Lecture Note #2A Chapter 2. Electrostatics (정전기학contents.kocw.net/KOCW/document/2015/inha/kimkyunghon/03.pdf · 2016-09-09 · Inha University 1 Chapter 2. Electrostatics

Inha University 1

Chapter 2. Electrostatics (정전기학)

2.1 The Electric Field

2.2 Divergence & Curl of Electrostatic Fields

2.2.1 Field Lines, Flux, and Gauss’s Law

2.2.2 The Divergence of E

2.2.3 Applications of Gauss’s Law

2.2.4 The Curl of E

2.3 Electric Potential

2.4 Work and Energy in Electrostatics

2.5 Conductors

Lecture Note #2A

+

Inha University 2

2.1 The Electric Field (전기장)

2.1.1 Electrostatics vs. Electrodynamics

S Q

“Test” charge

a,

v,a 0and/or 0 v If

“Electrodynamics”

,a 0and/or 0 v If

“Electrostatics”

Total forces acting on the test charge Q =

321 FFFF

(전하의속도, 가속도)

(속도) (가속도)

(속도) (가속도)

Inha University 3


2.1.2 Coulomb’s Law

r

Q

“Test” charge

x

2

212

0 10858mN

C.

where

'rr

r

The force acting on the test charge Q due to a single point charge q, which is at

rest, =

rr

ˆ4

12

0

qQF

y

z

r

'r

: the permittivity of free space.

: the separation vector (거리벡터) from the source q’s position r’ to the test charge Q’s position r.

If q and Q have the same sign, the Coulomb force is repulsive (반발력)

If q and Q have the opposite sign to each other, the Coulomb force is attractive (인력)

q

A “point”

charge

Inha University 4


2.1.3 The Electric Field

where

EQF

Total forces acting on the test charge Q =

EQ

qqqQ

QqQqQq

FFFF

ˆˆˆ4

ˆˆˆ4

1

323

322

2

212

1

1

0

323

322

2

212

1

1

0

321

rr

rr

rr

rr

rr

rr

Q

“Test” charge

x

y

z

r

'

ir

n

i

i

i

iqrE

12

0

ˆ4

1r

r

: the Electric Field

(전기장)

r

r Q

FE

Inha University 5


2.1.3 The Electric Field

since

z

dz

qzz

qz

zEEEEE zz

ˆ2

2

4

1ˆ

4

12

ˆ

23 220

310

2121

r

Total electric field at the point P =

1210

1 ˆ4

1r

r

qE

[Example 2.1] Find the electric field at a distance z above the midpoint between two equal charges

(q), a distance d apart.

r

x

z

z

P

q q

2

d

2

d

x

z

z

P

q q

2

d

2

d

1E

2E

(Solution)

3101

210

210

14

1

4

1cos

4

1

rrrr

qzzqqE z

1E

zE1

xE1

2E

zE2

xE2

021 xx EE

When z >> d,

zz

qz

z

qzE

2

0

23 20

2

4

12

4

1

Inha University 6


2.1.4 Continuous Charge DistributionsFor a continuous charge distribution, the electric field at the point P =

rr

ˆ4

12

0

dqrE

'dldq'dl

dq

L

q

r

r

For a line charge distribution, the electric field at the point P =

' ˆ4

12

0

dlrE rr

Line-charge density :

'da'rdq'da

dq'r

A

q ,

For a surface charge distribution, the electric field at the point P =

S

dar

rE ' ˆ'

4

12

0

rr

Surface-charge density :

'dV'rdq'dV

dq'r

V

q ,

For a volume charge distribution, the electric field at the point P =

V

dVr

rE ' ˆ'

4

12

0

rr

Volume-charge density :

r

r

: Coulomb’s LawBack

r'

r'rr

r

Inha University 7


2.1.4 Continuous Charge Distributions

2

0

2

0 4

12

4

1

z

q

z

LE

For points far from the line (z >> L), the electric field at the points:

zLzz

L

xzx

xzz

xzz

dxxz

xxdx

xzzz

dlxz

xxzz

xzdl

rE

L

Lx

L

Lx

L

Lx

L

Lx

L

Lx

L

Lx

ˆ2

4

1

1ˆˆ

4

ˆ1

ˆ 4

'ˆ ˆ

4

1'ˆ

'

