19. electric forces and electric fieldsphome.postech.ac.kr/user/genphys/download/phy102-19.pdf ·...
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19. Electric Forces and Electric Fields19. Electric Forces and Electric Fields
• Newton’s Law0=F
r⇒ 0=∆vr
amF rr=
원인
1. 관성의법칙
2. 운동의법칙
3. 작용반작용의법칙결과
2112 FFrr
=
• Force in NatureMass1. Gravitational Force
2. Electromagnetic Force
3. Strong Force
4. Weak Force
Charge
19-1. Historical Overview
• Ancient Period ---- Charge
---- Current vs Magnetic Field • Oested
---- Electromagnetism --- Phenomenology • Faraday
---- Electromagnetism 의완성• Maxwell
Maxell’s Equations
cf) Newton’s Law
19-2. Properties of Electric Charges
• Two kinds of charges (positive and negative) in Nature
+ + : Repulsive– – : Repulsive+ – : Attractive– + : Attractive
• The force : proportional to the inverse square of the separation.
21 rF ∝ ⇒ Coulomb’s Law
• Charge : Conserved & Quantized
Sum of Charges = 0
Charge:
⋅⋅⋅⋅±±±= ,e,e,eQ 32
C.e 191061 −×=Electron has charge of -e.
19-3. Insulators and Conductors
• Materials
Metals (Conductors) : free movement of electric chargese.g. Fe, Cu, Au, Ag, ···
Insulators : no free movement of electric chargese.g. Rubber, Ceramic, Cotton, ···
Semiconductor : Insulator but the electric charges can be created either by the thermal energy or by dopings
e.g. Si, GaAs, ···
• Charging by Induction
Grounded
Conductor
Charges are uniformly distributed at the surface of conductors.
Induced Charge due to polarization in insulators.
19-4. Coulomb’s Law
• Gravitational Force : between two massive objects
221
rmmGFG −=2
21
rmmFG −∝ G = 6.67×10-11N·m2/kg2⇒
Always attractive
• Electrical Force (Coulomb’s Law) : between two charged particles
221
rqqFC ∝ 2
21
rqqkF eC =⇒ ke= 8.99×10-11N·m2/C2
041πε
=ek2212
0 1085428 mNC. ⋅×=ε −
Permittivity of vacuum
Charge of an electron or proton : |e| = 1.602 × 10-19 C= (6.25 × 1018)-1C
• Vector notation of Coulomb’s Law
r̂rqqkF eC 2
21=r
12221
21 r̂rqqkF e=
r
21221
12 r̂rqqkF e=
r
1221 r̂r̂ −=
1221 FFrr
−= : Newton’s third law
The magnitude is the same but the direction is opposite.
• Coulomb Force due to many charges
+ –
+−
q1 q2
q3q4
12Fr13F
r
14Fr
Vector Sum
∑=++=i
iFFFFF 11413121
rrrrr
Example 19.2 Hydrogen Atom
+- e
p
Average separation r = a0 = 5.3 × 10-11m|e| = 1.60 × 10-19Cmp = 1.67 × 10-27kg , me = 9.11 × 10-31kg
( )( )211
219
2
29
221
10351060110998
m.C.
CmN.
rqqkF ee −
−
×
×⋅×==
Electrical force
N. 81028 −×=
Gravitational force
( )211
2731
2
211
221
103510671101191076
m.kg.kg.
kgmN.
rmmGFG −
−−−
×
×⋅×⋅×==
N. 471063 −×=
19-5. Electric Fields
Electric Field : Electric force per unit charge acting on a test charge
0qFEr
r≡
SI unit : N/C
For a point charge q:
Electric force on a test charge
r̂r
qqkF e 20=
r
r̂rqkE e 2=
rElectric Field:
Force ⇒ Field : Often much easier to describe the physical phenomena
For a group of charges : Vector sum
∑=i
ii
ie r̂
rqkE 2
rElectric Field:
Example 19.3: Electric Field of a dipole
221221 r̂rqkr̂
rqkEEE ee −=+=
rrr
21 EE =
( )θ−+θ= sinEsinEEy 21
011 =θ−θ= sinEsinE
θ=θ+θ== cosEcosEcosEEEx 121 2
θ= cosrqkE e 22
( ) ( )
( ) 2322
212222
2
2
ayqak
aya
ayqk
e
e
+=
+⋅
+=
For r >> a:
3
2yqakE e=
3
1r
∝
• Electric Field Due to Continuous Charge Distributions
Electric field at P due to charge ∆q
r̂rqkE e 2
∆=∆
r
Total Electric field at P
∑∆
=i
ii
ie r̂
rqkE 2
r
∆qi → 0 for the continuous charge distributions:
r̂rdqkr̂
rqkE e
ii
i
i
qe lim
i
∫∑ =∆
=→∆
220
r
• Charge Density for uniform distributions
- Charge desity , ρ : Charge per unit volumes
r̂rdVkE e ∫
ρ= 2
r
VQtot≡ρ ,
- Surface charge density, σ : Charge per unit area
r̂rdSkE e ∫
σ= 2
r
AQtot≡σ ,
- Linear charge density, λ : Charge per unit length
r̂rdlkE e ∫
λ= 2
r
LQtot≡λ ,
Example 19.4 Electric Field due to a Charged Rod
Total Charge Q
Linear charge density
lQ
=λ
∫∫+ λ
=λ
=ld
dee x
dxkrdlkE 22
( )dldlk
dldk
xk e
e
dlx
dxe +
λ=
+−λ=
−λ=
+=
=
111
( )dldQkE e
+=
2dQkE e≈If d >> l, : like a point charge
Example 19.5 Electric Field due to an Uniform Ring Charge
aQπ
=λ2
Total Charge Q Linear charge density
∫∫λ
== r̂rdlkEdE e 2
rr
xx dEdE 21 =,dEdE ⊥⊥ −= 21
( ) ( ) ( ) 23222122222 axxdqk
axx
axdqkcos
rdqkdE ee
ex+
=+
⋅+
=θ=
( ) ( ) Qaxxkdq
axxkE ee
x 23222322 +=
+= ∫
If x = 0, E = 0.
