PHYS1112 - Electricity and Magnetism
Lecture Notes
Dr. Jason Chun Shing Pun
Department of Physics
The University of Hong Kong
January 2005
Contents
1 Vector Algebra
1.1 Definitions . . . . . . . . . . . . . . .
1.2 Vector Algebra . . . . . . . . . . . .
1.3 Components of Vectors . . . . . . . .
1.4 Multiplication of Vectors . . . . . . .
1.5 Vector Field (Physics Point of View)
1.6 Other Topics . . . . . . . . . . . . .
2 Electric Force & Electric Field
2.1 Electric Force . . . . . . . . . .
2.2 The Electric Field . . . . . . . .
2.3 Continuous Charge Distribution
2.4 Electric Field Lines . . . . . . .
2.5 Point Charge in E-field . . . . .
2.6 Dipole in E-field . . . . . . . . .
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3 Electric Flux and Gauss’ Law
3.1 Electric Flux . . . . . . . . . . . .
3.2 Gauss’ Law . . . . . . . . . . . . .
3.3 E-field Calculation with Gauss’ Law
3.4 Gauss’ Law and Conductors . . . .
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4 Electric Potential
4.1 Potential Energy and Conservative Forces . .
4.2 Electric Potential . . . . . . . . . . . . . . . .
4.3 Relation Between Electric Field E and Electric
4.4 Equipotential Surfaces . . . . . . . . . . . . .
5 Capacitance and DC Circuits
5.1 Capacitors . . . . . . . . . .
5.2 Calculating Capacitance . .
5.3 Capacitors in Combination .
5.4 Energy Storage in Capacitor
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Potential V
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1
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2
4
6
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8
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12
18
21
22
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25
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28
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31
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36
36
40
45
48
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51
51
51
54
55
5.5
5.6
5.7
5.8
5.9
5.10
Dielectric Constant . . . .
Capacitor with Dielectric .
Gauss’ Law in Dielectric .
Ohm’s Law and Resistance
DC Circuits . . . . . . . .
RC Circuits . . . . . . . .
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6 Magnetic Force
6.1 Magnetic Field . . . . . . .
6.2 Motion of A Point Charge in
6.3 Hall Effect . . . . . . . . . .
6.4 Magnetic Force on Currents
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Magnetic Field
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7 Magnetic Field
7.1 Magnetic Field . . . . . . . . . . . . .
7.2 Parallel Currents . . . . . . . . . . . .
7.3 Ampère’s Law . . . . . . . . . . . . . .
7.4 Magnetic Dipole . . . . . . . . . . . . .
7.5 Magnetic Dipole in A Constant B-field
7.6 Magnetic Properties of Materials . . .
8 Faraday’s Law of Induction
8.1 Faraday’s Law . . . . . . .
8.2 Lenz’ Law . . . . . . . . .
8.3 Motional EMF . . . . . .
8.4 Induced Electric Field . .
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9 Inductance
9.1 Inductance . . . . . . . . . . . . . . . . .
9.2 LR Circuits . . . . . . . . . . . . . . . .
9.3 Energy Stored in Inductors . . . . . . . .
9.4 LC Circuit (Electromagnetic Oscillator) .
9.5 RLC Circuit (Damped Oscillator) . . . .
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57
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81
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98
98
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100
104
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107
107
110
112
113
115
10 AC Circuits
10.1 Alternating Current (AC) Voltage . . . . .
10.2 Phase Relation Between i, V for R,L and C
10.3 Single Loop RLC AC Circuit . . . . . . . .
10.4 Resonance . . . . . . . . . . . . . . . . . .
10.5 Power in AC Circuits . . . . . . . . . . . .
10.6 The Transformer . . . . . . . . . . . . . .
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116
116
117
119
121
121
123
ii
11 Displacement Current and
11.1 Displacement Current . .
11.2 Induced Magnetic Field .
11.3 Maxwell’s Equations . .
Maxwell’s
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iii
Equations
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125
125
127
128
Chapter 1
Vector Algebra
1.1
Definitions
A vector consists of two components: magnitude and direction .
(e.g. force, velocity, pressure)
A scalar consists of magnitude only.
(e.g. mass, charge, density)
1.2
Vector Algebra
Figure 1.1: Vector algebra
~a + ~b = ~b + ~a
~ = (~a + ~c) + d~
~a + (~c + d)
1.3. COMPONENTS OF VECTORS
1.3
2
Components of Vectors
Usually vectors are expressed according to coordinate system. Each vector can
be expressed in terms of components.
The most common coordinate system: Cartesian
~a = ~ax + ~ay + ~az
Magnitude of ~a = |~a| = a,
a=
q
a2x + a2y + a2z
~a = ~ax + ~ay
q
a2x + a2y
a =
ax = a cosφ; ay = a sinφ
ay
tanφ =
ax
Figure 1.2: φ measured anti-clockwise
from position x-axis
Unit vectors have magnitude of 1
â =
~a
= unit vector along ~a direction
|~a|
î ĵ
l l
x y
k̂
l
z
are unit vectors along
directions
~a = ax î + ay ĵ + az k̂
Other coordinate systems:
1.3. COMPONENTS OF VECTORS
3
1. Polar Coordinate:
~a = ar r̂ + aθ θ̂
Figure 1.3: Polar Coordinates
2. Cylindrical Coordinates:
~a = ar r̂ + aθ θ̂ + az ẑ
r̂ originated from nearest point on
z-axis (Point O’)
Figure 1.4: Cylindrical Coordinates
3. Spherical Coordinates:
~a = ar r̂ + aθ θ̂ + aφ φ̂
r̂ originated from Origin O
Figure 1.5: Spherical Coordinates
1.4. MULTIPLICATION OF VECTORS
1.4
4
Multiplication of Vectors
1. Scalar multiplication:
~b=m~a
~b,~a are vectors; m is a scalar
If
then
b=m a
(Relation between magnitude)
o
bx =m ax
Components also follow relation
by =m ay
i.e.
~a =
ax î +
ay ĵ +
az k̂
m~a = max î + may ĵ + maz k̂
2. Dot Product (Scalar Product):
~a · ~b = |~a| · |~b| cosφ
Result is always a scalar. It can be positive or negative depending on φ.
~a · ~b = ~b · ~a
Notice: ~a · ~b = ab cosφ = ab cosφ0
i.e. Doesn’t matter how you measure
angle φ between vectors.
Figure 1.6: Dot Product
î · î = |î| |î| cos0◦ = 1 · 1 · 1 = 1
î · ĵ = |î| |ĵ| cos90◦ = 1 · 1 · 0 = 0
î · î = ĵ · ĵ = k̂ · k̂ = 1
î · ĵ = ĵ · k̂ = k̂ · î = 0
~a = ax î + ay ĵ + az k̂
~b = bx î + by ĵ + bz k̂
then ~a · ~b = ax bx + ay by + az bz
~a · ~a = |~a| · |~a| cos0◦ = a · a = a2
If
1.4. MULTIPLICATION OF VECTORS
5
3. Cross Product (Vector Product):
If
~c = ~a × ~b,
then c = |~c| = a b sinφ
~a × ~b 6= ~b × ~a !!!
~a × ~b = −~b × ~a
Figure 1.7: Note: How angle φ is measured
• Direction of cross product determined from right hand rule.
• Also, ~a × ~b is ⊥ to ~a and ~b, i.e.
~a · (~a × ~b) = 0
~b · (~a × ~b) = 0
• IMPORTANT:
~a × ~a = a · a sin0◦ = 0
|î × î| = |î| |î| sin0◦ = 1 · 1 · 0 = 0
|î × ĵ| = |î| |ĵ| sin90◦ = 1 · 1 · 1 = 1
î × î = ĵ × ĵ = k̂ × k̂ = 0
î × ĵ = k̂; ĵ × k̂ = î; k̂ × î = ĵ
¯ î
¯
~a × ~b = ¯¯ ax
¯
bx
ĵ k̂ ¯¯
¯
(ay bz − az by ) î
ay az ¯¯ =
+(az bx − ax bz ) ĵ
by bz
+(ax by − ay bx ) k̂
1.5. VECTOR FIELD (PHYSICS POINT OF VIEW)
6
4. Vector identities:
~a × (~b + ~c) = ~a × ~b + ~a × ~c
~a · (~b × ~c) = ~b · (~c × ~a) = ~c · (~a × ~b)
~a × (~b × ~c) = (~a · ~c) ~b − (~a · ~b) ~c
1.5
Vector Field (Physics Point of View)
~
A vector field F(x,
y, z) is a mathematical function which has a vector output
for a position input.
~ y, z))
(Scalar field U(x,
1.6
Other Topics
Tangential Vector
Figure 1.8: d~l is a vector that is always tangential to the curve C with infinitesimal
length dl
Surface Vector
Figure 1.9: d~a is a vector that is always perpendicular to the surface S with
infinitesimal area da
1.6. OTHER TOPICS
Some uncertainty!
7
(d~a versus − d~a)
Two conventions:
• Area formed from a closed curve
Figure 1.10: Direction of d~a determined from right-hand rule
• Closed surface enclosing a volume
Figure 1.11: Direction of d~a going from inside to outside
Chapter 2
Electric Force & Electric Field
2.1
Electric Force
The electric force between two charges
q1 and q2 can be described by
Coulomb’s Law.
F~12 = F orce on q1 exerted by q2
F~12 =
where r̂12 =
1
4π²0
· qr12q2 · r̂12
12
~r12
is the unit vector which locates particle 1 relative to particle 2.
|~r12 |
i.e.
~r12 = ~r1 − ~r2
• q1 , q2 are electrical charges in units of Coulomb(C)
• Charge is quantized
Recall 1 electron carries 1.602 × 10−19 C
• ²0 = Permittivity of free space = 8.85 × 10−12 C 2 /N m2
COULOMB’S LAW:
(1) q1 , q2 can be either positive or negative.
2.2. THE ELECTRIC FIELD
9
(2) If q1 , q2 are of same sign, then the force experienced by q1 is in direction
away from q2 , that is, repulsive.
(3) Force on q2 exerted by q1 :
F~21 =
1
q2 q1
· 2 · r̂21
4π²0 r21
BUT:
r12 = r21 = distance between q1 , q2
~r2 − ~r1
−~r12
~r21
=
=
= −r̂12
r̂21 =
r21
r21
r12
∴
F~21 = −F~12 Newton’s 3rd Law
SYSTEM WITH MANY CHARGES:
The total force experienced by charge
q1 is the vector sum of the forces on q1
exerted by other charges.
F~1 = Force experienced by q1
= F~1,2 + F~1,3 + F~1,4 + · · · + F~1,N
PRINCIPLE OF SUPERPOSITION:
F~1 =
N
X
F~1,j
j=2
2.2
The Electric Field
While we need two charges to quantify the electric force, we define the electric
field for any single charge distribution to describe its effect on other charges.
2.2. THE ELECTRIC FIELD
10
Total force F~ = F~1 + F~2 + · · · + F~N
The electric field is defined as
F~
~
=E
q0 →0 q0
lim
(a) E-field due to a single charge qi :
From the definitions of Coulomb’s Law, the
force experienced at location of q0 (point P)
F~0,i =
1
q0 qi
· 2 · r̂0,i
4π²0 r0,i
where r̂0,i is the unit vector along the direction from charge qi to q0 ,
r̂0,i = Unit vector from charge qi to point P
= r̂i (radical unit vector from qi )
~
~ = lim F
Recall E
q0 →0 q0
∴ E-field due to qi at point P:
~i =
E
1
qi
· 2 · r̂i
4π²0 ri
where ~ri = Vector pointing from qi to point P,
thus r̂i = Unit vector pointing from qi to point P
Note:
(1) E-field is a vector.
(2) Direction of E-field depends on both position of P and sign of qi .
(b) E-field due to system of charges:
Principle of Superposition:
In a system with N charges, the total E-field due to all charges is the
vector sum of E-field due to individual charges.
2.2. THE ELECTRIC FIELD
i.e.
11
~ =
E
X
~i =
E
i
1 X qi
r̂i
4π²0 i ri2
(c) Electric Dipole
System of equal and opposite charges
separated by a distance d.
Figure 2.1: An electric dipole. (Direction of
d~ from negative to positive charge)
Electric Dipole Moment
p~ = q d~ = qddˆ
p = qd
Example:
~ due to dipole along x-axis
E
Consider point P at distance x along the perpendicular axis of the dipole p~ :
~
E
Notice:
=
~+
E
↑
(E-field
due to +q)
+
~−
E
↑
(E-field
due to −q)
~ + and E
~ − cancel out.
Horizontal E-field components of E
∴ Net E-field points along the axis oppo-