4

1

220

222220

232223220

22220

1210

r

r

[Example 2.2]

r

dx

,ˆ ˆ ' xxzzrr

r

(Solution) ,ˆ zzr

,xx'r

dx'dl

,22 xz r 22

ˆ ˆ ˆ

xz

xxzz

r

rr

Lq 2

In the limit L , the electric field of an infinite straight wire:z

E

2

4

1

0

(2.9)

<적분과정은 다음 쪽에>

Inha University 8

Mathematical Information

xcosxsec

dxxsecxtand

xtanxcos

dx

xcosdxxsin

xsindxxcos

xsecxcosxcos

xsinxcos

xcos

xsinxtan

xsin

xcosxcot

xcos

xsinxtan

1

111

2

2

2

22

22

2

22

222222

2323

2

23222

2

2322

11

1

xzz

x

z

sindcos

zsec

d

z

tanz

dsecz

tanzz

dsecz

xz

dx

dseczdxtanzx 2

z

x

z

xtan

22 xz

22 xz

xsin

dydxxyxz 2 22

22

21

232322

1

2 xzy

y

dy

xz

xdx /

Inha University 9

2.2 Divergence and Curl of Electrostatic Fields


The Electric field from a point charge q located at the origin : rr

qE

2

04

1

High flux (선속) density Strong electric field

Low flux density Weak electric field The Electric field flux through

a surface S :

SS

E cosdaEadE

Inha University 10

+


2.2.1 Gauss’s Law

The Electric field flux through any closed surface = the total electric charge inside.

0

QadE

Gauss surface

The charge outside the surface has no contribution to the total flux

through the Gauss surface

0

0

2

00

2

2

0

sin4

ˆ sinˆ4

1

qdd

qrddrr

r

qadE

SE

2q -q

r2 –terms cancels out.

Thus, the integral has no radius dependency.

When there are a bunch of scattered charges inside the

Gauss surface,

n

i

iEE1

n

i

in

i

iE

qadEadE

1 01

Total charge

Q

n

i

iqQ1

Gauss’s Law (integral form)

Inha University 11

+++ +

++

+

+

++

+

+


2.2.1 Gauss’s Law

The Gauss’s Law (a differential form) :

0

E

VVS

E dVQ

dVEadE 1

00

where is the charge density, i.e.,

(continued)

Q VdVQ

VV

dVdVE 0

: Gauss’s Law (a differential form)

Inha University 12


2.2.2 The Divergence of the Electric Field

From the Coulomb’s law (see page 8)

rr

qE

2

04

1

'ˆ'

4

112

10

dVr

rE rr

V

dVrrE ''ˆ

4

1

21

1

0

r

r

rr

r 3

24

ˆ

From the 3-dimensional Delta function theorem (chapter 1)

00

3

0

44

14

4

1

rr'dV'r'rrrE

V

To recover the integral form of the Gauss’s law from the differential form

0

E

VSV

QdVadEdVE

00

1

Page 8

0

QadE

S

r

[Appendix]

'rr

r

Inha University 13



Find the field outside a uniformly charged solid sphere of radius R and

total charge q.

rr

qrEoutside 2

04

1

For a Gaussian spherical surface of a radius r > R, the Gauss’s law becomes

0

qadE

S

[Example 2.3]

0

22

00

2

S

4 sin

ˆ ˆ

qrEddrE

daErdarEadE

rr

Srr

S

2

04

1

r

qEr

(Solution)

Inha University 14



How about the field inside a uniformly charged solid sphere of radius R and

total charge q ? (Assume that the total charge is uniformly distributed over the

entire volume of the sphere.)

rr

Einside

03

For a Gaussian spherical surface of a radius r < R, the Gauss’s law becomes

'