2xQkE e≈If x >> a, : like a point charge
19-6. Electric Field Lines
• Electric Field lines : Visualize the electric field patterns
1. E is tangent to the electric field lines
Er E
r
2. Strength of E ∝ the number of lines per unit area
• Rules for drawing Electric field lines
1. Begin on positive charges and terminate on negative charges.
2. The number of lines ∝ the magnitude of charges
3. No two lines cross each otherQAE ∝⋅
19-7. Electric FluxElectric Flux Φ
AE ⋅=Φ
: ∝ the number of the electric field lines through a surface
Electric Flux Φ is defined as
θ⋅=⋅≡Φ cosAEAErr
Ar
:Directed to the surface normalA
r
Electric flux, ∆Φi, through a small surface ∆Ai
iii AErr
∆⋅=∆Φ
A general definition of Φ :
∫∑ ⋅=∆⋅≡Φ→∆ Surfacei
iiA
AdEAElimi
rrrr
0
19-8. Gauss Law
• Electric Flux through a closed surface
∫ ⋅=Φ AdEC
rr
(i) No Charge inside
Er
Er
2Ar
1Ar 21 AEAEC
rrrr⋅+⋅=Φ
0=⋅+⋅−= AEAE
0=⋅=Φ ∫ AdEC
rr
(i) A charge inside
∫∫ ⋅=⋅=Φ21 SS
C AdEAdErrrr
S1
S2
qkrrqk ee π=π⋅= 44 2
2
041πε
=ek
0ε=Φ
qC
For an empirical surface, ΦC is the same. i.e. ΦC is independent of the surface contour.
• Gauss’s Law
0ε=⋅=Φ ∫ inc
CqAdE
rr
The net flux through any closed surface is proportional to the net charge inside of the surface.
19-9. Applications of Gauss’s Law
• Gauss’s law is very useful to determine the electric field due to charge distributions with high degree of symmetries, such as spherical, cylindrical, or long plane shapes.
Example 19.7 Electrical Field due to a point charge
0ε=⋅=Φ ∫
qAdEC
rr
0
24ε
=⋅πqEr
204
1rqE
πε= r̂
rqE 2
041πε
=⇒r
Example 19.8 A Spherically Symmetric Charge Distribution
Charge density ρ
343
aQ
π=ρ⇒Qa =ρπ 3
34
(a) r >a
00 ε=
ε=⋅=Φ ∫
QqAdE incC
rr
0
24ε
=⋅πQEr
2204
1rQk
rQE e=
πε=⇒ for r > a
(b) r <a
0ε=⋅=Φ ∫ inc
CqAdE
rr
3
0
0
34
1
r
dV
περ
=
⋅ρε
= ∫
3
3
0
24arQEr
ε=π 33
04 aQrk
arQE e=
πε=⇒
π=ρ 34
3aQ for r < a
Example 19.9 A Cylindrically Symmetric Charge Distribution+++++++++
xy
dy
θ
( ) 232222 yxxdykcos
yxdykdE e
ex+
λ=θ
+λ
=
( )∫∫∞+
∞−
∞+
∞− +
λ== dy
yxxkdyEE e
x 2322
0ε=⋅=Φ ∫ inc
CqAdE
rr
0
2ελ
=⋅πlErl
rk
rE e
λ=
λπε
= 22
1
0
Example 19.10 A Non-conducting Plane Sheet of Charge
0ε=⋅=Φ ∫ inc
CqAdE
rr
0
2ε⋅σ
=⋅AEA
02εσ
=E Constant i.e. Uniform field
19-10. Conductors in Electrostatic Equilibrium
• Charges freely move inside Conductors. • Excess charges entirely on its surface • E = 0 inside of Conductors
0εσ
=⋅=⋅=Φ ∫AAEAdEC
rr
0εσ
=E
For r < R, + +
++
++
+
++
++
++
+ ++
+++
R
r
Gauss Surface
0=⋅=Φ ∫ AdEC
rr
0=⇒ E
For r > R,
0ε=⋅=Φ ∫
QAdEC
rr
0
24ε
=π⋅QrE 2
041
rQE
πε=⇒