23. 가우스 법칙 (Gauss 가우 법칙 ( ’ law) )

23. 가우스 법칙 (Gauss 가우 법칙 ( ’ law) )

• Gauss’ Law: Motivation & Definition

∫

_E^r _⋅dA^r _{≡ Φ =} ^qencl

• Coulomb’s Law as a consequence of Gauss’ Law εo

∫

• Charges on Conductors:

? – Where are they?

A li ti f G ’ L

• Applications of Gauss’ Law – Uniform Charged Sphere

Infinite Line of Charge – Infinite Line of Charge – Infinite Sheet of Charge

– Two infinite sheets of chargeTwo infinite sheets of charge

지난 시간에 … 지난 시간에 …

전기장 (Electric field)

2

4 o

1 r

q q

E F

o ⁼ πε

= E = E1 + E2 +L+ E_n

o o

d q p =

전기 쌍극자 (Electric dipole)

E p×

τ = U = −p ⋅E

dVrˆ k

E^r ∫ ^{ρ Er = k} ∫ ^{σdSrˆ Er = k} ∫ ^λdlrˆ

연속전하의 전기장

r r k

E ⁼ _e∫ 2 r

k r

E ⁼ _e∫ 2 r

k r

E ⁼ _e∫ 2

Fundamental Law

of Electrostatics (정전기학)

• Coulomb’s Law

Force between two point charges Force between two point charges

E q q

F q⁰ _{2 0}

4 =

=

OR

r q

πε_o ^{2 0} 4

• Gauss’ Law

Relationship between Electric Fields

and charges

_q

∫

^{r r}

and charges

o

qencl

A d

E ^{⋅ ≡ Φ =} ε

∫

^{r r}

23-2 다발 (flux) 23 2. 다발 (flux)

다발 (flux)

Φ = (면에 수직인 속도성분)×(면의 넓이) Φ = (면에 수직인 속도성분)×(면의 넓이)

( ^v ) ^{A = v r • A r}

≡

Φ ( ^v cos θ ) ^{A v A}

Φ cos θ

22-3. 전기장 다발 (Electric flux)

• Consider flux through two surfaces ^{case 1}

전기장 다발 ( )

Consider flux through two surfaces that “intercept different numbers of field lines”

E-field surface area E A Flux:

case 2 case 1 Er = E_oyˆ ₂

w E_ow²

Case 2 is smaller!

case 2 Er = E_oyˆ E_ow² ⋅cosθ θ

smaller!

w2

22-3. 전기장 다발 (Electric flux)전기장 다발 ( )

Gauss 폐 곡면 (Gaussian surface)

Electric Flux ^{Φ ≡} ∫ ^{E r ⋅ d A}

Electric Flux

•What does this new quantity mean?

∫

≡

Φ E d A

q y

• The integral is over a CLOSED SURFACE

• Since is a SCALAR product, the electric flux is a SCALAR quantity

• The integration vectorThe integration vector is normal to the surface and points OUT is normal to the surface and points OUT of the surface. is interpreted as the component of E which is NORMAL to the SURFACE

• Therefore, the electric flux through a closed surface is the sum of the normal components of the electric field all over the surface.

• The sign matters!!The sign matters!!

Pay attention to the direction of the normal component as it penetrates the surface… is it “out of” or “into” the surface?

“O t f” i “+” “i t ” i “ ”

• “Out of” is “+” “into” is “-”

dS dS

질문

dS dS

1 2

A positive charge is contained inside a spherical shell.

How does the electric flux d

Ф

_E through the surface

element dS change hen the charge is mo ed from position element dS change when the charge is moved from position 1 to position 2?

a) d

Ф

_E ^increases

a) d

Ф

_E ^increases

b) d

Ф

_E ^decreases

c) d

Ф

_E doesn’t change

dS dS

1 2

A positive charge is contained inside a spherical shell.

How does the flux

Ф

_E through the entire surface change when the charge is moved from position 1 to position 2?

Ф

a)

Ф

_E ^increases

b)

Ф

_EE ^decreases

c)

Ф

_E doesn’t change

질문

• Consider 2 spheres (of radius R and 2R) drawn around a single charge as shown.