site to the dipole moment vector.
2.3. CONTINUOUS CHARGE DISTRIBUTION
12
Magnitude of E-field = 2E+ cos θ
E+ or E− magnitude!
µ z
∴E =2
}|
But r =
s
³ d ´2
cos θ =
∴E =
{ ¶
1
q
· 2
4π²0 r
2
cos θ
+ x2
d/2
r
p
1
·
4π²0 [x2 + ( d2 )2 ] 32
(p = qd)
Special case:
When x À d
3
d 3
d
[x2 + ( )2 ] 2 = x3 [1 + ( )2 ] 2
2
2x
• Binomial Approximation:
(1 + y)n ≈ 1 + ny
E-field of dipole +
• Compare with
if y ¿ 1
1
p
1
· 3 ∼ 3
4π²0 x
x
1
E-field for single charge
r2
• Result also valid for point P along any axis with respect to dipole
2.3
Continuous Charge Distribution
E-field at point P due to dq:
~ =
dE
dq
1
· 2 · r̂
4π²0 r
2.3. CONTINUOUS CHARGE DISTRIBUTION
13
∴ E-field due to charge distribution:
ˆ
~ =
E
ˆ
~ =
dE
V olume
V olume
1
4π²0
·
dq
r2
· r̂
(1) In many cases, we can take advantage of the symmetry of the system to
simplify the integral.
(2) To write down the small charge element dq:
1-D
2-D
3-D
Example 1:
dq = λ ds
dq = σ dA
dq = ρ dV
λ = linear charge density
σ = surface charge density
ρ = volume charge density
ds = small length element
dA = small area element
dV = small volume element
Uniform line of charge
charge per
unit length
=λ
(1) Symmetry considered: The E-field from +z and −z directions cancel along
z-direction, ∴ Only horizontal E-field components need to be considered.
(2) For each element of length dz, charge dq = λdz
∴ Horizontal E-field at point P due to element dz =
~ cos θ =
|dE|
1
λdz
· 2 cos θ
4π²0 r
|
{z
}
dEdz
∴ E-field due to entire line charge at point P
ˆL/2
E =
−L/2
λdz
1
· 2 cos θ
4π²0 r
ˆL/2
= 2
0
dz
λ
· 2 cos θ
4π²0 r
2.3. CONTINUOUS CHARGE DISTRIBUTION
14
To calculate this integral:
• First, notice that x is fixed, but z, r, θ all varies.
• Change of variable (from z to θ)
(1)
z = x tan θ
x = r cos θ
∴ dz = x sec2 θ dθ
∴ r 2 = x2 sec2 θ
z=0 ,
(2) When
θ = 0◦
z = L/2 θ = θ0
λ
E = 2·
4π²0
= 2·
λ
4π²0
λ
4π²0
λ
= 2·
4π²0
λ
= 2·
4π²0
= 2·
E=
where tan θ0 =
ˆθ0
0
ˆθ0
0
L/2
x
x sec2 θ dθ
· cos θ
x2 sec2 θ
1
· cos θ dθ
x
¯θ0
1
· (sin θ)¯¯
0
x
1
· · sin θ0
x
1
L/2
· ·q
x
x2 + ( L2 )2
·
1
λL
· q
4π²0 x x2 + ( L )2
2
along x-direction
Important limiting cases:
1
λL
· 2
4π²0 x
But λL = Total charge on rod
∴ System behave like a point charge
1. x À L :
E+
2. L À x :
E+
1
λL
·
4π²0 x · L2
Ex =
λ
2π²0 x
ELECTRIC FIELD DUE TO INFINITELY LONG LINE OF CHARGE
2.3. CONTINUOUS CHARGE DISTRIBUTION
Example 2:
15
Ring of Charge
E-field at a height z above a ring of
charge of radius R
(1) Symmetry considered: For every charge element dq considered, there exists
~ field components cancel.
dq0 where the horizontal E
⇒ Overall E-field lies along z-direction.
(2) For each element of length dz, charge
dq
=
λ
↑
Linear
charge density
·
ds
↑
Circular
length element
dq = λ · R dφ,
where φ is the angle
measured on the ring plane
∴ Net E-field along z-axis due to dq:
dE =
dq
1
· 2 · cos θ
4π²0 r
2.3. CONTINUOUS CHARGE DISTRIBUTION
16
ˆ
Total E-field
=
dE
ˆ
2π
=
0
Note:
Here in this case,
convert r, θ to R, z.
1
λR dφ
·
· cos θ
4π²0
r2
θ, R and r are fixed as φ varies! BUT we want to
λRz
1
· 3
E=
4π²0 r
E=
BUT:
Example 3:
z
(cos θ = )
r
1
λ(2πR)z
· 2
4π²0 (z + R2 )3/2
ˆ
2π
dφ
0
along z-axis
λ(2πR) = total charge on the ring
E-field from a disk of surface charge density σ
We find the E-field of a disk by
integrating concentric rings of
charges.
2.3. CONTINUOUS CHARGE DISTRIBUTION
17
Total charge of ring
dq = σ · ( 2πr
| {z dr} )
Area of the ring
Recall from Example 2:
E-field from ring: dE =
1
dq z
· 2
4π²0 (z + r2 )3/2
ˆ R
1
2πσr dr · z
E =
4π²0 0 (z 2 + r2 )3/2
ˆ R
r dr
1
2πσz 2
=
4π²0 0
(z + r2 )3/2
∴
• Change of variable:
⇒
u = z 2 + r2
du = 2r dr
⇒ (z 2 + r2 )3/2 = u3/2
⇒ r dr = 21 du
• Change of integration limit:
(
r = 0 , u = z2
r = R , u = z 2 + R2
1
∴ E =
· 2πσz
4π²0
ˆ
BUT:
u−3/2 du =
ˆ
z 2 +R2
z2
1 −3/2
u
du
2
u−1/2
= −2u−1/2
−1/2
∴ E
¯z 2 +R2
1
¯
σz (−u−1/2 )¯ 2
z
2²0
Ã
!
1
1
−1
σz √ 2
+
=
2²0
z + R2 z
=
σ
E=
2²0
"
z
1− √ 2
z + R2
#
2.4. ELECTRIC FIELD LINES
18
VERY IMPORTANT LIMITING CASE:
If R À z, that is if we have an infinite sheet of charge with charge density σ:
"
σ
z
1− √ 2
E =
2²0
z + R2
¸
·
σ
z
'
1−
2²0
R
E≈
#
σ
2²0
E-field is normal to the charged surface
Figure 2.2: E-field due to an infinite sheet of charge, charge density = σ
Q: What’s the E-field belows the charged sheet?
2.4
Electric Field Lines
To visualize the electric field, we can use a graphical tool called the electric
field lines.
Conventions:
1. The start on position charges and end on negative charges.
2. Direction of E-field at any point is given by tangent of E-field line.
3. Magnitude of E-field at any point is proportional to number of E-field lines
per unit area perpendicular to the lines.
2.4. ELECTRIC FIELD LINES
19
2.4. ELECTRIC FIELD LINES
20
2.5. POINT CHARGE IN E-FIELD
2.5
21
Point Charge in E-field
~ the force experienced by the charge is
When we place a charge q in an E-field E,
~ = m~a
F~ = q E
Applications:
Ink-jet printer, TV cathoderay tube.
Example:
Ink particle has mass m, charge q (q < 0 here)
Assume that mass of inkdrop is small, what’s the deflection y of the charge?
Solution:
First, the charge carried by the inkdrop is negtive, i.e. q < 0.
Note:
Horizontal motion:
~ points in opposite direction of E.
~
qE
Net force = 0
∴ L = vt
(2.1)
2.6. DIPOLE IN E-FIELD
Vertical motion:
22
~ À |m~g |,
|q E|
q is negative,
∴ Net force = −qE = ma
∴ a=−
Vertical distance travelled:
y=
2.6
(Newton’s 2nd Law)
qE
m
(2.2)
1 2
at
2
Dipole in E-field
Consider the force exerted on the dipole in an external E-field:
Assumption: E-field from dipole doesn’t affect the external E-field.
• Dipole moment:
p~ = q d~
• Force due to the E-field on +ve
and −ve charge are equal and
opposite in direction. Total external force on dipole = 0.
BUT:
There is an external torque on
the center of the dipole.
Reminder:
Force F~ exerts at point P.
The force exerts a torque
~τ = ~r × F~ on point P with
respect to point O.
Direction of the torque vector ~
τ is determined from the right-hand rule.
2.6. DIPOLE IN E-FIELD
Reference:
23
Halliday Vol.1 Chap 9.1 (Pg.175)
Chap 11.7 (Pg.243)
torque
work done
Net torque ~τ
• direction:
torque
clockwise
• magnitude:
τ = τ+ve + τ−ve
d
d
= F · sin θ + F · sin θ
2
2
= qE · d sin θ
= pE sin θ
~
~τ = p~ × E
Energy Consideration:
When the dipole p~ rotates dθ, the E-field does work.
Work done by external E-field on the dipole:
dW = −τ dθ
Negative sign here because torque by E-field acts to decrease θ.
BUT: Because E-field is a conservative force field
potential energy (U ) for the system, so that
1 2
dU = −dW
∴ For the dipole in external E-field:
dU = −dW = pE sin θ dθ
ˆ
∴ U (θ) =
ˆ
dU =
pE sin θ dθ
= −pE cos θ + U0
1
2
more to come in Chap.4 of notes
ref. Halliday Vol.1 Pg.257, Chap 12.1
, we can define a
2.6. DIPOLE IN E-FIELD
24
set U (θ = 90◦ ) = 0,
∴ 0 = −pE cos 90◦ + U0
∴ U0 = 0
∴ Potential energy:
~
U = −pE cos θ = −~p · E
Chapter 3
Electric Flux and Gauss’ Law
3.1
Electric Flux
Latin: flux = ”to flow”
Graphically:
Electric flux ΦE represents the number of E-field lines
crossing a surface.
Mathematically:
~ is perpendicular to the area A.
Reminder: Vector of the area A
~ is not
For non-uniform E-field & surface, direction of the area vector A
uniform.
~ = Area vector for
dA
small area element
dA
3.1. ELECTRIC FLUX
∴ Electric flux
~ through surface S:
Electric flux of E
26
~ · dA
~
dΦE = E
ˆ
~ · dA
~
ΦE =
E
S
ˆ
= Surface integral over surface S
S
= Integration of integral over all area elements on surface S
Example:
~ =
E
1
−2q
−q
· 2 r̂ =
r̂
4π²0 r
2π²0 R2
~ = dA r̂
For a hemisphere, dA
ˆ
−q
ΦE =
r̂ · (dA r̂)
2
S 2π²0 R ˆ
q
= −
dA
2π²0 R2 S
(∵ r̂ · r̂ = 1)
| {z }
2πR2
=
−q
²0
For a closed surface:
~
Recall: Direction of area vector dA
goes from inside to outside of closed
surface S.
3.1. ELECTRIC FLUX
27
˛
~ · dA
~
E
Electric flux over closed surface S: ΦE =
S
˛
= Surface integral over closed surface S
S
Example:
Electric flux of charge q over closed
spherical surface of radius R.
~ =
E
1 q
q
· 2 r̂ =
r̂
4π²0 r
4π²0 R2
at the surface
~ = dA · r̂
Again, dA
˛ z
∴ ΦE
~
E
}|
{
~
dA
z }| {
q
=
r̂
·
dA r̂
2
S 4π²0 R
˛
q
dA
=
4π²0 R2 S
| {z }
ΦE
Total surface area of S = 4πR2
q
=
²0
IMPORTANT POINT:
If we remove the spherical symmetry of closed surface S, the total number of
E-field lines crossing the surface remains the same.
∴ The electric flux ΦE
3.2. GAUSS’ LAW
28
˛
˛
~ · dA
~=
E
ΦE =
S
3.2
S0
~ · dA
~= q
E
²0
Gauss’ Law
˛
ΦE =
~ · dA
~= q
E
²0
S
for any closed surface S
And q is the net electric charge enclosed in closed surface S.
• Gauss’ Law is valid for all charge distributions and all closed surfaces.
(Gaussian surfaces)
• Coulomb’s Law can be derived from Gauss’ Law.
• For system with high order of symmetry, E-field can be easily determined if
we construct Gaussian surfaces with the same symmetry and applies Gauss’
Law
3.3
E-field Calculation with Gauss’ Law
(A) Infinite line of charge
Linear charge density: λ
Cylindrical symmetry.
E-field directs radially outward from the
rod.
Construct a Gaussian surface S in the
shape of a cylinder, making up of a
curved surface S1 , and the top and
bottom circles S2 , S3 .
˛
Gauss’ Law:
~ · dA
~ = Total charge = λL
E
²0
²0
S
3.3. E-FIELD CALCULATION WITH GAUSS’ LAW
˛
ˆ
ˆ
~ · dA
~=
E
ˆ
~ · dA
~+
E
S
|
S1
{z
}
~ A
~
Ekd
∴
29
~ · dA
~+
E
|
S2
{z
~ · dA
~
E
S3
}
~
~
= 0 ∵E⊥d
A
ˆ
E
dA =
S
| 1{z }
λL
²0
Total area of surface S1
λL
E(2πrL) =
²0
∴ E =
λ
2π²0 r
(Compare with Chapter 2 note)
~ =
E
λ
r̂
2π²0 r
(B) Infinite sheet of charge
Uniform surface charge density:
σ
Planar symmetry.
E-field directs perpendicular to
the sheet of charge.
Construct Gaussian surface S in
the shape of a cylinder (pill
box) of cross-sectional area A.
˛
Gauss’ Law:
ˆ
~ · dA
~ = Aσ
E
²0
S
~ · dA
~=0
E
ˆ
S1
ˆ
~ · dA
~+
E
S2
~ ⊥ dA
~ over whole surface S1
∵E
~ · dA
~ = 2EA (E
~ k dA
~ 2, E
~ k dA
~ 3)
E
S3
3.3. E-FIELD CALCULATION WITH GAUSS’ LAW
Note: For S2 ,
For S3 ,
∴
2EA =
both
both
Aσ
²0
30
~ and dA
~ 2 point up
E
~ and dA
~ 3 point down
E
⇒
E=
σ
2²0
(Compare with Chapter 2 note)
(C) Uniformly charged sphere
Total charge = Q
Spherical symmetry.
(a) For r > R:
Consider a spherical Gaussian surface S of
radius r:
~ k dA
~ k r̂
E
˛
~ · dA
~=Q
E
Gauss’ Law:
²0
S
˛
Q
E · dA =
²0
S
˛
Q
E
dA =
²0
S
| {z }
surface area of S = 4πr2
∴
~ =
E
Q
r̂ ;
4π²0 r2
for r > R
(b) For r < R:
Consider a spherical Gaussian surface S 0 of
radius r < R, then total charge included q is
proportional to the volume included by S 0
∴
Volume enclosed by S 0
q
=
Q
Total volume of sphere
3.4. GAUSS’ LAW AND CONDUCTORS
˛
Gauss’ Law:
q
4/3 πr3
=
Q
4/3 πR3
31
⇒
q=
r3
Q
R3
~ · dA
~= q
E
²0
S0
˛
dA =
E
0
| S{z }
r3 1
·Q
R3 ²0