0

' 1

VSdVadE

[Example 2.3]

(Solution)

2R

q

r 1

'

2

0

2

VS

r dd'drsin'rddsinrE

2

000

2

0

2

00

2 dφθdθsin'dr'rdφθdθsinrEr

r

3

3

00

2

0

2 r'dr'rrE

r

r

03

rEr

Since ,

R

q

3

3

4

rR

rqr

rrEinside 3

00 4

3

2

04

R

qRE

3

04

R

rqrEinside

2

04

1

r

qrEoutside

q

Inha University 15


Application Guides for Gaussian Surfaces

1) Spherical symmetry:

Make your Gaussian surface a concentric sphere.

Use the symmetry property in choosing the Gaussian surface

2) Cylindrical symmetry:

Make your Gaussian surface a coaxial cylinder.

3) Plane symmetry:

Make a thin Gaussian “rectangular box” that straddles

the surface.(표면의아래와위를감싸는직사각형가우스면을선택)

(Assume infinitely long cylinders.)

(Assume infinitely large planes.)

Inha University 16



A long cylinder carries a charge density that is proportional to the distance from

the axis: = ks, for some constant k. Find the electric field inside this cylinder.

sksE ˆ3

1 2

0

For a Gaussian cylinder of length l and radius s, the Gauss’s law becomes

0enclosed

S

qadE

[Example 2.4]

slEdzsdE

daEdaEsdasEadE

s

l

zs

Ss

Ss

Ss

S

2

ˆ ˆ

0

2

0

3

03

22 klsslEs

(Solution)

33

0

2

00'

2

'

3

22

3

1 ''

klslskdzddssk

dzds'ds'ks'dVq

l

z

s

s

VVenlosed

Inha University 17



An infinite plane carriers a uniform surface charge . Find its electric field.

nE02

For a Gaussian rectangular box extending equal distances above and below the plane,

the Gauss’s law becomes

0enclosed

S

qadE

[Example 2.5]

0

2

AAEAEAEadE

S

02

E

(Solution)

Aqenlosed

where A is the are of the box surface.

Note that there is no distance dependence

for this infinite plane charge distribution.

Inha University 18



Two infinite parallel planes carry equal but opposite uniform charge densities

. Find its electric field in each of three regions: (i) to the left of both, (ii)

between them, (iii) to the right of both.

022 00

xxEEE

For a Gaussian rectangular box extending equal distances

above and below the plane, the Gauss’s law becomes

[Example 2.6]

(Solution)

In the region (i),

xxxEEE000 22

In the region (ii),x

0

E

022 00

xxEEE

In the region (iii),

Inha University 19


2.2.4 The Curl of EFor a point charge q at the origin, the electric field at a distance r from the origin becomes

rr

qE

2

04

1

Let us calculate the line integral of the electric field from a point a to

another point b.

b

aldE

In the spherical coordinates, ˆdsinrˆdrrdrld

drr

qldE

2

04

1

.r

q

r

q

r

qdr

r

qldE

ba

r

r

b

a

b

a

b

a

00

2

0 4

1

4

1

4

1

When the path is a close case (ra = rb),

When there are many charges, the principle of superposition

states holds (중첩의 원리가 적용) x

y

z

ab

.ldE 0

LoopS

ldad

vv

: Stoke’s Theorem

: Curl Theorem

SLoop

adEldE

0 E

21 EEE

02121 EEEEE

: always true for

static charges

Inha University 20


2.3.1 Introduction to Potential

The potential difference between two points a and b is

where is a standard reference point (typically, ).

