Ch02_Coulomb's Law and Electric Field Intensity

7/26/2019 Ch02_Coulomb's Law and Electric Field Intensity

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20

21

R 4

QQF

πε=

Chapter 2. CoulombChapter 2. Coulomb’’s Law and Electric Field Intensitys Law and Electric Field Intensity

The Experimental Law of CoulombThe Experimental Law of Coulomb

CoulombCoulomb’ ’ s Laws Law

The magnitude of force between two very small objectsseparated in free space by a distance which is large compared totheir size is given by

Q1 and Q2: positive or negative quantities of charge (coulomb)

R: separation (meters)

ε0: permittivity of free space = 8.854 x 10-12 F/m

= 1/36π x 10-9 F/m

1/4πε0 ≈ 9 x 109



In vector form:In vector form:

If QIf Q11 is inis in rr11 and Qand Q22 is inis in rr22, then, then R R 1212 = r= r22 -- rr11 is the directed lineis the directed line

segment from Qsegment from Q11 to Qto Q22..

FF22 = force experienced by Q= force experienced by Q22::

122 aF2

120

21

R 4

QQ

πε=

aa1212 is the unit vector in the direction ofis the unit vector in the direction of R R 1212 ==|| 12

12

rrrr

−−

211221 aaFF2

210

122

120

21

R 4

QQ

R 4

QQ

πε

=

πε

−=−=

r1

Q1 Q

2

x

y

z

r2

F2

F1

R12



Electric Field IntensityElectric Field Intensity

Consider a chargeConsider a charge QQii located at Plocated at Pii, and another charge Q, and another charge Qtt situated insituated in

the vicinity ofthe vicinity of QQii. The position of Q. The position of Qtt with respect towith respect to QQii isis PPitit. At. At

any positionany position PPitit, Q, Qtt experiences a force due toexperiences a force due to QQii

itt aF2

it0

ti

P4

QQ

πε=

The forceThe force exerted by Qexerted by Qii per unitper unit

charge ischarge is

itt a

F2

it0

i

t P4

Q

Q πε

=

The right side of the equation is a vector field called theThe right side of the equation is a vector field called the electric fieldelectric field

intensity due tointensity due to QQii..

The unit of electric field intensityThe unit of electric field intensity EE is N/C, or V/mis N/C, or V/m

Qi

Qt

Pit

Ft



For a charge Q at the origin and using the spherical coordinateFor a charge Q at the origin and using the spherical coordinate

system, the electric fieldsystem, the electric field EE at any given point is given byat any given point is given by

raE 20r4

Qπε

=

For a charge atFor a charge at rrmm,, EE atat rr isis

m

marrE 2

0 ||4

Q

−πε=

aamm is a unit vector in the directionis a unit vector in the direction

ofof rr -- rrmm

x

y

z

Q

rm

r

Note: If Q is positive,Note: If Q is positive, EE “ “radiatesradiates” ” fromfrom

the charge. If Q is negative,the charge. If Q is negative, EE pointspoints

into the charge.into the charge.



In there an n charges, the electric field atIn there an n charges, the electric field at rr isis

∑= −πε

=n

1i

i2

i0

i

||4

Q)( a

rr

rE



Example:Example:

A 1uC charge is at the origin and a 2uC charge is at P(0,10,0). A 1uC charge is at the origin and a 2uC charge is at P(0,10,0). WhatWhat

is the electric field at Q(6,4,5)? If ais the electric field at Q(6,4,5)? If a – –0.5 uC charge is placed at Q,0.5 uC charge is placed at Q,

how much force would it experience?how much force would it experience?