surface area of S 0 = 4πr2
∴
3.4
~ =
E
1
Q
· 3 r r̂ ;
4π²0 R
for r ≤ R
Gauss’ Law and Conductors
For isolated conductors, charges are free
to move until all charges lie outside the
surface of the conductor. Also, the Efield at the surface of a conductor is perpendicular to its surface. (Why?)
Consider Gaussian surface S of shape of cylinder:
˛
~ · dA
~ = σA
E
²0
S
3.4. GAUSS’ LAW AND CONDUCTORS
32
ˆ
BUT
ˆS1
~ · dA
~=0 (∵ E
~ ⊥ dA
~)
E
~ · dA
~=0
E
~ = 0 inside conductor )
(∵ E
S3
ˆ
ˆ
~ · dA
~ = E
E
S2
~ k dA
~)
(∵ E
dA
S2
| {z }
Area of S2
= EA
∴ Gauss’ Law
∴
⇒
EA =
σA
²0
On conductor’s surface E =
σ
²0
BUT, there’s no charge inside conductors.
∴
Notice:
Inside conductors E = 0
Always!
Surface charge density on a conductor’s surface is not uniform.
Example: Conductor with a charge inside
Note: This is not an isolated system (because of the charge inside).
Example:
3.4. GAUSS’ LAW AND CONDUCTORS
33
I. Charge sprayed on a conductor sphere:
First, we know that charges all move
to the surface of conductors.
(i) For r < R:
Consider Gaussian surface S2
˛
~ · dA
~ = 0 ( ∵ no charge inside )
E
S2
⇒ E = 0 everywhere.
(ii) For r ≥ R:
Consider Gaussian surface S1 :
˛
~ · dA
~ = Q
E
²0
S1
˛
~ = Q
E
dA
²0
S1
For a conductor
z }| {
~ k dA
~ k r̂ )
( E
| {z }
4πr2
E =
II. Conductor sphere with hole inside:
Q
4π²0 r2
|{z}
Spherically symmetric
3.4. GAUSS’ LAW AND CONDUCTORS
34
Consider Gaussian surface S1 :
charge included = 0
Total
∴ E-field = 0 inside
The E-field is identical to the case of a
solid conductor!!
III. A long hollow cylindrical conductor:
Example:
Inside hollow cylinder ( +2q )
(
Inner radius
Outer radius
a
b
Outside hollow cylinder ( −3q )
(
Question:
Inner radius
Outer radius
c
d
Find the charge on each surface of the conductor.
For the inside hollow cylinder, charges distribute only on the surface.
∴
Inner radius a surface, charge = 0
and Outer radius b surface, charge = +2q
For the outside hollow cylinder, charges do not distribute only on
outside.
∵
It’s not an isolated system. (There are charges inside!)
Consider Gaussian surface S 0 inside the conductor:
E-field always = 0
∴
Need charge −2q on radius c surface to balance the charge of inner
cylinder.
So charge on radius d surface = −q.
(Why?)
IV. Large sheets of charge:
Total charge Q on sheet of area A,
3.4. GAUSS’ LAW AND CONDUCTORS
∴
Surface charge density σ =
35
Q
A
By principle of superposition
Region A:
Region B:
Region C:
E=0
Q
E=
²0 A
E=0
E=0
Q
E=
²0 A
E=0
Chapter 4
Electric Potential
4.1
Potential Energy and Conservative Forces
(Read Halliday Vol.1 Chap.12)
Electric force is a conservative force
Work done by the electric force F~ as a
charge moves an infinitesimal distance d~s
along Path A = dW
Note:
d~s is in the tangent direction of the curve of Path A.
dW = F~ · d~s
~ in moving the particle from Point 1 to Point 2
∴ Total work done W by force F
ˆ
2
W =
F~ · d~s
1
Path A
ˆ
2
= Path Integral
1
Path A
=
Integration over Path A from Point 1 to Point 2.
4.1. POTENTIAL ENERGY AND CONSERVATIVE FORCES
37
DEFINITION:
A force is conservative if the work done on a particle by
the force is independent of the path taken.
∴ For conservative forces,
ˆ
ˆ
2
2
F~ · d~s =
F~ · d~s
1
1
Path A
Path B
Let’s consider a path starting at point
1 to 2 through Path A and from 2 to 1
through Path C
ˆ
2
Work done =
ˆ
F~ · d~s
+
Path A
2
=
F~ · d~s
2
1
ˆ
1
Path C
ˆ
2
F~ · d~s −
F~ · d~s
1
1
Path A
Path B
DEFINITION: The work done by a conservative force on a particle when it
moves around a closed path returning to its initial position is zero.
~ × F~ = 0 everywhere for conservative force F~
MATHEMATICALLY, ∇
Conclusion: Since the work done by a conservative force F~ is path-independent,
we can define a quantity, potential energy, that depends only on the
position of the particle.
Convention: We define potential energy U such that
ˆ
dU = −W = − F~ · d~s
∴ For particle moving from 1 to 2
ˆ
ˆ
2
2
dU = U2 − U1 = −
1
F~ · d~s
1
where U1 , U2 are potential energy at position 1, 2.
4.1. POTENTIAL ENERGY AND CONSERVATIVE FORCES
38
Example:
Suppose charge q2
moves from point 1
to 2.
ˆ
2
From definition: U2 − U1 = −
ˆ
1
F~ · d~r
r2
= −
F dr
( ∵ F~ k d~r )
ˆr1r2
1 q1 q2
dr
2
r1 4π²0 r
¯r
ˆ
dr
1
1 q1 q2 ¯¯ 2
(∵
=− +C )
=
¯
r2
r
4π²0 r ¯r1
µ
¶
1
1
1
q1 q2
−
−∆W = ∆U =
4π²0
r2 r1
= −
Note:
(1) This result is generally true for 2-Dimension or 3-D motion.
(2) If q2 moves away from q1 ,
then r2 > r1 , we have
• If q1 , q2 are of same sign,
then ∆U < 0, ∆W > 0
(∆W = Work done by electric repulsive force)
• If q1 , q2 are of different sign,
then ∆U > 0, ∆W < 0
(∆W = Work done by electric attractive force)
(3) If q2 moves towards q1 ,
then r2 < r1 , we have
• If q1 , q2 are of same sign,
then ∆U 0, ∆W 0
• If q1 , q2 are of different sign,
then ∆U 0, ∆W 0
4.1. POTENTIAL ENERGY AND CONSERVATIVE FORCES
39
(4) Note: It is the difference in potential energy that is important.
REFERENCE POINT: U (r = ∞) = 0
¶
µ
1
1 1
∴ U∞ − U1 =
q1 q2
−
4π²0
r2 r1
↓
∞
U (r) =
q1 q2
1
·
4π²0
r
If q1 , q2 same sign,
then U (r) > 0 for all r
If q1 , q2 opposite sign, then U (r) < 0 for all r
(5) Conservation of Mechanical Energy:
For a system of charges with no external force,
E
=
K
.
(Kinetic Energy)
or
+
U
= Constant
&
(Potential Energy)
∆E = ∆K + ∆U = 0
Potential Energy of A System of Charges
Example: P.E. of 3 charges q1 , q2 , q3
Start: q1 , q2 , q3 all at r = ∞, U = 0
Step1:
Move q1 from ∞ to its position ⇒ U = 0
Move q2 from ∞ to new position ⇒
Step2:
U=
1 q1 q2
4π²0 r12
Move q3 from ∞ to new position ⇒ Total P.E.
Step3:
Step4: What if there are 4 charges?
1
U=
4π²0
·
q1 q2 q1 q3 q2 q3
+
+
r12
r13
r23
¸
4.2. ELECTRIC POTENTIAL
4.2
40
Electric Potential
Consider a charge q at center, we consider its effect on test charge q0
DEFINITION: We define electric potential V so that
∆V =
∆U
−∆W
=
q0
q0
( ∴ V is the P.E. per unit charge)
• Similarly, we take V (r = ∞) = 0.
• Electric Potential is a scalar.
• Unit:
V olt(V ) = Joules/Coulomb
• For a single point charge:
V (r) =
1
q
·
4π²0 r
• Energy Unit: ∆U = q∆V
electron − V olt(eV ) = 1.6
×{z10−19} J
|
charge of electron
Potential For A System of Charges
For a total of N point charges, the potential V at any point P can be derived
from the principle of superposition.
Recall that potential due to q1 at
q1
1
·
point P: V1 =
4π²0 r1
∴ Total potential at point P due to N charges:
V
= V1 + V2 + · · · + VN (principle of superposition)
·
¸
1
qN
q1 q2
=
+ + ··· +
4π²0 r1 r2
rN
4.2. ELECTRIC POTENTIAL
41
V =
N
1 X
qi
4π²0 i=1 ri
~ F~ , we have a sum of vectors
Note: For E,
For V, U , we have a sum of scalars
Example: Potential of an electric dipole
Consider the potential of
point P at distance x > d2
from dipole.
"
−q
1
+q
V =
d +
4π²0 x − 2
x + d2
Special Limiting Case:
1
x∓
xÀd
d
2
"
d
1
1
1
1±
= ·
d '
x 1 ∓ 2x
x
2x
"
#
#
1
q
d
d
∴
V =
·
1+
− (1 − )
4π²0 x
2x
2x
p
(Recall p = qd)
V =
4π²0 x2
1
1
For a point charge E ∝ 2 V ∝
r
r
For a dipole
E∝
1
r3
V ∝
1
r2
For a quadrupole
E∝
1
r4
V ∝
1
r3
Electric Potential of Continuous Charge Distribution
For any charge distribution, we write the electrical potential dV due to infinitesimal charge dq:
dV =
dq
1
·
4π²0 r
#
4.2. ELECTRIC POTENTIAL
42
ˆ
∴
V =
1
dq
·
4π²0 r
charge
distribution
Similar to the previous examples on E-field, for the case of uniform charge
distribution:
1-D
2-D
3-D
⇒
⇒
⇒
long rod
charge sheet
uniformly charged body
Example (1):
⇒ dq = λ dx
⇒ dq = σ dA
⇒ dq = ρ dV
Uniformly-charged ring
Length of the infinitesimal ring element
= ds = Rdθ
∴
dV =
charge dq = λ ds
= λR dθ
1
dq
1
λR dθ
·
=
·√ 2
4π²0 r
4π²0
R + z2
The integration is around the entire ring.
ˆ
∴
V =
dV
ring
ˆ
2π
λR dθ
1
·√ 2
4π²0
R + z2
0
ˆ 2π
λR
√
dθ
=
4π²0 R2 + z 2 | 0 {z }
=
2π
Total charge on the
ring = λ · (2πR)
LIMITING CASE:
V
=
zÀR ⇒ V =
Q
√
4π²0 R2 + z 2
Q
Q
√ =
2
4π²0 |z|
4π²0 z
4.2. ELECTRIC POTENTIAL
Example (2):
43
Uniformly-charged disk
Using the principle of superposition, we will find the potential
of a disk of uniform charge density by integrating the potential of
concentric rings.
ˆ
1
dq
∴
dV =
4π²0
r
disk
Ring of radius x:
dq = σ dA = σ (2πxdx)
ˆ
∴
V
V
R
1
σ2πx dx
·√ 2
x + z2
0 4π²0
ˆ R
σ
d(x2 + z 2 )
=
4²0 0 (x2 + z 2 )1/2
√
σ √ 2
=
( z + R2 − z 2 )
2²0
σ √ 2
=
( z + R2 − |z|)
2²0
=
Recall:
n
|x| =
+x;
−x;
x≥0
x<0
Limiting Case:
(1) If |z| À R
√
s
³
R2 ´
z2
³
R2 ´ 1
= |z| · 1 + 2 2
z
³
R2 ´
' |z| · 1 + 2
2z
z 2 + R2 =
z2 1 +
( (1 + x)n ≈ 1 + nx if x ¿ 1 )
(
|z|
1
=
)
2
z
|z|
σ R2
Q
·
=
(like a point charge)
2²0 2|z|
4π²0 |z|
where Q = total charge on disk = σ · πR2
∴ At large z, V '
4.2. ELECTRIC POTENTIAL
44
(2) If |z| ¿ R
√
∴
At z = 0, V =
σR
;
2²0
∴
³
z 2 ´ 12
R2
³
z2 ´
' R 1+
2R2
z 2 + R2 = R · 1 +
σ h
z2 i
V '
R − |z| +
2²0
2R
Let’s call this V0
σR h
|z|
z2 i
1−
+
2²0
R
2R2
h
|z|
z2 i
V (z) = V0 1 −
+
R
2R2
V (z) =
The key here is that it is the difference between potentials of two points
that is important.
⇒ A convenience reference point to compare in this example is the
potential of the charged disk.
∴
The important quantity here is
V (z) − V0 = −
z 2 ½½
|z|
V0 + ½2 V0
R
2R
½
neglected as z ¿ R
V (z) − V0 = −
V0
|z|
R
4.3. RELATION BETWEEN ELECTRIC FIELD E AND ELECTRIC
POTENTIAL V
4.3
45
Relation Between Electric Field E and Electric Potential V
~
(A) To get V from E:
Recall our definition of the potential V:
∆V =
W12
∆U
=−
q0
q0
where ∆U is the change in P.E.; W12 is the work done in bringing charge
q0 from point 1 to 2.
´2
− 1 F~ · d~s
∴
∆V = V2 − V1 =
q0
However, the definition of E-field:
~
F~ = q0 E
ˆ
∴
2
∆V = V2 − V1 = −
~ · d~s
E
1
Note: The integral on the right hand side of the above can be calculated
along any path from point 1 to 2. (Path-Independent)
ˆ P
~ · d~s
Convention: V∞ = 0 ⇒ VP = −
E
∞
~ from V :
(B) To get E
Again, use the definition of V :
∆U = q0 ∆V = −W
| {z }
Work done
However,
W =
~ · ∆~s
qE
0
|{z}
Electric force
= q0 Es ∆s
where Es is the E-field component along
the path ∆~s.
∴
q0 ∆V = −q0 Es ∆s
4.3. RELATION BETWEEN ELECTRIC FIELD E AND ELECTRIC
POTENTIAL V
∴
Es = −
46
∆V
∆s
For infinitesimal ∆s,
∴
Es = −
dV
ds
Note: (1) Therefore the E-field component along any direction is the negtive derivative of the potential along the same direction.
~ then ∆V = 0
(2) If d~s ⊥ E,
~
(3) ∆V is biggest/smallest if d~s k E
~
Generally, for a potential V (x, y, z), the relation between E(x,
y, z) and V
is
∂V
∂V
∂V
Ex = −
Ey = −
Ez = −
∂x
∂y
∂z
∂ ∂ ∂
, ,
are partial derivatives
∂x ∂y ∂z
∂
For
V (x, y, z), everything y, z are treated like a constant and we only
∂x
take derivative with respect to x.
Example:
If
∂V
∂x
=
∂V
∂y
=
∂V
∂z
=
V (x, y, z) = x2 y − z
For other co-ordinate systems
(1) Cylindrical:
V (r, θ, z)




