gauss’s law ∇⋅ d = ρv - hong kong polytechnic ...em/em06pdf/static electric fields.pdf ·...
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E.1
Static Electric Fields
Coulomb’s Law
Gauss’s Law
Electric Potential
Electrical Properties of Materials– Conductors– Dielectrics
Capacitance
24ˆ
Rq
πεRΕ =
vρ=⋅∇ D
V−∇=E
EJ σ=ED ε=
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E.2
Introduction
Example: Electric field due to a chargeFrom the field distribution, we can guess that the divergence ofthe electric field is non-zero and the curl of the electric field is zero.
vρ=⋅∇ D
0=×∇ E
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E.3
Coulomb’s Law
Coulomb’s law1. An isolated charge q induces an electric field E at every
point in space and is given by
ε is the electrical permittivity of the medium
mVR
q /4
ˆ2πε
RΕ =
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E.4
Coulomb’s Law
2. In the presence of an electric field E at a give point in space, the force acting on a test charge q’ is
For a material with electrical permittivity ε, we have
Nq ΕF '=
EED orεεε ==
mFo /10854.8 12−×=ε
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E.5
Electric Field due to Charges
The expression for the electric field due to a single charge canbe extended to find the field due to multiple point charges as well as continuous charge distributions.
The electric field at position vector R caused by charges q1, q2,…, qN located at points with position vectors R1, R2,…, RN is given by
Note: is the unit vector from qi to the observation point.
Demonstrations: D4.1, D4.2, M4.1 Q1-2, M4.2 Q1, M4.3 Q1-2
mVqN
i i
i
i
i /)(4
11
2∑= −
−
−=
RRRR
RRΕ
πε
ii RRRR −− )(
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E.6
Electric Field due to Charges
Electric field due to a volume charge distribution
∫∫ ==⇒
==
vv
v
v
Rdvd
mVRdv
Rdqd
2/
22
''ˆ
41
/'4'ˆ
'4'ˆ
ρπε
περ
πε
'RΕΕ
'RRΕ
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E.7
Electric Field due to Charges
The electric field due to charges on a surface is
The electric field due to charges on a line is
∫=' 2'
'ˆ4
1S
s
Rdsρ
πε'RΕ
∫=' 2'
'ˆ4
1l
l
Rdlρ
πε'RΕ
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E.8
Gauss’s Law
Gauss’s Law states that the outward flux of D through a surface is proportional to the enclosed charged Q.
∫ =⋅
=⋅∇
S
v
QdSD
D ρ Differential form
Integral form
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E.9
Example: point charge
Construct a spherical surface with centre at q and radius R. Electric field is the same everywhere on the surface. Applying integral form of Gauss’s law gives
RDE
sRsD
ˆ4
/
4
ˆ
2
2
Rq
qDR
dsD
dDd
R
s R
s Rs
πεε
π
==∴
==
=
⋅=⋅
∫∫∫
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E.10
Example: infinite long line of charge
Since the line of charge is infinite in extent and is along the z-axis, D must be in the radial r-direction and must not depend on or z.
rDE
rrsD
ˆ2
/
2
ˆˆ0
2
0
r
hrhD
dzrdDd
l
lr
h
z rs
περε
ρπ
φπ
φ
==∴
==
⋅=⋅ ∫ ∫∫ = =
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E.11
Example: Spherical charge distribution
Find the electric field due to a spherical charge distribution of radius a with uniform volume charge density ρv in free space.
Gaussian surface of radius r a
Case (i): For a Gaussian surface of radius r less than the radius a of the charge distribution, the total charge enclosed by the Gaussian surface is:
vrQ ρπ 3
34
=
Applying Gauss’s Law to the Gaussian surface,
rDE
srsD
ˆ3
/
4
ˆ2
o
vo
r
s Rs
rQDr
dDd
ερε
π
==∴
==
⋅=⋅ ∫∫
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E.12
Example: Spherical charge distribution
Case (ii): For a Gaussian surface of radius r greater than a :
vaQ ρπ 3
34
=
Applying Gauss’s law,
rDE
srsD
ˆ3
/
4
ˆ
2
3
2
ra
QDr
dDd
o
vo
r
s Rs
ερε
π
==∴
==
⋅=⋅ ∫∫
Demonstrations: D4.9, D4.10, M4.6 Q2-3
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E.13
Electric Potential
∫ =
=×∇
Cd 0
0
lE.