: Electric Potential (전위)

0r

: The fundamental

theorem of gradients

r

rldErV

0

x

y

z

b

a

r

a

b

r

a

r

b

r

ldEldEldE

ldEldEaVbV

0

0

00

0r

From the fundamental theorem for gradients, Eq. (1.55) in Chapter 1,

b

aldVaVbV

aTbTldTb

a

b

a

b

aldEldV

VE

Thus,

(Integral form)

: Electric Potential(Differential form)

Inha University 21


2.3.2 Comments on Potential

Potential (V) Potential Energy (U) (W)

In the most cases, V(r = ) = 0

rVKldEldEldErVrr

'' '

VE

= 0

qVU

0 E

for static charges,Since

0

y

E

x

Ez

x

E

z

Ey

z

E

y

Ex

EEE

zyx

zyx

E xyzxyz

zyx

and

= 0 = 0

Potential (V) is a scalar quantity.

Potential (V) is independent on the reference coordinates.

aVbVa'Vb'V V'V

Exception) The case of a uniformly charged infinite plane.

??? 22 00

zdzzVz

x

y'

z

rV

'

0

x'

y

z'

'0

r

rV

z

In real world there is no charged infinite plane.

Inha University 22


2.3.2 Comments on Potential

The total force acting on the test charge Q is Superposition Principle

Unit of potential (V) :

VoltC/JC/mN

S Q

“Test” charge

x

y

z

r

'

ir EQ

FFFF

321

The total electric field acting on the test charge Q is

V

EEEQ

FE

321

The total potential acting on the test charge Q is

321 VVVV

r

r

r

rldEEEldErV

00

321

: vector sum

: vector sum

: scalar sum

(단위: N)

(단위: N/C)

(단위: Nm/C=Volt)

Inha University 23


rr

qEout 2

04

1

.r

q

'r

q

'r

'drqldErV

rrr

00

2

0 4

11

44

[Example 2.7]

From Gauss’s law, the field inside is

(Solution)

00

0

inSin adE

q

q

R

From Gauss’s law, the field outside is

0 inE

0

qadE

outSout

For a point outside the sphere (r > R), the potential

becomes

For a point inside the sphere (r < R), the potential becomes

.R

q

'r

q'dr

'r

'drqldErV

Rr

R

Rr

00

2

0 4

10

1

40

4

Rr

V(r)

R

q

04

1

r

rV1

Rr

E(r)

2

04

1

R

q

2

1

rrE

0

Inha University 24


2.3.3 Poisson’s Equations and Laplace’s Equation

Thus,

Since

: Poisson’s equation

0

E

In region, where there is no charge, so = 0,

,VE

From the curl of E relation,

(divergence of E) and 0 E

(curl of E),

0

2

VVE

0

2

V

02 V : Lapalce’s equation

0 VE

Inha University 25


2.3.4 The Potential of a Localized Charge Distribution

q

The electric field from a point charge q at the distance r is

Since V() = 0,

rr

qE

2

04

1

1) A Point Charge

P The linear infinitesimal displacement in the spherical coordinates is

ˆdsinrˆdrrdrld

drr

qldE

2

04

1

r

q

r

q

'r

q

'r

'drqldEVrV

rrr

000

2

0 4

10

4

11

44

r

qrV

04

1

r

qi

P

riq1

q2

2) Multiple discrete charges

n

i i

i

r

qrV

104

1

With the superposition principle

P

3) A continuous charge distribution

r

dqrV

04

1

'V

'dVr

'rrV

04

1

Inha University 26


The Electric Potential & Electric Field of a Localized Charge Distribution

Electric Field

'dlr

rV

04

1

'V

'dVr

'rrV

04

1

r

'dlrr

rE 4

12

0

S

'darr

'rrE

4

12

0

V

'dVrr

'rrE

4

12

0

Electric Potential

S

'dar

'rrV

04

1

r

r

'dV

for the case of V() = 0.