Er









= −
∂V
∂r
1 ∂V
·
E
=
−
θ


r ∂θ









 Ez
= −
∂V
∂z
4.3. RELATION BETWEEN ELECTRIC FIELD E AND ELECTRIC
POTENTIAL V
47
(2) Spherical:



Er










= −
∂V
∂r
1 ∂V
Eθ = − ·

r ∂θ


V (r, θ, φ)








 Eφ
= −
1
∂V
·
r sin θ ∂φ
Note: Calculating V involves summation of scalars, which is easier than
adding vectors for calculating E-field.
∴
To find the E-field of a general charge system, we first calculate
~ from the partial derivative.
V , and then derive E
Example: Uniformly charged disk
From potential calculations:
σ √ 2
V =
( R + z 2 − |z| )
2²0
For
z > 0,
∴
for a point along
the z-axis
|z| = z
Ez = −
i
∂V
σ h
z
=
1− √ 2
∂z
2²0
R + z2
(Compare with
Chap.2 notes)
Example: Uniform electric field
(e.g. Uniformly charged +ve and −ve plates)
Consider a path going from the −ve
plate to the +ve plate
Potential at point P, VP can be deduced
from definition.
ˆ
i.e.
s
VP − V− = −
~ · d~s
E
ˆ0 s
= −
(−E ds)
ˆ0 s
= E
ds = Es
0
Convenient reference:
V− = 0
∴
VP = E · s
(V− = Potential of
−ve plate)
~ d~s pointing
∵ E,
opposite directions
4.4. EQUIPOTENTIAL SURFACES
4.4
48
Equipotential Surfaces
Equipotential surface is a surface on which the potential is constant.
⇒ (∆V = 0)
V (r) =
1
+q
·
= const
4π²0 r
⇒
r = const
⇒
Equipotential surfaces are
circles/spherical surfaces
Note: (1) A charge can move freely on an equipotential surface without any
work done.
(2) The electric field lines must be perpendicular to the equipotential
surfaces. (Why?)
On an equipotential surface, V = constant
~ · d~l = 0, where d~l is tangent to equipotential surface
⇒ ∆V = 0 ⇒ E
~ must be perpendicular to equipotential surfaces.
∴
E
Example: Uniformly charged surface (infinite)
Recall
V = V0 −
↑
σ
|z|
2²0
Potential at z = 0
Equipotential surface means
σ
|z| = C
2²0
⇒ |z| = constant
V = const ⇒ V0 −
4.4. EQUIPOTENTIAL SURFACES
49
Example: Isolated spherical charged conductors
Recall:
(1) E-field inside = 0
(2) charge distributed on the
outside of conductors.
(i) Inside conductor:
E = 0 ⇒ ∆V = 0 everywhere in conductor
⇒ V = constant everywhere in conductor
⇒ The entire conductor is at the same potential
(ii) Outside conductor:
Q
4π²0 r
∵
Spherically symmetric (Just like a point charge.)
BUT not true for conductors of arbitrary shape.
V =
Example: Connected conducting spheres
Two conductors connected can be seen as a
single conductor
4.4. EQUIPOTENTIAL SURFACES
∴
50
Potential everywhere is identical.
Potential of radius R1 sphere
Potential of radius R2 sphere
q1
4π²0 R1
q2
V2 =
4π²0 R2
V1 =
V1 = V2
q1
q2
⇒
=
R1
R2
⇒
q1
R1
=
q2
R2
Surface charge density
σ1 =
q1
4πR12
| {z }
Surface area of radius R1 sphere
∴
∴
σ1
q1 R22
R2
=
· 2 =
σ2
q2 R 1
R1
If R1 < R2 , then σ1 > σ2
And the surface electric field E1 > E2
For arbitrary shape conductor:
At every point on the conductor,
we fit a circle. The radius of this
circle is the radius of curvature.
Note:
Charge distribution on a conductor does not have to be uniform.
Chapter 5
Capacitance and DC Circuits
5.1
Capacitors
A capacitor is a system of two conductors that carries equal and opposite
charges. A capacitor stores charge and energy in the form of electro-static field.
We define capacitance as
C=
Q
V
Unit: Farad(F)
where
Q = Charge on one plate
V = Potential difference between the plates
Note: The C of a capacitor is a constant that depends only on its shape and
material.
i.e. If we increase V for a capacitor, we can increase Q stored.
5.2
5.2.1
Calculating Capacitance
Parallel-Plate Capacitor
5.2. CALCULATING CAPACITANCE
52
(1) Recall from Chapter 3 note,
Q
σ
=
²0
²0 A
~ =
|E|
(2) Recall from Chapter 4 note,
ˆ
+
∆V = V+ − V− = −
~ · d~s
E
−
Again, notice that this integral is independent of the path taken.
~
∴ We can take the path that is parallel to the E-field.
ˆ
∴
∆V
−
=
~ · d~s
E
+
ˆ −
=
+
=
Q
²0 A
E · ds
ˆ −
ds
+
| {z }
Length of path taken
=
(3) ∴
5.2.2
Q
·d
²0 A
C=
Q
²0 A
=
∆V
d
Cylindrical Capacitor
Consider two concentric cylindrical wire
of innner and outer radii r1 and r2 respectively. The length of the capacitor
is L where r1 < r2 ¿ L.
5.2. CALCULATING CAPACITANCE
53
(1) Using Gauss’ Law, we determine that the E-field between the conductors
is (cf. Chap3 note)
~ =
E
1
λ
1
Q
· r̂ =
·
r̂
2π²0 r
2π²0 Lr
where λ is charge per unit length
(2)
ˆ
−
∆V =
~ · d~s
E
+
~
Again, we choose the path of integration so that d~s k r̂ k E
ˆ r2
ˆ r2
Q
dr
∴
∆V =
E dr =
2π²0 L r1 r
r1
| {z }
r
ln( r2 )
1
∴
5.2.3
C=
Q
L
= 2π²0
∆V
ln(r2 /r1 )
Spherical Capacitor
For the space between the two conductors,
E=
ˆ
∆V
1
Q
· 2;
4π²0 r
−
=
~ · d~s
E
ˆ+r2
Choose d~
s k r̂
Q
1
· 2 dr
r1 4π²0 r
·
¸
Q
1
1
=
−
4π²0 r1 r2
=
·
C = 4π²0
r1 r2
r2 − r1
¸
r 1 < r < r2
5.3. CAPACITORS IN COMBINATION
5.3
54
Capacitors in Combination
(a) Capacitors in Parallel
In this case, it’s the potential difference
V = Va − Vb that is the same across the
capacitor.
BUT: Charge on each capacitor different
Total charge Q = Q1 + Q2
= C1 V + C2 V
Q = (C1 + C2 ) V
{z
}
|
Equivalent capacitance
∴
For capacitors in parallel: C = C1 + C2
(b) Capacitors in Series
The charge across capacitors are
the same.
BUT: Potential difference (P.D.) across capacitors different
∴
Q
C1
Q
= Vc − Vb =
C2
∆V1 = Va − Vc =
P.D. across C1
∆V2
P.D. across C2
Potential difference
∆V
= Va − Vb
= ∆V1 + ∆V2
1
1
Q
= Q(
+
)=
C1 C2
C
∆V
where C is the Equivalent Capacitance
∴
1
1
1
=
+
C
C1 C2
5.4. ENERGY STORAGE IN CAPACITOR
5.4
55
Energy Storage in Capacitor
In charging a capacitor, positive charge
is being moved from the negative plate
to the positive plate.
⇒ NEEDS WORK DONE!
Suppose we move charge dq from −ve to +ve plate, change in potential energy
dU = ∆V · dq =
q
dq
C
Suppose we keep putting in a total charge Q to the capacitor, the total potential
energy
ˆ
ˆ Q
q
U = dU =
dq
0 C
U=
∴
Q2
1
= C∆V 2
2C
2
(∵ Q=C∆V )
The energy stored in the capacitor is stored in the electric field between the
plates.
Note : In a parallel-plate capacitor, the E-field is constant between the plates.
∴
We can consider the E-field energy
density u =
∴
Total energy stored
Total volume with E-field
u=
U
Ad
|{z}
Rectangular volume
Recall