– Which of the following statementsWhich of the following statements about the net electric flux through the 2 surfaces (Φ_2R ^and Φ_R^{) is true?}

(a) Φ_R< Φ_2R ^(b) Φ_R⁼ Φ_2R ^(c) Φ_R> Φ_2R

• Look at the lines going out through each circle -- each circle has the b f li

same number of lines.

• The electric field is different at the two surfaces, because E is

proportional to 1 / r ², but the surface areas are also different. The surface area of a sphere is proportional to r ².

surface area of a sphere is proportional to r .

• Since flux = , the r ²and 1/r ²terms will cancel, and the two circles have the same flux!

circles have the same flux!

질문

A cube is placed in a uniform electric field. Find the flux through the bottom g surface of the cube.

a) Ф

_bottom

< 0

b) Ф = 0

b) Ф

_bottom

= 0

c) Ф

_bottom

> 0

Lecture 3, ACT 2

질문

Imagine a cube of side a positioned in a region of constant electric field as shown.g

Which of the following statements about the net electric flux Φ_E through the

surface of this cube is true?

a

surface of this cube is true?

(a) Φ_E = 0 (b) Φ_E ∝ 2a² (c) Φ_E∝ 6a²

a

(a) Φ_E 0 (b) Φ_E ∝ 2a (c) _E 6a

• The electric flux through the surface is defined by:^{g y}

^{Φ ≡} ∫ ^{E r ⋅ d A}

• on the bottom face is negative. (dS is out; E is in)

• is ZERO on the four sides that are parallel to the electric field.

^Φ ∫ ^{E d A}

∫

^{Er⋅ Ad}

∫

^{Er⋅ Ad}

• on the top face is positive. (dS is out; E is out)

• Therefore, the total flux through the cube is:

∫ ∫

^{Er⋅ Ad}

g

보기문제 23-1

고른 전기장 E 속에 놓인 반지름 R 인 원통꼴의 가우스 곡면에서의 전기장 플럭스 Φ의 값?

보기문제 23-2

그림과 같은 정육면체의 왼쪽, 오른쪽, 위쪽 면에서의 전기장 플럭스? 단, 전기장은

E = ( 3.0 x) i + ( 4.0) j (N/C).

E ( 3.0 x) i ( 4.0) j (N/C).

1) 왼쪽 면

2) 오른쪽 면 2) 오른쪽 면

3) 위쪽 면

23-4. Gauss 법칙법칙

o

q

encl

A d

E ^{⋅ ≡ Φ =} ε

∫ ^{r r}

encl o

Φ = q

ε

OR,

The net electric flux through any closed surface is encl

o

q

The net electric flux through any closed surface is proportional to the charge enclosed by that surface.

It is very useful in finding E

when the physical situation exhibits massive SYMMETRY.

23-5. Gauss 법칙과 Coulomb 법칙법칙과 법칙

q

encl

A d Φ ∫ E ^{r r}

o

q

encl

A d

E ^{⋅ =} ε

= Φ ∫

가우스 법칙에서 쿨롱 법칙 끌어내기 점 전하 q 가 원점에 있을 때

반지름 r 인 공 표면에서의 전기장?

구 대칭성 때문에 공표면 어디에서나 구 대칭성 때문에 공표면 어디에서나

전기장의 크기는 같고, 방향은 밖으로 나아 가는 반지름 방향

Gauss 폐곡면의 선택 Gauss 폐곡면의 선택

S₁ ^{Φ =} ∫ ^{⋅ =} ∫ ^⋅

2

1 S

S

C Er dAr Er dAr q k r r

k_e q ⋅ π = π _e

= ₂ 4 ² 4 k 1

S₂ ⁰

= 4πε ke

ε

0

=

Φ q

C

0

For any empirical surface, Φ_C is the same.

i.e. Φ_C is independent of the surface contour.

23-6 전하를 띤 고립된 도체전하를 띤 립된 체

고립된 도체에 들어 있는 잉여전하 (excess charge)는 도체의 표면에 퍼져 있게 된다

도체의 표면에 퍼져 있게 된다.

◂ 서로 밀어내는 정전기력

◂ 도체 속에서는 전기장이 Æ E = 0 (왜?)

E 0 E = 0

E = 0 E = 0

보기문제 23-4

속빈 쇠공: 속 반지름 R, 전기적으로 중성 점전하: - 5.0 μC, 중심에서 R/2인 곳

1) 쇠공의 안팎의 면에 쌓인 전하량?