Inha University 27

x'

y'

z'

r

z


'd'd'sinR'da 2

[Example 2.8]

Let us set the point P on the z axis. Then,

(Solution)

'cosRzzRr 2222

q=4R2

R

Then, the potential equation becomes

S

'dar

'rrV

04

1

The infinitesimal surface area on the sphere is

0 22

2

2

00 22

2

0

22

24

'cosRzzR

'd'sinR

'd'cosRzzR

'd'sinRzV

Let .u'cosRzzR 222 du'd'sinRz 2

2221

0 22

11

2

1

2

2

2

2

2zRzR

Rzu

Rzduu

Rz'cosRzzR

'd'sin zR

zR

zR

zR

24 R

q

<다른풀이방법은[Example 2.7] 참조>

Inha University 28

x'

y'

z

' r

z


[Example 2.8] (continued)

q=4R2

R

22

222

0 22

2

0

2

12

224

zRzRz

R

zRzRRz

R

'cosRzzR

'd'sinRzV

For points outside the sphere, z > R & RzzR 2

For points inside the sphere, z < R & zRzR 2

z

RRzzR

z

RzV

0

2

02

002

RzRzR

z

RzV

Since the total charge on the shell is q = 4R2 ,

Rr,R

q

Rr,r

q

rV

4

1

4

1

0

0

Rr

V(r)

R

q

04

1

r

rV1

Rr

E(r)

2

04

1

R

q

2

1

rrE

0

0

Inha University 29


2.3.5 Boundary Conditions

[Problem 2.11] case

Relationship among the charge density , electric field and electric potential

ldEV

Boundary condition at a discontinuity

Field discontinuity at the surface of the conducting sphere

'V'dV

r

'rV

04

1

V E

V

'dVrr

'rrE

4

12

0

VE

0

2

V

0 ;0

EE

• Let us take a Gaussian surface of very thin rectangular-

type box shape overlapped over the conducting surface.

Gauss’s law for the conducting surface

.Aq

adE enclosed

S00

Inha University 30



The normal component of the electric field is discontinuous by an amount of /0 at the boundary.

• In the limit of the thickness of the rectangular-box-shaped Gaussian surface close to zero,

.A

AEEadE belowaboveS

0

(continued -1)

.EE belowabove

0

• The tangential component of the electric field is always continuous.

.ldE 0

For a very thin rectangular loop covering the surface

.lEElElEldE //

below

//

above

//

below

//

above 0

//

below

//

above EE

.nEEEEEE //

belowbelow

//

aboveabovebelowabove

0

where is a unit vector perpendicular to the surface from “below” to “above”n

(2.33)

A

b

aldEaVbV

Inha University 31

downn

upn

a

b



As the path length shrinks to zero,

• The potential is continuous across any boundary, since

(continued - 2)

• The gradient of V inherits the discontinuity in E, since E = -V,

: the normal derivative of V

0

n

V

n

V belowabove

nVn

V

where

belowabove VV (2.34)

(2.36)

(2.37)

0

b

abelow

b

aabovebelowabove ldEldEVV

//

below

//

above EE

nVV belowabove

0

(2.35)

.ˆ0

nEE belowabove

Inha University 32

Next Class

Chapter 2. Electrostatics

2.1 The Electric Field

2.2 Divergence & Curl of Electrostatic Fields


2.2.2 The Divergence of E


2.2.4 The Curl of E


2.4 Work and Energy in Electrostatics

2.5 Conductors

Inha University 33

Appendix

zzzyyyxxxrr ˆ'ˆ'ˆ''

r

222'''' zzyyxxrr

r

zzzyyxxz

yzzyyxxy

xzzyyxxx

ˆ '''

ˆ '''ˆ '''1

21-222

21-22221-222

r

zzzyyxxzz

yzzyyxxyyxzzyyxxxx

ˆ ''''22

1

ˆ ''''22

1ˆ ''''2

2

1

23-222

23-22223-222

23-222'''ˆ 'ˆ 'ˆ ' zzyyxxzzzyyyxxx

233

ˆ1ˆ

1

r

r

rrr

rr

Proof of2

ˆ1

r

r

r

Back

rr

r 3

24

ˆ

rr

2

ˆ1

r

r

r

rrr

32 411

rr

r 1ˆ2

Documents

Lecture Note #2A Chapter 2. Electrostatics (정전기학contents.kocw.net/KOCW/document/2015/inha/kimkyunghon/03.pdf · 2016-09-09 · Inha University 1 Chapter 2. Electrostatics