C







 E
=
²0 A
d
=
∆V
d
C
z }| {
∴
⇒
∆V = Ed
(∆V )2
1
V olume
z}|{
z}|{
1 ²0 A
1
u= (
) · ( Ed )2 ·
2 d
Ad
5.4. ENERGY STORAGE IN CAPACITOR
1
²0 E 2
2
↑
can be generally applied
u=
56
Energy per unit volume
of the electrostatic field
Example : Changing capacitance
(1) Isolated Capacitor:
Charge on the capacitor plates remains constant.
²0 A
1
BUT: Cnew =
= Cold
2d
2
Q2
Q2
∴
Unew =
=
= 2Uold
2Cnew
2Cold /2
∴
In pulling the plates apart, work done W > 0
Summary :
(V = Q
) ⇒
C
1
² E2 =
2 0
Q
V
u
→ Q
→ 2V
→ u
C
E
U
→
→
→
C/2
E
2U
(2) Capacitor connected to a battery:
Potential difference between capacitor plates remains constant.
1
1 1
1
Unew = Cnew ∆V 2 = · Cold ∆V 2 = Uold
2
2 2
2
∴
In pulling the plates apart, work done by battery < 0
Summary :
Q
V
u
→
→
→
Q/2
V
u/4
C
E
U
→
→
→
C/2
E/2
U/2
(E = Vd )
(U = u · volume)
5.5. DIELECTRIC CONSTANT
5.5
57
Dielectric Constant
We first recall the case for a conductor being placed in an external E-field E0 .
In a conductor, charges are free to move
inside so that the internal E-field E 0 set
up by these charges
E 0 = −E0
so that E-field inside conductor = 0.
Generally, for dielectric, the atoms and
molecules behave like a dipole in an E-field.
Or, we can envision this so that in the absence of E-field, the direction of dipole
in the dielectric are randomly distributed.
5.6. CAPACITOR WITH DIELECTRIC
58
The aligned dipoles will generate an induced E-field E 0 , where |E 0 | < |E0 |.
We can observe the aligned dipoles in the form of induced surface charge.
Dielectric Constant : When a dielectric is placed in an external E-field E0 ,
the E-field inside a dielectric is induced.
E-field in dielectric
E=
1
E0
Ke
Ke = dielectric constant
≥1
Example :
Vacuum
Porcelain
Water
Perfect conductor
Air
5.6
Ke
Ke
Ke
Ke
Ke
=1
= 6.5
∼ 80
=∞
= 1.00059
Capacitor with Dielectric
Case I :
Again, the charge remains constant after dielectric is inserted.
1
BUT: Enew =
Eold
Ke
1
∆Vold
∴
∆V = Ed ⇒ ∆Vnew =
Ke
Q
∴
C=
⇒ Cnew = Ke Cold
∆V
For a parallel-plate capacitor with dielectric:
C=
Ke ²0 A
d
5.6. CAPACITOR WITH DIELECTRIC
We can also write
C=
² = Ke ²0
²A
d
59
in general with
(called permittivity of dielectric)
(Recall ²0 = Permittivity of free space)
Q2
Energy stored U =
;
2C
1
∴
Unew =
Uold < Uold
Ke
∴
Work done in inserting dielectric < 0
Case II : Capacitor connected to a battery
Voltage across capacitor plates remains constant after insertion of dielectric.
In both scenarios, the E-field inside capacitor remains constant
(∵ E = V /d)
BUT: How can E-field remain constant?
ANSWER: By having extra charge on capacitor plates.
Recall: For conductors,
σ
²0
Q
E =
²0 A
E =
⇒
(Chapter 3 note)
(σ = charge per unit area = Q/A)
After insertion of dielectric:
E0 =
E
Q0
=
Ke
Ke ²0 A
But E-field remains constant!
∴
E0 = E ⇒
Q0
Q
=
Ke ²0 A
²0 A
⇒ Q0 = Ke Q > Q
5.7. GAUSS’ LAW IN DIELECTRIC
Capacitor C = Q/V
Energy stored U = 21 CV 2
(i.e. Unew > Uold )
∴
∴
5.7
60
⇒
⇒
C 0 → Ke C
U 0 → Ke U
Work done to insert dielectric > 0
Gauss’ Law in Dielectric
The Gauss’ Law we’ve learned is applicable in vacuum only. Let’s use the capacitor as an example to examine Gauss’ Law in dielectric.
Free charge
on plates
±Q
±Q
0
∓Q0
Induced charge
on dielectric
˛ Gauss’ Law
˛ Gauss’ Law: 0
Q
~ · dA
~=
~ 0 · dA
~ = Q−Q
E
E
²0
²0
S
S
0
Q
Q
−
Q
⇒ E0 =
(1)
∴
E0 =
(2)
²0 A
²0 A
E0
However, we define
E0 =
(3)
Ke
Q
Q0
Q
=
−
From (1), (2), (3)
∴
Ke ²0 A
²0 A ²0 A
∴
Induced charge density σ 0 =
³
Q0
1 ´
=σ 1−
<σ
A
Ke
where σ is free charge density.
Recall Gauss’ Law in Dielectric:
˛
~ 0 · dA
~
²0 E
=
S
Q
−
Q0
↑
↑
↑
E-field in dielectric
free charge
induced charge
5.8. OHM’S LAW AND RESISTANCE
61
˛
h
i
~ 0 · dA
~ =Q−Q 1− 1
E
Ke
˛S
Q
~ 0 · dA
~=
⇒ ²0 E
Ke
S
⇒ ²0
˛
~ 0 · dA
~=Q
Ke E
²0
S
Gauss’ Law
in dielectric
Note :
E0
for dielectric
Ke
(2) Only free charges need to be considered, even for dielectric where there
are induced charges.
(1) This goes back to the Gauss’ Law in vacuum with E =
(3) Another way to write:
where
˛
~ · dA
~=Q
E
²
S
~ is E-field in dielectric,
E
² = Ke ²0 is Permittivity
Energy stored with dielectric:
Total energy stored:
With dielectric, recall
1
CV 2
2
Ke ²0 A
C=
d
U=
V = Ed
∴ Energy stored per unit volume:
ue =
U
1
= Ke ²0 E 2
Ad
2
and udielectric = Ke uvacuum
∴
More energy is stored per unit volume in dielectric than in vacuum.
5.8
Ohm’s Law and Resistance
ELECTRIC CURRENT is defined as the flow of electric charge through a
cross-sectional area.
5.8. OHM’S LAW AND RESISTANCE
i=
dQ
dt
62
Unit: Ampere (A)
= C/second
Convention :
(1) Direction of current is the direction of flow of positive charge.
(2) Current is NOT a vector, but the current density is a vector.
~j = charge flow per unit time per unit area
ˆ
~
~j · dA
i=
Drift Velocity :
Consider a current i flowing through
a cross-sectional area A:
∴
In time ∆t, total charges passing through segment:
∆Q = q A(Vd ∆t) n
| {z }
Volume of charge
passing through
where q is charge of the current carrier,
per unit volume
∴
Current:
Current Density:
i=
n is density of charge carrier
∆Q
= nqAvd
∆t
~j = nq~vd
Note : For metal, the charge carriers are the free electrons inside.
∴ ~j = −ne~
vd for metals
∴
Inside metals, ~j and ~vd are in opposite direction.
We define a general property, conductivity (σ), of a material as:
~
~j = σ E
5.8. OHM’S LAW AND RESISTANCE
63
Note : In general, σ is NOT a constant number, but rather a function of position
and applied E-field.
A more commonly used property, resistivity (ρ), is defined as
∴
ρ=
1
σ
~ = ρ~j
E
Unit of ρ : Ohm-meter (Ωm)
where Ohm (Ω) = Volt/Ampere
OHM’S LAW:
Ohmic materials have resistivity that are independent of the applied electric field.
i.e. metals (in not too high E-field)
Example :
Consider a resistor (ohmic material) of
length L and cross-sectional area A.
∴
Electric field inside conductor:
ˆ
~ · d~s = E · L
∆V = E
Current density:
j=
⇒
E=
∆V
L
i
A
∴
E
j
∆V
1
ρ =
·
L i/A
ρ =
∆V
L
=R=ρ
i
A
where R is the resistance of the conductor.
Note: ∆V = iR is NOT a statement of Ohm’s Law. It’s just a definition for
resistance.
5.9. DC CIRCUITS
64
ENERGY IN CURRENT:
Assuming a charge ∆Q enters
with potential V1 and leaves with
potential V2 :
∴
Potential energy lost in the wire:
∆U = ∆Q V2 − ∆Q V1
∆U = ∆Q(V2 − V1 )
∴
Rate of energy lost per unit time
∆U
∆Q
=
(V2 − V1 )
∆t
∆t
Joule’s heating
For a resistor R,
5.9
P = i2 R =
P = i · ∆V =
Power dissipated
in conductor
∆V 2
R
DC Circuits
A battery is a device that supplies electrical energy to maintain a current in a
circuit.
In moving from point 1 to 2, electric potential energy increase by
∆U = ∆Q(V2 − V1 ) = Work done by E
Define
E = Work done/charge = V2 − V1
5.9. DC CIRCUITS
65
Example :
Va = Vc
Vb = Vd
)
assuming(1) perfect conducting wires.
By Definition: Vc − Vd = iR
Va − Vb = E
∴
Also, we have assumed(2)
E
R
zero resistance inside battery.
E = iR
⇒
i=
Resistance in combination :
Potential differece (P.D.)
Va − Vb = (Va − Vc ) + (Vc − Vb )
= iR1 + iR2
∴
Equivalent Resistance
R = R1 + R2
1
1
1
=
+
R
R1 R2
for resistors in series
for resistors in parallel
5.9. DC CIRCUITS
66
Example :
For real battery, there is an
internal resistance that
we cannot ignore.
∴
E = i(R + r)
E
i =
R+r
Joule’s heating in resistor R :
P = i · (P.D. across resistor R)
= i2 R
E 2R
P =
(R + r)2
Question: What is the value of R to obtain maximum Joule’s heating?
Answer: We want to find R to maximize P.
dP
E2
E 2 2R
=
−
dR
(R + r)2 (R + r)3
Setting
dP
E2
=0 ⇒
[(R + r) − 2R] = 0
dR
(R + r)3
⇒ r−R=0
⇒ R=r
5.9. DC CIRCUITS
67
ANALYSIS OF COMPLEX CIRCUITS:
KIRCHOFF’S LAWS:
(1) First Law (Junction Rule):
Total current entering a junction equal to the total current leaving the
junction.
(2) Second Law (Loop Rule):
The sum of potential differences around a complete circuit loop is zero.
Convention :
(i)
Va > Vb
⇒
Potential difference = −iR
i.e. Potential drops across resistors
(ii)
Vb > Va
⇒
Potential difference = +E
i.e. Potential rises across the negative plate of the battery.
Example :
5.9. DC CIRCUITS
68
By junction rule:
i1 = i2 + i3
(5.1)
Loop A ⇒ 2E0 − i1 R − i2 R + E0 − i1 R = 0
Loop B ⇒ −i3 R − E0 − i3 R − E0 + i2 R = 0
Loop C ⇒ 2E0 − i1 R − i3 R − E0 − i3 R − i1 R = 0
(5.2)
(5.3)
(5.4)
By loop rule:
BUT:
(5.4) = (5.2) + (5.3)
General rule: Need only 3 equations for 3 current
i1 = i2 + i3
3E0 − 2i1 R − i2 R = 0
−2E0 + i2 R − 2i3 R = 0
(5.1)
(5.2)
(5.3)
Substitute (5.1) into (5.2) :
3E0 − 2(i2 + i3 )R − i2 R = 0
⇒ 3E0 − 3i2 R − 2i3 R = 0
Subtract (5.3) from (5.4), i.e. (5.4)−(5.3)
3E0 − (−2E0 ) − 3i2 R − i2 R = 0
⇒
i2 =
5 E0
·
4 R
Substitute i2 into (5.3) :
−2E0 +
E0 ´
R − 2i3 R = 0
4 R
³5
·
(5.4)
5.10. RC CIRCUITS
69
3 E0
i3 = − ·
8 R
⇒
Substitute i2 , i3 into (5.1) :
i1 =
³5
4
−
3 ´ E0
7 E0
= ·
8 R
8 R
Note: A negative current means that it is flowing in opposite direction from the
one assumed.
5.10
RC Circuits
(A) Charging a capacitor with battery:
Using the loop rule:
+E0 −
Q
iR −
=0
|{z}
C
|{z}
P.D.
across R P.D.
across C
Note: Direction of i is chosen so that the current represents the rate at
which the charge on the capacitor is increasing.
i
z}|{
∴
⇒
dQ Q
1st order
+
differential eqn.
dt C
dt
dQ
=
EC − Q
RC
E =R
Integrate both sides and use the initial condition:
t = 0, Q on capacitor = 0
ˆ Q
ˆ t
dQ
dt
=
0 EC − Q
0 RC
5.10. RC CIRCUITS
70
¯Q
− ln(EC − Q)¯¯ =
0
t ¯¯t
¯
RC 0
⇒ − ln(EC − Q) + ln(EC) =
³
1 ´
t
=
Q
RC
1 − EC
1
t/RC
⇒
Q = e
1 − EC
Q
⇒
= 1 − e−t/RC
EC
⇒ Q(t) = EC(1 − e−t/RC )
t
RC
⇒ ln
Note: (1) At t = 0 , Q(t = 0) = EC(1 − 1) = 0
(2) As t → ∞ , Q(t → ∞) = EC(1 − 0) = EC
= Final charge on capacitor (Q0 )
(3) Current:
dQ
i =
dt
³ 1 ´
= EC
e−t/RC
RC
E −t/RC
i(t) =
e
R