ROQ R

PQP

Q

1 uC 2 uC

R R OQOQ = <6,4,5>= <6,4,5> R R PQPQ = <6,4,5>= <6,4,5> -- <0,10,0> = <6,<0,10,0> = <6,--6,5>6,5>

775.877546R 222OQ ==++=

849.9975)6(6R 222PQ ==+−+=



zyxzyx

OQOQ 570.0456.0684.0

77.8

546

R αaa

αaaR a

OQ++=

++==

zyx

zyx

PQPQ 508.0609.0609.085.9

566

R αaa

αaaR

a

PQ

+−=

+−

==

V/m53.6623.5384.79

)570.0456.0684.0()77(4

101

zyx

zyx0

6

QO

aaa

aaaE

++=

++πε

×=

−

V/m18.9491.11291.112

)508.0609.0609.0()97(4

102

zyx

zyx0

6

QP

aaa

aaaE

+−=

+−πε

×=

−

V/m71.16068.5975.192 zyx

QPQOQ

aaa

EEE

+−=

=+=



EPQ

EOQ P

-0.5 uC

1 uC 2 uC

EQ

FQ

uN36.8084.2938.96

)71.16068.5975.192(105.0F

:EQFthen,QFESince

zyx

zyx6

Q

aaa

aaa

−+−=

+−×−=

==

−v

v v

r

r



Cathode Ray TubeCathode Ray Tube

The electric fieldsThe electric fields

formed by theformed by theparallel platesparallel plates

deflect the electrondeflect the electron

beam generated atbeam generated at

the back of thethe back of the

tube.tube.



Electric Field Due to a Continuous Charge DistributionElectric Field Due to a Continuous Charge Distribution

DefineDefine ρρvv = volume charge density == volume charge density =

∆∆vv contains the chargecontains the charge ∆∆QQ

The total charge Q isThe total charge Q is

v

Qlim

0v ∆

∆

→∆

ρ==vol

vvol

dvdQQ

Due to an incremental chargeDue to an incremental charge ∆∆Q, the incremental electric fieldQ, the incremental electric field ∆∆E isE is

||||4

v)(

||||4

Q2

0

v2

0 r'r

r'r

r'r

r'

r'r

r'r

r'r ∆E

−

−

−πε

∆ρ=

−

−

−πε

∆=

rr = position of the point in question= position of the point in question

rr’ ’ = location of= location of ∆∆QQ

To get the field at the point, add theTo get the field at the point, add the

contribution of allcontribution of all ∆∆QsQs

r'

rorigin

∆Q

∆∆∆∆Er - r'



IfIf ∆∆vv is shrunk so it will approach zero (N will approach infinity),is shrunk so it will approach zero (N will approach infinity), thethe

summation becomes an integral:summation becomes an integral:

∑= −

−

−πε

∆ρ=

N

1i2

0

v

||||4

v)()(

r'r

r'r

r'r

r'rE

∫−

−

−πε

ρ=

vol2

0

v

||||4

dv)()(

r'r

r'r

r'r

r'rE

||||4

v)(

20

v

r'r

r'r

r'r

r'∆E

−

−

−πε

∆ρ=



Field of a Line ChargeField of a Line Charge

DefineDefine ρρLL = line charge density (C/m)= line charge density (C/m)

Consider a uniform line charge along the zConsider a uniform line charge along the z--axis:axis:

Due to symmetryDue to symmetry, the electric, the electric

fieldfield is a function ofis a function of ρρ and in theand in the

aaρρρρρρρρ direction only.direction only.

dE2

dQ1 = ρ

L dL

dQ2 = ρ

L dL

dE1

z-axis



r

r'R = r - r'

dE

dEρρρρ

dEz

x

y

z

(0, 0, z) dQ = ρL dz

rr == ρρ aaρρρρρρρρ

rr’ ’ = z= z aazz

R R == rr -- rr’ ’ == ρρ aaρρρρρρρρ -- zz aazz

2222

0

L

z

z

)z(4

dz

+ρ

−ρ

+ρπε

ρ=

zρ aadE

SinceSince EE is directed alongis directed along aaρρρρρρρρ only, theonly, the

zz--component may be ignoredcomponent may be ignored whenwhen

solving for the total electric field:solving for the total electric field:

ρ2 /3220

L

)z(4

dzd aE

+ρπε

ρρ=



ρaE2 /322

0

L

)z(4

dzd

+ρπε

ρρ= ∫

∞

∞− +ρπε

ρρ== ρρ aEE

2 /3220

L

)z(4

dz

Evaluate the integral using change of variable:Evaluate the integral using change of variable:

Let z =Let z = ρρ tantan αα -->> dzdz == ρρ secsec22 αα ddαα

ρρρ

ρρ

ρ

aaa

aa

aE

ρπε

ρ=α

ρπε

ρ=αα

ρπε

ρ=

αα

α

ρπε

ρ=α

α+ρ

α

πε

ρ=

αρ+ρπεααρρρ=

π

π−

π

π−

π

π−

π

π−

π

π−

∫

∫∫

∫

0

L2 /

2 /0

L2 /

2 /0

L

2 /

2 /3

2

0

L2 /

2 /2 /32

2

0

L

2 /

2 /2 /3222

0

2

L

2sin

4dcos

4

dsec

sec

4d

)tan1(

sec

4

)tan(4d)sec(

α

ρ

z

tan α = z/ρ



ρaEρπε

ρ=

0

L

2

Note:Note:

1. The electric field due to an infinite line of charge is dire1. The electric field due to an infinite line of charge is directed radiallycted radially

outward or into to the line charge.outward or into to the line charge.

2. The electric field varies inversely with the distance from t2. The electric field varies inversely with the distance from the linehe line

chargecharge3. If the line charge density is positive, the electric field3. If the line charge density is positive, the electric field “ “emanatesemanates” ”

from the charge. If the line charge density is negative, the elfrom the charge. If the line charge density is negative, the electricectric

fieldfield “ “convergeconvergess” ” to the line charge.to the line charge.

Therefore, the electric field due to aTherefore, the electric field due to a uniform uniform line chargeline charge alongalong

the z the z - - axis axis is equal tois equal to



Field of a Sheet of ChargeField of a Sheet of Charge

DefineDefine ρρSS = surface charge density= surface charge density

Due to symmetry, the electricDue to symmetry, the electric

field at a point is not a functionfield at a point is not a function

of y and z, and does not haveof y and z, and does not havecomponents parallel to the ycomponents parallel to the y--zz

plane.plane.

To simplify the solution, treatTo simplify the solution, treat

the vertical strip as athe vertical strip as a uniformuniformline charge. The sheet ofline charge. The sheet of

charge may now be regarded ascharge may now be regarded as

an infinite number of linean infinite number of line

charges placed beside eachcharges placed beside each

other.other.

(x, 0, 0)

y

dy

ρs

Ex

If the surface charge density isIf the surface charge density is ρρss, the line charge density of a, the line charge density of a

“ “verticalvertical” ” strip isstrip is ρρss dydy



For a line chargeFor a line charge in the zin the z--axisaxis,,

recall thatrecall that

ρaEρπε

ρ=0

L2

(x, 0, 0)

y

dy

ρs

dEx

β

Therefore:Therefore:

x

x

x x

a

a

adE

)yx(2xdy

yx

x

yx2

dy

cosyx2

dy

220

S

22220

S

220

S

+περ=

++πε

ρ=

β+πε

ρ

=



x x adE22

0

S

yx

xdy

2 +πε

ρ= dy

yx

1

2

x22

0

S∫∞

∞− +πε

ρ== x x aEE

Note:Note: Cautan

a1

uadu 1

22 +=

+

−∫

x x

aaE

π−−

π

πε

ρ=

πε

ρ=

+∞=

−∞=

− )2

(22x

ytan

x

1

2

x

0

Sy

y

1

0

S

xaE0

S

2ε

ρ−=

At the negative x At the negative x--axis (or at theaxis (or at the “ “back back ” ” of the sheet of charge),of the sheet of charge),

xaE0

S

2ε

ρ=



In general, the electric field due to an infinite sheet of chargIn general, the electric field due to an infinite sheet of charge is equal toe is equal to

