24장 읽고 있자 ……

24. 전기퍼텐셜 (Electric potential)

• Conservative Forces and Energy Conservation– Total energy is constant and is sum of kinetic and potential

• Electric Potential Energy (U)

• Electric Potential (V)

– A property of the space and sources as is the Electric Field– Potential differences drive all biological & chemical reactions,

as well as all electric circuits

• Path-independence of Potential

• (계산) Electric Potentials: : electric dipole, N point charges, continuous charges

cFU ↔

qEV ↔

지난 시간에 …

Q

Electric Flux ∫ ⋅≡Φ AdEr

Gauss 법칙o

enclqAdEε

=Φ≡⋅∫rr

enclo q=ΦεOR,s

2204

1rQk

rQE e=

πε=⇒

3304 a

QrkarQE e=

πε=⇒

복습 : 일 (Work) 이란 ? • Work done by the force given by:

Positive: Force is in direction movedNegative: Force is opposite direction movedZero: Force is perpendicular to direction moved

• Careful ask WHAT is doing work!

Opposite sign for work done by you!: 즉, 계(system)가한일은외부에서해준일과부호가반대!: ΔW = - ΔWapp

• Conservative Forces

Δ Potential Energy (U) = - Wconservative

θcosFddFW =•=rr

Conservation of Energy of a particle

• Kinetic Energy (K)– non-relativistic

• Potential Energy (U)– determined by force law

• for Conservative Forces: K+U is constant– total energy is always constant

• examples of conservative forces– gravity; gravitational potential energy – springs; coiled spring energy (Hooke’s Law): U(x)=kx2/2– electric; electric potential energy

• examples of non-conservative forces (heat)– friction– viscous damping (terminal velocity)

221 mvK =

),,( zyxU

전기력도 보존력이다 !• Consider a charged particle traveling

through a region of static electric field:

• A negative charge is attracted to the fixed positive charge

• negative charge has more potential energy and less kinetic energy far from the fixed positive charge, and…

• more kinetic energy and less potential energy near the fixed positive charge.

• But, the total energy is conserved

+

-

• 다음에 대해 알아보자

electric potential energy (U)electrostatic potential (V)

• A direct calculation of the work done to move a positive chargefrom point A to point B is not easy.

• Neither the magnitude nor the direction of the field is constant along the straight line from A to B.

• But, you DO NOT have to do the direct calculation.

• Remember: potential difference is INDEPENDENT OF THE PATH!!

• Therefore we can take any path we wish.

A positive charge Q is moved from A to Balong the path shown by the arrow. What is the sign of the work done to move the charge from A to B? A B

(a) WAB < 0 (b) WAB = 0 (c) WAB > 0

0=• ldErrChoose a path along the arc of a circle centered at

the charge. Along this path W = 0 at every point!!

질문

Electric potential energy• Imagine two positive charges, one with charge Q1,

the other with charge Q2:

• Initially the charges are very far apart, so we say that the initial energy Ui of interaction is zero

(we are free to define the energy zero somewhere).

• If we want to push the particles together, this will require work (since they want to repel).

the final energy Uf of the system will increase by the same amount: ΔU = Uf – Ui = ΔWapp

Q2Q1

Q1Q2

Electric potential energy• The initial energy Ui (∞) of interaction is zero

• The final energy Uf (r) of the system will increase by the same amount of the work applied:

ΔU = Uf – Ui ≡ U = ΔWapp

ΔWapp = - ΔWcoulomb= - ΔW∞ ≡ - W

즉, 외부에서해준일과두입자간작용하는힘이한일은

부호가반대 !!!

U = - W

Q2Q1r

확인문제 1, 2, 3

1. (a) 전기장이양성자에한일은 +, - ??

(b) 양성자의퍼텐셜에너지는증가, 감소 ??

2. (a) 힘(f)이양성자에한일은 +, - ??

(b) 양성자는더높은 (낮은) 퍼텐셜쪽으로움직이는가 ??

음의 일

증가

양의 일

높은 쪽으로

Electric potential energy (U)m]N[J ⋅=•−=•−=−= ∫∫ ∞∞

rr

q sdEqsdFWU rrrr

Electric potential (V) : 단위전하당 U

J/C] [V(volt) =•−=−=≡ ∫∞

rsdE

qW

qUV rr

Qr q q

즉, 전하 q가 무한대에서 r 위치까지 이동하는 동안 전기력 (전기장)이 한 일

즉, 단위전하가 무한대에서 r 위치까지 이동하는 동안 전기력 (전기장)이 한 일

F (E)

Q f

전위차 (electric potential difference : ΔV)

( ) ( )

J/C] [V(volt) =•−=−=

−−−=Δ

∫∞∞

f

iif

if

sdEVV

VVVVVrr

두 지점(f, i) 간의 전기 퍼텐셜차 (전위차 : DV)

i

★전자볼트 (electron volt, eV) (아주 작은 에너지의 단위)

≡전자하나의전하량( e ≡ | q전자|)을 1 volt 전위차옮기는데드는에너지

JC -19-19 10 1.60 J/C) 1()10 (1.60 V) (1 e eV 1 ×=××==

★전기장의 또 다른 차원

[ ] [ ] [ ]V/m(J/C)/m(J/m)/C[N/C] [E] ==⇒=

등퍼텐셜면 (등전위면 : Equipotential surface)

WI = 0WII = 0WIII = WIV

등퍼텐셜면 ≡ 전위가 같은 점들로 이루어진 면

각 이동 경로에서전기장이 한 일 ?

등퍼텐셜면은 항상 전기장에 수직 (왜?)

확인문제 3

3. (a) 등퍼텐셜면에대한전기장의방향은?