2) 안팎의 면의 전하분포?

2) 안팎의 면의 전하분포?

1) 쇠공의 안쪽 면의 전하량:

쇠 속은 전기장이 0

⇒ 쇠 속의 가우스 면에서도 전기장이 0

⇒ 가우스면 전체의 플럭스도 0

⇒ 가우스면 속의 알속전하는 0

쇠공 안쪽 면에 5 0 C 전하 분포

⇒ 쇠공 안쪽 면에 +5.0 μC 전하 분포 2) 쇠공의 바깥 면의 전하량:

쇠공은 전기적으로 중성, 전하는 보존

⇒ 쇠공 바깥 면에 - 5.0 μC 전하 분포

▸ 전하분포는 공 안쪽은 고르지 않으나 바깥쪽은 고름

23-7 Gauss 법칙의 적용: 선전하 - 원통대칭 23 7 Gauss 법칙의 적용: 선전하 원통대칭

o

qencl

A d

E^{⋅ =} ε

=

Φ

∫

^{r r}

23-8 Gauss 법칙의 적용: 면전하 - 원통대칭법칙의 적용 면전하 원통대칭

o

q

encl

A d

E ^{⋅ =} ε

=

Φ ∫ ^{r r}

^{Coulomb 법칙}

o

∫ ^{2r d}

E z

E σ ^R

( )

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

− +

=

= +

= ∫

2 2

0 2 2 32

2 1 4

R z

z r dr E z

E

o o z

ε σ

ε

(면부터의 거리에 무관하게 일정)

If RÆ infinite, (면부터의 거리에 무관하게 일정) σ

o

E ε σ

≈ 2

두 도체판

• Field outside must be zero

• Field outside must be zero.

Two ways to see:

E=0 σ^{+ -}σ E=0

– Superposition +

+

+ -

- -

– Gaussian surface encloses zero charge

+ +

- A -

+ +

-

• Field inside is NOT zero: -

+ -

+ -

-

– Superposition

G i f l A

A +

+ -

-

– Gaussian surface encloses non-zero charge

A

E

23-9 Gauss 법칙의 적용: 구 껍질 - 구대칭 23 9 Gauss 법칙의 적용: 구 껍질 구대칭

q

encl

A d

E ^{⋅ =} ε

=

Φ ∫ ^{r r}

ε

o

∫

(S )

구 껍질 q

(S₂) (S₁)

구 껍질

구대칭 전하분포 구대칭 전하분

Charge density ρ : ₃⁴ πa³ρ = Q ₃

4 3

a Q

= π ρ

⇒ 4 aπ Q

(a) r >a

0 0 = ε

= ε

⋅

=

Φ^C ∫ ^{Er dAr qinc Q}

Q

(a) r >a

0 0

0

4 2

= ε

⋅

π Q

E

r 2 2

4 0

1

r k Q r

E Q = _e

= πε

⇒ for r > a

(a) r >a

(b) r <a

ε0

=

⋅

=

Φ^C ∫ ^{Er dAr qinc}

1

3 0

4

1 dV π ρ

⋅ ε ρ

= ∫

Q

3 0 4 r3π ε

= ρ

⎟⎞

⎜⎛ 3Q ₂ Q r³ Qr

r k E Q

(b) r <a

⎟⎠

⎜ ⎞

⎝

⎛

= π

ρ ₃

4 3

a Q

3 0

4 2

a r E Q

r = ε

π _{3 3}

4 0 a

k Qr a

r

E Q = _e

= πε

⇒ for r < a

Q Q

1 Q

Q k

E = =

⇒

for r > a

k Qr r

E Q

⇒

for r < a

2 2

4 0 k r

E r = _e

= πε

3 ⇒

3

4 0 k a

E a = _e

= πε

⇒

23. SummarySu a y

Electric Flux

^{Φ ≡} ∫ ∫ ^{E r ⋅ d A}

G 법칙

∫

^{Er dAr Φ qencl}

^Φ

Gauss 법칙

o

qencl

A d

E ^{⋅ ≡ Φ =} ε

∫

^OR,

ε

o

Φ = q

encl S₁

s

1

S₂