N0

S

2aE

ε

ρ=

NoteNotess::1.1. aaNN is a unit vector perpendicular and pointing away from theis a unit vector perpendicular and pointing away from the

surfacesurface

2.2. The electric fieldThe electric field intensity is constantintensity is constant..

3.3. If theIf the surfacesurface charge density is positive, the electric fieldcharge density is positive, the electric field

“ “emanatesemanates” ” from thefrom the sheet of chargesheet of charge. If the line charge density is. If the line charge density is

negative, the electric fieldnegative, the electric field “ “is intois into” ” to theto the surfacesurface charge.charge.



Example:Example:

A line charge with charge density equal to 10 nC/m is at x = 4, A line charge with charge density equal to 10 nC/m is at x = 4, z = 3. Az = 3. A

sheet of charge with surface charge density equal tosheet of charge with surface charge density equal to --1 nC/m1 nC/m22 is at the xyis at the xy--

plane. What is the electric field at P(2,3,5)?plane. What is the electric field at P(2,3,5)?

P

EL

ES

x

y

z Side view:Side view: EL

ES

D

x

z



zxzx

D

22

zx

707.0707.0828.2

22

828.22)2(D

22

aaaa

a

aaD

+−=+−

=

=+−=

+−=E

L

ES

D

x

z

V/m939.44939.44)707.0707.0()828.2(2

1010zxzx

0

9

L aaaaE +−=+−πε

×=

−

V/m472.562

101 zz0

9

S aaE −=ε

×−=−

V/m53.1194.44 zxSLP aaEEE −−=+=



Streamlines and Sketches of FieldsStreamlines and Sketches of Fields

Given an expression ofGiven an expression of EE, how will one draw (sketch) the field?, how will one draw (sketch) the field?

Take, for example, a point charge.Take, for example, a point charge.



The arrows show the direction of the field at every point alongThe arrows show the direction of the field at every point along thethe

line, and the separation of the lines is inversely proportionalline, and the separation of the lines is inversely proportional to theto thestrength of the field.strength of the field.

The lines are calledThe lines are called streamlinesstreamlines..



Given a two dimensional field (Given a two dimensional field (EEzz = 0), the equation of a streamline is= 0), the equation of a streamline is

obtained by solving the differential equationobtained by solving the differential equation

dxdy

E

E

x

y =

E

Ex

Ey

∆y

∆x

x

y



Example.Example. The electric field intensity is given asThe electric field intensity is given as

EE = 5e= 5e--2x2x (sin 2y(sin 2y aa x x -- coscos 2y2y aa y y) V/m) V/m

Find the equation of the streamline passing through the point P(Find the equation of the streamline passing through the point P(0.5,0.5,

ππ /10, 0). /10, 0).

Solution:Solution: Solving the differential equation :Solving the differential equation :dx

dy

E

E

x

y=

y2coty2sine5

y2cose5

dx

dyx2

x2−=−=

−

−

y2cosKe

y2sece'K e

)y2ln(sec'Cx2

C)y2ln(sec

2

1x

ydy2tandx

ydy2tandx

x2

x2'Cx2

=

==

=+−

+=−

=−

=−

−+−

∫∫



To solve for K, use the fact that the streamline passes throughTo solve for K, use the fact that the streamline passes through P (0.5,P (0.5,

ππ /10, 0): /10, 0):

KeKe2(0.5)2(0.5) == coscos ππ /5 /5K = 0.298K = 0.298

Therefore, the equation of the streamline through P isTherefore, the equation of the streamline through P is

0.298 e0.298 e2x2x == coscos 2y2y

y2cosKe x2 =

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Ch02_Coulomb's Law and Electric Field Intensity