19 Electric Charge 1 static electricity applications: air fresheners

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19 Electric Charge
1 static electricity applications: air fresheners, xerography, painting cars, smokestack pollution
control, …
2 conductors, insulators, semiconductors, superconductors
3 Coulomb’s law: The force of attraction or repulsion between two point charges q1 and q 2 is
directly proportional to the product of their charges and inversely proportional to the square of the
distance between them.
F =K⋅
K=
q1 q 2
r2
(N ⋅ m C )
9
1
= 8.99 × 10
4πε 0
2
2
permittivity of vacuum
ε 0 = 8.85 × 10 −12
(C
2
N ⋅ m2
)
Principle of Superposition
v
v
v
v
F1,net = F12 + F13 + .... + F1n
electric potential
V =K⋅
q1
r
Coulomb’s law and the Principle of Superposition constitute the physical input for electrostatics.
For a conducting sphere of surface charge density ρ
R
dV = K ⋅
dq
r'
V = ∫ dV =
θ
dq = ρ ⋅ r 2 sin θ d θ dφ
r
K
∫∫ r ' ρr
2
sin θdθdφ
r’
φ : 0 → 2π
sin θdθ
r'
1 r'
dr '
= 2πKρr 2 ∫ ⋅
r ' Rr
R +r
2πkρr
dr '
=
R R∫− r
= Kρr 2 ⋅ 2π ∫
=
r ' 2 = R 2 + r 2 − 2 Rr cos θ
2 r ' dr ' = 2 Rr sin θ d θ
r'
sin θ d θ =
dr '
Rr
r ': R − r → R + r
4πKρr 2 K (4πr 2 )⋅ ρ Kq
=
=
R
R
R
Charge is quantized
Millikan oil drop experiment
When the Electric field is on, and the oil drop is
at rest.
qE = mg
When the Electric field is off, and the oil drop
reaches its terminal velocity.
mg =bυ
qE= bυ
bυ
q=
= ne
E
the elementary charge e can be obtained by the
{n1, n2, … .nm}
20 Electric Field
electric field
v v
E=Fq
Electric Field Due to a Point Charge
v
F=
1 q q0 ∧
r
4πε 0 r 2
v
v F
1
q ∧
E=
=
⋅ 2r
q0 4πε 0 r
+
The Electric Field Due to an Electric Dipole
E = E+ + E−
1 q
1 q
=
−
2
4πε 0 r+ 4πε 0 r−2


q 1
=
4πε 0  z 2


z 〉〉 d


1
1

−
2
2
d 
d  


z 2 1 +
 

1 −
2z 
2z  



2d 
2d
1 + 2 z −  1 − 2 z


2d
1
q
p
=
=
⋅ 3
2
4πε 0 z z
2πε 0 z
for
θ ≠0
1
p
⋅ 3 ⋅ f (θ )
E=
2πε 0 r
≅
q
4πε 0 z 2
dipole moment p = qd
d
2
d
r− = z +
2
r+ = z −



f (θ ) = 1
for
θ =0
The Electric Field Due to a Line of Charge
λ:the charge per unit length
ds = Rd φ
v
λ ds
1
⋅ 2
dE =
4 πε 0 r
v
v
= d E z + d E ||
dE z =
1 λ ds
cos θ
4πε 0 r 2
1 λ ds
sin θ
∫ dE || = 0
4 πε 0 r 2
v
v
v
λ R cos θ
E = ∫ d E = ∫ dE z =
dφ
4πε 0 r 2 ∫
dE || =
=
=
λ R cos θ ⋅ 2π
q cos θ
=
2
2
4πε 0 (z + R ) 4πε 0 (z 2 + R 2 )
qz
4πε 0 (z + R
2
for
v
E =
)
3
2 2
zˆ
z 〉〉 R
q
zˆ
4πε 0 z 2
v
z =0
E =0
point charge
net force = 0
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