E

= Initial current = i0
i(t = 0)
=
R

i(t → ∞) = 0
(4) At time = 0, the capacitor acts like short circuit when there is
zero charge on the capacitor.
(5) As time → ∞, the capacitor is fully charged and current = 0, it
acts like a open circuit.
5.10. RC CIRCUITS
71
(6) τc = RC is called the time constant. It’s the time it takes for
the charge to reach (1 − 1e ) Q0 ' 0.63Q0
(B) Discharging a charged capacitor:
Note: Direction of i is chosen so that the current represents the rate at
which the charge on the capacitor is decreasing.
∴
i=−
dQ
dt
Loop Rule:
Vc − iR = 0
Q dQ
+
R=0
⇒
C
dt
dQ
1
⇒
=−
Q
dt
RC
Integrate both sides and use the initial condition:
t = 0, Q on capacitor = Q0
ˆ Q
ˆ t
dQ
1
=−
dt
RC 0
Q0 Q
t
⇒ ln Q − ln Q0 = −
RC
³Q´
t
⇒ ln
=−
Q0
RC
Q
= e−t/RC
⇒
Q0
⇒ Q(t) = Q0 e−t/RC
dQ
Q0 −t/RC
(i = − )
⇒ i(t) =
e
dt
RC
1 Q0
(At t = 0)
⇒ i(t = 0) = ·
R |{z}
C
Initial P.D. across capacitor
i0 =
V0
R
5.10. RC CIRCUITS
At t = RC = τ
72
Q(t = RC) =
1
Q0 ' 0.37Q0
e
Chapter 6
Magnetic Force
6.1
Magnetic Field
For stationary charges, they experienced an electric force in an electric field.
For moving charges, they experienced a magnetic force in a magnetic field.
Mathematically,
~ (electric force)
F~E = q E
~ (magnetic force)
F~B = q~v × B
Direction of the magnetic force determined from right hand rule.
~ : Unit = Tesla (T)
Magnetic field B
1T = 1C moving at 1m/s experiencing 1N
Common Unit: 1 Gauss (G) = 10−4 T ≈ magnetic field on earth’s surface
Example: What’s the force on a 0.1C charge moving at velocity ~v = (10ĵ −
~ = (−3î + 4k̂) × 10−4 T
20k̂)ms−1 in a magnetic field B
~
F~ = q~v × B
6.1. MAGNETIC FIELD
74
= +0.1 (10ĵ − 20k̂) × (−3î + 4k̂) × 10−4 N
= 10−5 (−30 · −k̂ + 40î + 60ĵ + 0)N
Effects of magnetic field is usually quite small.
~
F~ = q~v × B
|F~ | = qvB sin θ,
∴
~
where θ is the angle between ~v and B
~
Magnetic force is maximum when θ = 90◦ (i.e. ~v ⊥ B)
~
Magnetic force is minimum (0) when θ = 0◦ , 180◦ (i.e. ~v k B)
Graphical representation of B-field: Magnetic field lines
Compared with Electric field lines:
Similar characteristics :
(1) Direction of E-field/B-field indicated by tangent of the field lines.
(2) Magnitude of E-field/B-field indicated by density of the field lines.
Differeces :
(1) F~E k E-field lines; F~B ⊥ B-field lines
(2) E-field line begins at positive charge and ends at negative charge; Bfield line forms a closed loop.
Example : Chap35, Pg803 Halliday
Note:
Isolated magnetic monopoles do not exist.
6.2. MOTION OF A POINT CHARGE IN MAGNETIC FIELD
6.2
75
Motion of A Point Charge in Magnetic Field
Since F~B ⊥ ~v , therefore B-field only changes the direction of the velocity but not
its magnitude.
~ = q v⊥ B ,
Generally, F~B = q~v × B
∴ We only need to consider the motion
component ⊥ to B-field.
We have circular motion. Magnetic
force provides the centripetal force on the
moving charge particles.
∴
v2
r
v2
|q| vB = m
r
mv
∴
r =
|q|B
FB = m
where r is radius of circular motion.
Time for moving around one orbit:
T =
2πr
2πm
=
v
qB
Cyclotron Period
(1) Independent of v (non-relativistic)
(2) Use it to measure m/q
Generally, charged particles with constant velocity moves in helix in the presence of constant B-field.
6.3. HALL EFFECT
76
Note :
(1) B-field does NO work on particles.
(2) B-field does NOT change K.E. of particles.
Particle Motion in Presence of E-field & B-field:
~ + q~v × B
~
F~ = q E
Special Case :
Lorentz Force
~ ⊥B
~
E
When |F~E | = |F~B |
qE = qvB
⇒
v=
E
B
∴
For charged particles moving at v = E/B, they will pass through the
crossed E and B fields without vertical displacement.
⇒ velocity selector
Applications :
• Cyclotron (Lawrence & Livingston 1934)
• Measuring e/m for electrons (Thomson 1897)
• Mass Spectrometer (Aston 1919)
6.3
Hall Effect
Charges travelling in a conducting wire will be pushed to one side of the wire by
the external magnetic field. This separation of charge in the wire is called the
Hall Effect.
6.3. HALL EFFECT
77
The separation will stop when FB experienced by the current carrier is balanced
by the force F~H caused by the E-field set up by the separated charges.
Define :
∆VH = Hall Voltage
= Potential difference across the conducting strip
∆VH
E-field from separated charges: EH =
W
where W = width of conducting strip
∴
In equilibrium:
~ H + q~vd × B
~ = 0, where ~vd is drift velocity
qE
∆VH
= vd B
W
∴
Recall from Chapter 5,
i = nqAvd
where n is density of charge carrier,
A is cross-sectional area = width × thickness = W · t
∴
⇒
n=
∆VH
i
=
B
W
nqW t
iB
qt∆VH
To determine density
of charge carriers
Suppose we determine n for a particular metal (∴ q = e), then we can measure
B-field strength by measuring the Hall voltage:
B=
net
∆VH
i
6.4. MAGNETIC FORCE ON CURRENTS
6.4
78
Magnetic Force on Currents
Current = many charges moving together
Consider a wire segment, length L,
carrying current i in a magnetic field.
~ )·
Total magnetic force = ( q~vd × B
| {z }
force on one
charge carrier
n
| A
{z L}
Total number of
charge carrier
Recall i = nqvd A
∴
~ ×B
~
Magnetic force on current F~ = iL
~ = Vector of which: |L|
~ = length of current segment; direction =
where L
direction of current
For an infinitesimal wire segment d~l
~
dF~ = i d~l × B
Example 1: Force on a semicircle current loop
d~l = Infinitesimal
~
arc length element ⊥ B
∴
dl = R dθ
∴
dF = iRB dθ
By symmetry argument, we only need to consider vertical forces, dF · sin θ
ˆ π
∴
Net force F =
dF sin θ
0
ˆ π
= iRB
sin θ dθ
0
F = 2iRB
(downward)
6.4. MAGNETIC FORCE ON CURRENTS
79
Method 2: Write d~l in î, ĵ components
∴
d~l = −dl sin θ î + dl cos θ ĵ
= R dθ (− sin θ î + cos θ ĵ)
~ = −B k̂ (into the page)
B
~
dF~ = i d~l × B
= −iRB sin θ dθ ĵ − iRB cos θ î
ˆ
∴
π
F~ =
dF~
0
"ˆ π
= −iRB
ˆ
π
sin θ dθ ĵ +
0
#
cos θ dθ î
0
= −2iRB ĵ
Example 2: Current loop in B-field
For segment2:
F2 = ibB sin(90◦ + θ) = ibB cos θ
(pointing downward)
For segment4:
F2 = ibB sin(90◦ − θ) = ibB cos θ
(pointing upward)
6.4. MAGNETIC FORCE ON CURRENTS
80
For segment1: F1 = iaB
For segment3: F3 = iaB
∴
Net force on the current loop = 0
But, net torque on the loop about O
= τ1 + τ3
b
b
= iaB · sin θ + iaB · sin θ
2
2
= i |{z}
ab B sin θ
A = area of loop
Suppose the loop is a coil with N turns of wires:
Total torque τ = N iAB sin θ
Define: Unit vector n̂ to represent the area-vector (using right hand rule)
Then we can rewrite the torque equation as
~
~τ = N iA n̂ × B
Define:
N iA n̂ = ~µ = Magnetic dipole moment of loop
~
~τ = ~µ × B
Chapter 7
Magnetic Field
7.1
Magnetic Field


 experiences magnetic force in B-field.
A moving charge 

can generate B-field.
~ due to moving point charge:
Magnetic field B
~ = µ0 · q~v × r̂ = µ0 · q~v × ~r
B
4π
r2
4π
r3
where
µ0 = 4π × 10−7 Tm/A (N/A2 )
Permeability of free space (Magnetic constant)
~ = µ0 · qv sin θ
|B|
4π
r2
(
maximum when θ = 90◦
minimum when θ = 0◦ /180◦
~ at P0 = 0 = B
~ at P1
B
~
~
B at P2 < B at P3
However, a single moving charge will NOT generate a steady magnetic field.
stationary charges generate steady E-field.
steady currents
generate steady B-field.
7.1. MAGNETIC FIELD
82
Magnetic field at point P can be
obtained by integrating the contribution from individual current segments.
(Principle of Superposition)
∴
Notice:
dq ~v = dq ·
~ =
dB
µ0 dq ~v × r̂
·
4π
r2
d~s
= i d~s
dt
~ =
dB
µ0 i d~s × r̂
·
4π
r2
Biot-Savart Law
For current around a whole circuit:
ˆ
ˆ
~ =
~ =
B
dB
entire
circuit
entire
circuit
µ0 i d~s × r̂
·
4π
r2
Biot-Savart Law is to magnetic field as
Coulomb’s Law is to electric field.
Basic element of E-field: Electric charges dq
Basic element of B-field: Current element i d~s
Example 1 : Magnetic field due to straight current segment
7.1. MAGNETIC FIELD
83
∴ |d~
s × r̂|
= dz sin φ
= dz sin(π − φ)
(Trigonometry Identity)
d · dz
d
= dz · = √ 2
r
d + z2
µ0 i dz d
µ0 i
d
dz
· 2 · =
· 2
4π r
r
4π (d + z 2 )3/2
ˆ L/2
ˆ
µ0 id +L/2
dz
B=
dB =
2
4π −L/2 (d + z 2 )3/2
−L/2
dB =
∴
µ0 i
4πd
µ0 i
B =
4πd
B =
¯
¯+L/2
z
¯
(z 2 + d2 )1/2 ¯−L/2
L
· L2
( 4 + d2 )1/2
·
Limiting Cases : When L À d (B-field due to long wire)
³ L2
4
∴
Recall : E =
B=
+ d2
´−1/2
≈
³ L2 ´−1/2
4
=
2
L
µ0 i
direction of B-field determined
; from right-hand screw rule
2πd
λ
for an infinite long line of charge.
2π²0 d
Example 2 : A circular current loop
7.1. MAGNETIC FIELD
84
~1
Notice that for every current element id~s1 , generating a magnetic field dB
at point P , there is an opposite current element id~s2 , generating B-field
~ 2 so that
dB
~ 1 sin α = −dB
~ 2 sin α
dB
∴
Only vertical component of B-field needs to be considered at point P .
∵d~
s⊥r̂
z}|{
µ0 i ds sin 90◦
dB =
·
4π
r2
∴
B-field at point P :
ˆ
B=
dB
cos
| {z α}
consider vertical
component
around
circuit
ˆ2π
∴
µ0 i cos α
· |{z}
ds
4πr2
Rdθ
0
ˆ 2π
µ0 i R
=
·
ds
4π r3
0
B =
| {z }
Integrate around circumference of circle = 2πR
∴
B=
B =
µ0 iR2
2r3
µ0 iR2
direction of B-field determined
; from right-hand screw rule
2
2
3/2
2(R + z )
Limiting Cases :
(1) B-field at center of loop:
z=0
⇒
B=
µ0 i
2R
(2) For z À R,
B=
³
µ0 iR2
2z 3 1 +
R2
z2
´3/2
≈
µ0 iR2
1
∝ 3
3
2z
z
p
4π²0 x3
∴ A circular current loop is also called a magnetic dipole.
Recall E-field for an electric dipole:
E=
7.1. MAGNETIC FIELD
85
(3) A current arc:
ˆ
B =
dB cos
| {z α}
around
circuit
z=0 ⇒
α = 0 here.
Rθ = length of arc
z }| {
µ0 i R
=
· 3 ·
4π |{z}
r
R=r
when α = 0
B =
ˆ
θ
0
ds
|{z}
Rdθ
µ0 i θ
4πR
Example 3 : Magnetic field of a solenoid
Solenoid is used to produce a strong and uniform magnetic field inside its
coils.
Consider a solenoid of length L consisting of N turns of wire.
N
Define: n = Number of turns per unit length =
L
Consider B-field at distance d from the
center of the solenoid:
For a segment of length dz, number of
current turns = ndz
∴
Total current = ni dz
7.2. PARALLEL CURRENTS
86
Using the result from one coil in Example 2, we get B-field from coils of
length dz at distance z from center:
dB =
µ0 (ni dz)R2
2r3
q
However r =
R2 + (z − d)2
ˆ
∴
+L/2
B =
(Integrating over the
entire solenoid)
dB
−L/2
µ0 niR2
=
2