E
According to vector identity, for any scalar V
0)( ≡∇×∇ V
The electrostatic field is irrotational or conservative, i.e.
So we can define a scalar electric potential V
∫∞ ⋅−=⇒
−∇=P
dV
V
lE
EThe integral is independent of the path taken
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E.14
Electric Potential
Electric potential due to point charges:
Electric potential due to volume distribution
Electric potential due to surface distribution
Electric potential due to line distribution
∑= −
=N
i i
iqV14
1RRπε
∫='
''4
1v
v dvR
V ρπε
∫='
''4
1S
s dsR
V ρπε
∫='
''4
1l
l dlR
V ρπε
Demonstrations: M4.1 Q.3 M4.2 Q.2, M4.6 Q.1
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E.15
Electrical Properties of Materials
The electromagnetic constitutive parameters of a material medium are – Electrical permittivity ε– Electrical permeability µ– Conductivity σ
Throughout this subject, we assume the materials are– Homogenous
• Constitutive parameters do not vary from point to point– Isotropic
• Constitutive parameters are independent of direction
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E.16
Conductors
A conductor has a large number of loosely attached electrons in the outermost shells of the atoms.
Upon applying an external electric field, the electrons migrate from one atom to the next along a direction opposite that of theexternal field.
Their movement gives rise to a conduction current
– Point form of Ohm’s law
EJ σ=
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E.17
Conductors
Conductivity of some common materials
Perfect dielectric: σ=0 and then J = 0 regardless of E
Perfect conductor:σ=∞ and then E = 0 regardless of J
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E.18
Dielectrics
Dielectrics are insulating materials, and contain bound charges which cannot move freely to generate currents.
Bound charges can move short distances under an electric field to form electric dipoles.
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E.19
Dielectrics
A dielectric medium polarized by an external electric field
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E.20
Dielectrics
The polarization (or bound) charges can also contribute to the electric field E as the free charges.
It can be shown that D is related to E and P according to the following equation:
PED += oεIn most materials, P is in the same direction and proportional to E so that
EE εεε == ro
where εo is the permittivity of vacuum, εr is the relative permittivity (or dielectric constant) of the material, ε is the permittivity of the material, and χ is the susceptibility of the material.
EED χεε oo +=
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E.21
Dielectric Strength
If the applied electric field exceeds the dielectric strength of the material, it will free the electronics completely from the molecules in the form of a conduction current. – Spark may occur– Dielectric breakdown
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E.22
Capacitance
Capacitance between two conductors is defined as:
)(FVQC =
where Q is the charge on the conductor and V is the potential difference between the conductors.
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E.23
Example 4-11: Parallel-Plate Capacitor
Because of the applied voltage difference, charge +Qaccumulates uniformly on the top plate and –Q accumulates uniformly on the lower plate.In the dielectric medium between the plates, the charges induces a uniform electric field (fringing field is ignored).
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E.24
Example 4-11: Parallel-Plate Capacitor
Applying Gauss’s law, the electric field can be determined:
AQzzEz
s
εερ
/ˆ/ˆ
ˆ
−=−=−=E
( )
dA
dAQQ
EdQ
d
QVQC
d
εε
=
==
⋅−=
=
∫
/
0lE
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E.25
Example: 4-12: Coaxial Line
Construct a Gaussian cylindrical surface of unit length and radius r. The E-field at radius r is:
rlQ
QrlDd rs
πεε
π
2ˆ/
2
rDE
sD
−==⇒
−==⋅∫
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E.26
Example: 4-12: Coaxial Line
( )abl
Q
drrl
Q
dV
b
a
b
a
/ln2
ˆ2
ˆ
πε
πε
=
⋅
−−=
⋅−=
∫
∫rr
lE
( )abl
VQC
/ln2πε
=
=∴
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E.27
Exercise