(b) 각경로에대해한일은 +, - ?

(c) 한일이큰순서대로

------

1, 2, 3, 5 : + , 4 : -

3, 1=2=5, 4

24-6. 점 전하가 만드는 퍼텐셜

rqE

oπε41

=

점전하 q에서임의의 r 만큼떨어진위치에서의전기장

점전하 q에서 R 만큼떨어진위치에서의퍼텐셜 (V)

rqds

ssdErV

rr

o2

o 411

4q- )(

πεπε==•−= ∫∫ ∞∞

rr

rqrV

o41)(πε

=

J/C] [V(volt) =•−=−=≡ ∫∞

rsdE

qW

qUV rr

24-7. 점전하무리가만드는퍼텐셜

∑∑==

==N

n n

N

nn r

qVV1o1 4

1πε

확인문제 4 : P 점에서의알짜전기퍼텐셜의크기순서는?

a = b = c

• The potential energy of an added test charge q0 at point P is just

0

31 20

1 2 3of q at P

p p p

QQ QU q k k kr r r

⎛ ⎞= + +⎜ ⎟⎜ ⎟

⎝ ⎠

Q1

Q2Q3

q0

r1p

r2pr3p

Q1

Q2Q3

P

질문

Which of the following charge distributions produces V (x) = 0 for all points on the x-axis?

(a)

x

+2μC

-2μC

+1μC

-1μC(b)

x

+2μC +1μC

-2μC-1μC (c)

x

+2μC -2μC

+1μC-1μC

The key here is to realize that to calculate the total potential at a point, we must only make an ALGEBRAIC sum of the individual contributions.

Therefore, to make V(x)=0 for all x, we must have the +Q and -Qcontributions cancel, which means that any point on the x-axis must be equidistant from +2μC and -2μC and also from +1μC and -1μC.

This condition is met only in case (a)!

24-8. 전기 쌍극자가 만드는 퍼텐셜

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=⎟

⎟⎠

⎞⎜⎜⎝

⎛ −+==

−+

+−

−+=∑

)()(

)()(

o)()(o

2

1 4q

41

rrrr

rq

rqVV

nn πεπε

P

r >> d )( , cos4

1

cos4

q

cos

2o

2o

2)()(

)()(

qdpr

pr

dV

rrr

drr

=⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛≈

≈

≈−

−+

+−

θπε

θπε

θ

* 유도 쌍극자 모멘트 (induced dipole moment)

원자 (분자)

E 에의해분극이일어남원자 (분자)에서의유도쌍극자모멘트발생

(참고 : 물질의 굴절률 근원)

E

24-9. 연속 전하분포가 만드는 퍼텐셜

rdqdV

oπε41

= ∫=⇒r

dqVoπε4

1

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ++=

+=

+==

=

∫ ddLL

dxdxV

dxdx

rdqdV

dxdq

o

L

o

oo

22

0 22

22

ln44

41

41

πελ

πελ

λπεπε

λ

L

[ ]zRzRz

dRRV

RzdRR

rdqdV

dRRdq

o

R

o

oo

−+=+

=

+==

=

∫ 22

0 22

22

2'''

2

'''2

41

41

''2

εσ

εσ

πσπεπε

πσ

선 전하

원판 전하

24-10. 퍼텐셜에서 전기장 구하기

sdEVVf

iifrr

⋅−=− ∫sdEdV rr

⋅−=

, , ,

ˆˆˆ

zVE

yVE

xVE

zzVy

yVx

xVV

sddVE

zyx ∂∂

−=∂∂

−=∂∂

−=⇒

⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+∂∂

−=∇−≡−=r

rr

If the E-field has only one component, Ex,

x̂dxdVx̂EE x −==

r

If the charge distribution has spherical symmetry, the E-field is radial, Er,

r̂drdVr̂EE r −==

r

예제 : 쌍극자의 전기퍼텐셜과 전기장

-q +q2a •

x

P

22

2ax

qakax

qax

qkrqkV e

ei

ie −

=⎟⎠⎞

⎜⎝⎛

+−

−== ∑

If x >> a , 2

2x

qakV e≈

( )222

22ax

xqakkdxdVE e

ex−

⋅=−=

3

4x

qakdxdVE e

x ≈−=

24-11. 점 전하계의 전기 퍼텐셜 에너지

• 고정된 점전하계의 전기적 퍼텐셜(위치) 에너지

≡ 주어진 전하분포가 되게 하는데 드는 에너지 (외부에서 해야 하는 일)

rqqVqWU

oapp

212 4

1πε

===

보기문제 24-6

그림과 같은 세 점 전하의 전기적 위치 에너지?

24-12. 고립된 대전 도체의 퍼텐셜

Qr̂rQkE e 2=

r( )Rr >

0= ( )Rr <

rQk

rdEsdErV

e

r

r

=

⋅=⋅−= ∫ ∫∞

∞ rrrr)(

RQk

rdErdEsdErV

e

R

r R

r

=

⋅+⋅=⋅−= ∫∫ ∫∞

∞

rrrrrr)(

( )Rr >

( )Rr <

0

24. Summary

Electric potential energy (U)

m]N[J ⋅=•−=•−=−= ∫∫ ∞∞

rr

q sdEqsdFWU rrrr

Electric potential (V) : 단위전하당 U

J/C] [V(volt) =•−=−=≡ ∫∞

rsdE

qW

qUV rr

∑∑==

==N

n n

N

nn r

qVV1o1 4

1πε

∫=r

dqVoπε4

1

점 전하

연속 전하

⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+∂∂

−=∇−≡−= zzVy

yVx

xVV

sddVE ˆˆˆ

rr

rV E