ˆ
+L/2
−L/2
dz
[R2 + (z − d)2 ]3/2

L
L
+d
−d
µ0 ni 
2
2

q
q
B =
+
2
R2 + ( L2 + d)2
R2 + ( L2 − d)2
along negative z direction
Ideal Solenoid :
then
∴
LÀR
µ0 ni
B=
[1 + 1]
2
direction of B-field determined
B = µ0 ni ; from right-hand screw rule
Question : What is the B-field at the end of an ideal solenoid?
7.2
Parallel Currents
~ due to two
Magnetic field at point P B
currents i1 and i2 is the vector sum of
~ fields B
~ 1, B
~ 2 due to individual curthe B
rents. (Principle of Superposition)
B=
µ0 ni
2
7.2. PARALLEL CURRENTS
87
Force Between Parallel Currents :
Consider a segment of length L on i2 :
~ 1 = µ0 i1 (pointing down)
B
2πd
Force on i2 coming from i1 :
∴
~ 2 = µ0 i1
B
2πd
(pointing up)
~ ×B
~ 1 = µ0 Li1 i2 = |F~12 |
|F~21 | = i2 L
(Def ’n of ampere, A)
2πd
Parallel currents attract, anti-parallel currents repel.
Example : Sheet of current
Consider an infinitesimal wire of width dx at position x, there exists another
~
~ +x and B
~ −x cancel.
element at −x so that vertical B-field
components of B
∴
Magnetic field due to dx wire:
dB =
∴
µ0 · di
2πr
³ dx ´
where di = i
a
Total B-field (pointing along −x axis) at point P :
+a/2
ˆ
B=
+a/2
ˆ
dB cos θ =
−a/2
−a/2
µ0 i dx
·
· cos θ
2πa r
7.3. AMPÈRE’S LAW
88
Variable transformation (Goal: change r, x to d, θ, then integrate over θ):
(
d = r cos θ ⇒ r = d sec θ
x = d tan θ ⇒ dx = d sec2 θ dθ
Limits of integration: −θ0 to θ0 ,
∴
µ0 i
B =
2πa
where tan θ0 =
ˆ
θ0
ˆ
−θ0
θ0
a
2d
d sec2 θ dθ
· cos θ
d sec θ
µ0 i
dθ
2πa −θ0
³ a ´
µ0 iθ0
µ0 i
B =
=
tan−1
πa
πa
2d
=
Limiting Cases :
(1) d À a
a
2d
µ0 i
B=
2πa
tan θ =
∴
⇒
θ≈
a
2d
B-field due to
infinite long wire
(2) d ¿ a (Infinite sheet of current)
tan θ =
∴
a
→ ∞
2d
B=
µ0 i
2a
⇒
θ=
π
2
Constant!
Question : Large sheet of opposite flowing currents.
What’s the B-field between & outside the sheets?
7.3
Ampère’s Law
In our study of electricity, we notice that the inverse square force law leads
to Gauss’ Law, which is useful for finding E-field for systems with high level of
symmetry.
For magnetism, Gauss’ Law is simple
7.3. AMPÈRE’S LAW
89
‹
˛
∵ There is no magnetic monopole.
~=0
~ · dA
B
S
S
A more useful law for calculating B-field for highly symmetric situations is the
Ampère’s Law:
˛
˛
~ · d~s = µ0 i
B
C
C
˛
= Line intefral evaluated around a closed loop C
(Amperian curve)
C
i = Net current that penetrates the area bounded by curve C ∗ (topological property)
Convention : Use the right-hand screw rule to determine the sign of current.
˛
~ · d~s = µ0 (i1 − i3 + i4 − i4 )
B
C
= µ0 (i1 − i3 )
Applications of the Ampere’s Law :
(1) Long-straight wire
Construct an Amperian
curve of radius d:
~
By symmetry argument, we know B-field
only has tangential component
˛
~ · d~s = µ0 i
∴
B
C
7.3. AMPÈRE’S LAW
90
Take d~s to be the tangential vector around the circular path:
∴
~
B
˛ · d~s = B ds
B
ds = µ0 i
| C{z }
Circumference
of circle = 2πd
∴
B-field due to long,
straight current
B=
B(2πd) = µ0 i
µ0 i
2πd
(Compare with 7.1 Example 1)
(2) Inside a current-carrying wire
Again, symmetry argument
~ is tangential
implies that B
to the Amperian curve and
~ → B(r)θ̂
B
Consider an Amperian curve of radius r(< R)
˛
˛
~
B · d~s = B ds = B(2πr) = µ0 iincluded
C
But iincluded ∝ cross-sectional area of C
∴
∴
∴
Recall: Uniformly charged infinite long rod
(3) Solenoid (Ideal)
Consider the rectangular
Amperian curve 1234.
iincluded
πr2
=
i
πR2
r2
iincluded =
i
R2
B=
µ0 i
·r ∝r
2πR2
7.3. AMPÈRE’S LAW
91
˛
ˆ
~ · d~s =
B
C
ˆ
=
(
=0
∵
4
ˆ
3
½
~ ·½
B
s+
½ d~
½
~ · d~s +
B
1
ˆ
2
ˆ
=0
˛
∴
2
½
ˆ
½
~ ·½
B
s+
½ d~
½
3
½
ˆ
½
~ ·½
B
s
½ d~
½
4
½
~ · d~s = 0 inside solenoid
B
~ = 0 outside solenoid
B
~ = 0 outside solenoid
B
ˆ
~ · d~s = B
~ · d~s = Bl = µ0 itot
B
∵
C
1
But itot = |{z}
nl ·i
Number of coils included
∴ B = µ0 ni
Note :
~ = 0 outside the ideal solenoid is only
(i) The assumption that B
approximate. (Halliday, Pg.763)
(ii) B-field everywhere inside the solenoid is a constant. (for ideal
solenoid)
(4) Toroid (A circular solenoid)
By symmetry argument, the B-field lines form concentric circles inside
the toroid.
Take Amperian curve C to be a circle of radius r inside the toroid.
˛
˛
~
B · d~s = B ds = B · 2πr = µ0 (N i)
C
C
∴
B=
µ0 N i
2πr
inside toroid
7.4. MAGNETIC DIPOLE
92
Note :
(i) B 6= constant inside toroid
(ii) Outside toroid:
Take Amperian curve to be circle of radius r > R.
˛
˛
~
B · d~s = B ds = B · 2πr = µ0 · iincl = 0
C
C
∴
B=0
Similarly, in the central cavity B = 0
7.4
Magnetic Dipole
Recall from §6.4, we define the magnetic dipole moment of a rectangular
current loop
~µ = N iAn̂
where n̂ = area unit vector with direction
determined by the right-hand rule
N = Number of turns in current loop
A = Area of current loop
This is actually a general definition of a magnetic dipole, i.e. we use it for
current loops of all shapes.
A common and symmetric example: circular current.
Recall from §7.1 Example 2, magnetic
field at point P (height z above the ring)
~ =
B
µ0 iR2 n̂
µ0 ~µ
=
2
2
3/2
2
2(R + z )
2π(R + z 2 )3/2
7.5. MAGNETIC DIPOLE IN A CONSTANT B-FIELD
93
At distance z À R,
~ = µ0 ~µ
B
2πz 3
due to magnetic dipole
(for z À R)
p~
4π²0 z 3
due to electric dipole
(for z À d)
~ =
E
"
Also, notice
~µ = magnetic dipole moment
Unit: Am2
J/T
#
µ0 = Permeability of free space
= 4π × 10−7 Tm/A
7.5
Magnetic Dipole in A Constant B-field
In the presence of a constant magnetic field, we have shown for a rectangular
~ . It applies to any magnetic
current loop, it experiences a torque ~τ = ~µ × B
dipole in general.
7.6. MAGNETIC PROPERTIES OF MATERIALS
94
External magnetic field aligns the magnetic
dipoles.
∴
Similar to electric dipole in a E-field, we can consider the work done in rotating the magnetic dipole. (refer to Chapter 2)
dW = −dU,
where U is potential energy of dipole
~
U = −~µ · B
Note :
(1) We cannot define the potential energy of a magnetic field in general.
However, we can define the potential energy of a magnetic dipole in a
constant magnetic field.
(2) In a non-uniform external B-field, the magnetic dipole will experience
a net force (not only net torque)
7.6
Magnetic Properties of Materials
Recall intrinsic electric dipole in molecules:
Intrinsic dipole (magnetic) in atoms:
In our classical model of atoms, electrons
revolve around a positive nuclear.
∴
”Current” i =
e
,
P
where P is period of one orbit around nucleus
7.6. MAGNETIC PROPERTIES OF MATERIALS
95
2πr
, where v is velocity of electron
v
∴
Orbit magnetic dipole of atom:
³ ev ´
erv
µ = iA =
(πr2 ) =
2πr
2
Recall: angular momentum of rotation l = mrv
e
·l
∴
µ=
2m
In quantum mechanics, we know that
P =
l is quantized, i.e. l = N ·
h
2π
where N = Any positive integer (1,2,3, ... )
h = Planck’s constant (6.63 × 10−34 J · s)
∴
Orbital magnetic dipole moment
µl =
eh
·N
|4mπ
{z }
Bohr’s magneton µB =9.27×10−24 J/T
There is another source of intrinsic magnetic dipole moment inside an atom:
Spin dipole moment: coming from the intrinsic ”spin” of electrons.
Quantum mechanics suggests that e− are always spinning and it’s either an ”up”
spin or a ”down” spin
µe = 9.65 × 10−27 ≈ µB
So can there be induced magnetic dipole?
7.6. MAGNETIC PROPERTIES OF MATERIALS
Recall: for electric field
Edielectric = Ke Evacuum ;
Ke ≥ 1
For magnetic field in a material:
~ net = B
~0
B
↑
+
~M
B
↑
B-field produced
by induced dipoles
applied
B-field
In many materials (except ferromagnets),
~M ∝ B
~0
B
Define :
~ M = χm B
~0
B
χm is a number called magnetic susceptibility.
∴
~ net = B
~ 0 + χm B
~0
B
~0
= (1 + χm ) B
~ net = κm B
~0 ;
B
κm = 1 + χm
Define : κm is a number called relative permeability.
One more term ......
Define : the Magnetization of a material:
~ = d~µ
M
dV
where µ
~ is magnetic dipole
moment, V is volume
(or, the net magnetic dipole moment per unit volume)
In most materials (except ferromagnets),
~ M = µ0 M
~
B
Three types of magnetic materials:
(1) Paramagnetic:
κm ≥ 1
,
(χm ≥ 0)
induced magnetic dipoles aligned
with the applied B-field.
96
7.6. MAGNETIC PROPERTIES OF MATERIALS
.
e.g. Al (χm = 2.2 × 10−5 ), Mg (1.2 × 10−5 ), O2 (2.0 × 10−6 )
(2) Diamagnetic:
κm ≤ 1
,
(χm ≤ 0)
induced magnetic dipoles aligned
opposite with the applied B-field.
e.g. Cu (χm ≈ −1 × 10−5 ), Ag (−2.6 × 10−5 ), N2 (−5 × 10−9 )
(3) Ferromagnetic:
e.g. Fe, Co, Ni
Magnetization not linearly proportional
to applied field.
Bnet
⇒
not a constant (can be as
Bapp
big as ∼ 5000 − 100, 000)
Interesting Case : Superconductors
χm = −1
A perfect diamagnetic.
NO magnetic field inside.
97
Chapter 8
Faraday’s Law of Induction
8.1
Faraday’s Law
In the previous chapter, we have shown that steady electric current can give
steady magnetic field because of the symmetry between electricity & magnetism.
We can ask:
Steady magnetic field can give steady electric current.
×
OR Changing magnetic field can give steady electric current. X
Define :
(1) Magnetic flux through surface S:
ˆ
~ · dA
~
B
Φm =
S
Unit of Φm :
Weber (Wb)
1Wb = 1Tm2
(2) Graphical:
Φm = Number of magnetic field lines passing through surface S
Faraday’s law of induction:
¯
¯
¯ dΦ ¯
m¯
¯
Induced emf |E| = N ¯¯
¯
dt ¯
where
N = Number of coils in the circuit.
8.2. LENZ’ LAW
~ = Constant
B
~ = Constant
A
E =0
99
~ = Constant
B
 = Constant
dA/dt 6= 0
∴ |E| > 0
B̂ = Constant
dB/dt 6= 0
~
A = Constant
∴ |E| > 0
~ = Constant
B
A = Constant
dÂ/dt 6= 0
∴ |E| > 0
Note : The induced emf drives a current throughout the circuit, similar to the
function of a battery. However, the difference here is that the induced emf
is distributed throughout the circuit. The consequence is that we cannot
define a potential difference between any two points in the circuit.
Suppose there is an induced current in the loop, can we
define ∆VAB ?
Recall:
∆VAB = VA − VB = iR > 0
⇒ VA > VB
Going anti-clockwise (same as i),
If we start from A, going to B, then we get VA > VB .
If we start from B, going to A, then we get VB > VA .
∴ We cannot define ∆VAB !!
This situation is like when we study the interior of a battery.


