Relation between E and V

0=⋅=+

−=

∫ ldEVVVV

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:

dzEdyEdxEldEdV

zyx −−−=

⋅−=!

0=×∇ E!

dzzVdy

yVdx

xVdV

∂

∂+

∂

∂+

∂

∂=

A

B Condition for Conservative field = independent of path of integration

VE −∇=

Example Given the potential Find D at .

φθ cossin102rV =

( )VED oo ∇−== εε!!

⎟⎠

⎞⎜⎝

⎛ 0,2,2 π

2/ˆ1.22 mCaD r=!

⎥⎦

⎤⎢⎣

⎡

∂

∂+−−= φθ φ

θφθφθ aVr

ar

ar

E r ˆsin10ˆcoscos10ˆcossin20 333

!

In spherical coordinates:

⎟⎠

⎞⎜⎝

⎛ +−+== φθπεε aaaED roo ˆ0ˆ0ˆ820

)0,2/,2(

!!

⎥⎦

⎤⎢⎣

⎡

∂

∂+

∂

∂+

∂

∂−= φθ φθθ

aVr

aVr

arVE r ˆsin

1ˆ1ˆ!



P.E. 4.12 Given that E=(3x2+y)ax +x ay kV/m, find the work done in moving a -2µC charge from (0,5,0) to (2,-1,0) by taking the straight-line path.

( )[ ]∫ ∫ ++=⋅=− xdydxyxdlEQW 23

( ) ∫∫−

+=− 1

5

2

0

23 xdydxyxQW

( )[ ]∫ ∫ =−+−+=⋅=− )3(353 2 dxxdxxxdlEQW

( )∫ +−=− 2

0

2 563 dxxxQW

)1218)(( −−= QW

610128 =+−=−QW

a) (0,5,0)→(2,5,0) →(2,-1,0)

b) y = 5-3x dxdy 3−=

mJW 12)2(6 µ−=

mJW 12=

Electric Dipole

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

⎥⎦

⎤⎢⎣

⎡ −=⎥

⎦

⎤⎢⎣

⎡−=

21

12

21 411

4 rrrrQ

rrQV

oo πεπε

2

cos4 r

dQVo

θπε

=

P

y

r1

r2 r

z

d

Q+

Q-

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

Energy Density in Electrostatic fields

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

∫∫ =⋅=v

o

vE dvEdvEDW

2

221 ε!!

Electrostatic fieldsece.uprm.edu/~pol/pdf/Electrostatics.pdf · R More Charge distributions Find E from n Point charge (we just saw this one) n Line charge n Surface charge nVolume

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