 chemical reactions.
provides the energy needed to drive the 

 charge carriers around the circuit by 

A battery 

The loop
sources of emf
8.2
changing magnetic flux.
non-electric means
Lenz’ Law
(1) The flux of the magnetic field due to induced current opposes the change
in flux that causes the induced current.
8.3. MOTIONAL EMF
100
(2) The induced current is in such a direction as to oppose the changes that
produces it.
(3) Incorporating Lentz’ Law into Faraday’s Law:
E = −N
dΦm
dt
dΦm
> 0, Φm ↑
dt
⇒
E appears
⇒
~
⇒ B-field due to
induced current
⇒
change in Φm
so that
If
=⇒
Induced current
appears.
Φm ↓
(4) Lenz’ Law is a consequence from the principle of conservation of energy.
8.3
Motional EMF
Let’s try to look at a special case when the changing magnetic flux is carried by
motion in the circuit wires.
Consider a conductor of length L moving
~
with a velocity v in a magnetic field B.
8.3. MOTIONAL EMF
101
Hall Effect for the charge carriers in the rod:
F~E + F~B = 0
~ + q~v × B
~ =0
⇒ qE
~ = −~v × B
~
⇒ E
~ is Hall electric field)
(where E
Hall Voltage inside rod:
ˆ
L
∆V
= −
∆V
= −EL
~ · d~s
E
0
∴
Hall Voltage : ∆V = vBL
Now, suppose the moving wire slides without
friction on a stationary U-shape conductor.
The motional emf can drive an electric current i in the U-shape conductor.
⇒ Power is dissipated in the circuit.
⇒ Pout = V i (Joule’s heating)
(see Lecture Notes Chapter 4)
What is the source of this power?
Look at the forces acting on the conducting rod:
• Magnetic force:
~ ×B
~
F~m = iL
Fm = iLB (pointing left)
• For the rod to continue to move at constant velocity v, we need to apply
an external force:
F~ext = −F~m = iLB
∴
(pointing right)
Power required to keep the rod moving:
Pin = F~ext · ~v
= iBLv
dx
= iBL
dt
d(xL)
= iB
dt
d(BA)
= i
dt
( xL = A, area
enclosed by circuit)
( BA = Φm , magnetic flux)
8.3. MOTIONAL EMF
102
Since energy is not being stored in the system,
∴
We ”prove” Faraday’s Law
Pin + Pout = 0
dΦm
iV + i
= 0
dt
⇒
V =−
dΦm
dt
Applications :
(1) Eddy current:
moving conductors in presence of magnetic field
Induced current
⇒ Power lost in Joule’s heating
³E2 ´
R
⇒ Extra power input to keep moving
To reduce Eddy currents:
(2) Generators and Motors:
Assume that the circuit loop is rotating at a constant angular velocity
ω, (Source of rotation, e.g. steam produced by burner, water falling
from a dam)
8.3. MOTIONAL EMF
103
Magnetic flux through the loop
Number of coils
↓
´
~ · dA
~ = N BA cos θ
ΦB = N B
loop
↓
changes with time! θ = ωt
∴
ΦB = N BA cos ωt
Induced emf: E = −
dΦB
dt
Induced current: i =
E
R
d
(cos ωt)
dt
= N BAω sin ωt
= −N BA
=
N BAω
sin ωt
R
Alternating current (AC) voltage generator
Power has to be provided by the source of rotation to overcome the
torque acting on a current loop in a magnetic field.
µ
~
z }| {
~ ×B
~
~τ = N iA
∴ τ = N iAB sin θ
8.4. INDUCED ELECTRIC FIELD
104
The net effect of the torque is to oppose the rotation of the coil.
An electric motor is simply a generator
operating in reverse.
⇒ Replace the load resistance R with
a battery of emf E.
With the battery, there is a current in the coil, and it experiences a
torque in the B-field.
⇒
Rotation of the coil leads to an induced emf, Eind , in
the direction opposite of that of the battery.
(Lenz’ Law)
∴
i=
E − Eind
R
⇒
As motor speeds up, Eind ↑, ∴ i ↓
mechanical power delivered = torque delivered = N iAB sin θ ↓
In conclusion, we can show that
∴
Pelectric
=
i2 R +
Electric power input
8.4
Pmechanical
Mechanical power delivered
Induced Electric Field
So far we have discussed that a change in magnetic flux will lead in an induced emf distributed
in the loop, resulting from an induced E-field.
However, even in the absence of the loop (so that there is no induced current),
the induced E-field will still accompany a change in magnetic flux.
8.4. INDUCED ELECTRIC FIELD
105
∴ Consider a circular path in a region
with changing magnetic field.
The induced E-field only has tangential components. (i.e. radial E-field = 0)
Why?
Imagine a point charge q0 travelling around the circular path.
Work done by induced E-field = q0 Eind · 2πr
|{z}
| {z }
f orce
distance
Recall work done also equals to q0 E, where E is induced emf
∴
E = Eind 2πr
Generally,
˛
E=
~ ind · d~s
E
¸
~ ind is induced E-field, ~s is
where
is line integral around a closed loop, E
tangential vector of path.
∴
Faraday’s Law becomes
˛
ˆ
d
~ ind · d~s = −
~ · dA
~
E
B
dt
C
S
L.H.S. = Integral around a closed loop C
R.H.S. = Integral over a surface bounded by C
~ determined by direction of line integration C (Right-Hand Rule)
Direction of dA
8.4. INDUCED ELECTRIC FIELD
106
”Regular” E-field
Induced E-field
created by charges
created by changing B-field
E-field lines start from +ve and end
on −ve charge
E-field lines form closed loops
can define electric potential so that
we can discuss potential difference
between two points
Electric potential cannot be defined
(or, potential has no meaning)
⇓
Conservative force field
⇓
Non-conservative force field
The classification of electric and magnetic effects depend on the frame of reference
of the observer. e.g. For motional emf, observer in the reference frame of the
moving loop, will NOT see an induced E-field, just a ”regular” E-field.
(Read: Halliday Chap.33-6, 34-7)
Chapter 9
Inductance
9.1
Inductance
An inductor stores energy in the magnetic field just as a capacitor stores energy
in the electric field.
We have shown earlier that a changing B-field will lead to an induced emf in
a circuit.
Question : If a circuit generates a changing magnetic field, does it lead to an
induced emf in the same circuit?
YES! Self-Inductance
The inductance L of any current element is
EL = ∆VL = −L
Unit of L: Henry(H)
di
dt
The negative sign
comes from Lenz Law.
1H=1· Vs
A
• All circuit elements (including resistors) have some inductance.
• Commonly used inductors: solenoids, toroids
• circuit symbol:
Example : Solenoid
EL = VB − VA = −L
∴
VB < VA
di
<0
dt
EL = VB − VA = −L
VB > VA
di
>0
dt
9.1. INDUCTANCE
108
Recall Faraday’s Law,
EL = −N
dΦB
d
= − (N ΦB )
dt
dt
where ΦB is magnetic flux, N ΦB is flux linkage.
Alternative definition of Inductance:
∴
−
∴
d
di
(N ΦB ) = −L
dt
dt
⇒
L=
N ΦB
i
Inductance is also flux linkage per unit current.
Calculating Inductance:
(1) Solenoid:
To first order approximation,
B = µ0 ni
where n = N/L = Number of
coils per unit length.
Consider a subsection of length l of the solenoid:
Flux linkage = N ΦB
= nl · BA
where A is
cross-sectional area
N ΦB
L=
= µ0 n2 lA
i
∴
L
= µ0 n2 A = Inductance per unit length
l
Notice :
(i) L ∝ n2
(ii) The inductance, like the capacitance, depends only on geometric
factors, not on i.
9.1. INDUCTANCE
109
(2) Toroid:
Recall: B-field lines are concentric circles.
Inside the toroid:
µ0 iN
2πr
(NOT a constant)
B=
where r is the distance from center.
Outside the toroid:
B=0
Flux linkage through the toroid
ˆ
~ · d~a
N ΦB = N B
(
ˆ
µ0 iN 2 b h dr
=
2π
r
a
2
³
µ0 iN h
b´
=
ln
2π
a
∴
Inductance
Again,
L=
~ k d~a
Notice B
Write da = h dr
N ΦB
µ0 N 2 h ³ b ´
=
ln
i
2π
a
L ∝ N2
Inductance with magnetic materials :
We showed earlier that for capacitors:
(
~ → E/κ
~ e
E
C → κe C
(after insertion of
dielectric κe > 1)
For inductors, we first know that
~ → κm B
~
B
Inductance
However
(after insertion of
magnetic material)
N ΦB
ˆi
~ · dA
~ → κm ΦB
ΦB = B
L=
+
KEY
9.2. LR CIRCUITS
110
∴
L → κm L
(after insertion of
magnetic material)
To increase inductance, fill the interior of inductor with ferromagnetic
materials. (×103 − 104 )
∴
9.2
LR Circuits
(A) ”Charging” an inductor
When the switch is adjusted to position a,
By loop rule (clockwise) :
E0 − ∆VR + ∆VL = 0
↓
↓
di
E0 − iR − L
= 0
dt
di R
E0
First Order Differ∴
+ i=
ential Equation
dt L
L
Similar to the equation for charging a capacitor!
(Chap5)
´
E0 ³
1 − e−t/τL
R
where τL = Inductive time constant = L/R
Solution: i(t) =
∴
iR = E0 (1 − e−t/τL )
di
E0 1
|∆VL | = L
= L·
·
· e−t/τL = E0 e−t/τL
dt
R τL
|∆VR | =
9.2. LR CIRCUITS
111
(B) ”Discharging” an inductor
When the switch is adjusted at position b after the inductor has been
”charged” (i.e. current i = E0 /R is flowing in the circuit.).
By loop rule:
∆VL − ∆VR = 0
↓
↓
di
−L
− iR = 0
dt
(Treat inductor as source of emf)
∴
di R
+ i=0
dt L
Discharging a capacitor
(Chap5)
i(t) = i0 e−t/τL
where i0 = i(t = 0) = Current when the circuit just switch to position b.
Summary : During charging of inductor,
1. At t = 0, inductor acts like open circuit when current flowing is zero.
2. At t → ∞, inductor acts like short circuit when current flowing is
stablized at maximum.
3. Inductors are used everyday in switches for safety concerns.
9.3. ENERGY STORED IN INDUCTORS
9.3
112
Energy Stored in Inductors
Inductors stored magnetic energy through the magnetic field stored in the circuit.
Recall the equation for charging inductors:
E0 − iR − L
di
=0
dt
Multiply both sides by i :
E0 i
|{z}
Power input by emf
(Energy supplied to
one charge = qE0 )
i2 R
|{z}
=
di
| {zdt}
+
Joule’s heating
(Power dissipated
by resistor)
Li
Power stored
in inductor
dUB
di
Power stored in inductor =
= Li
dt
dt
Integrating both sides and use initial condition
∴
At t = 0,
∴
i(t = 0) = UB (t = 0) = 0
Energy stored in inductor: UB =
1 2
Li
2
Energy Density Stored in Inductors :
Consider an infinitely long solenoid of cross-sectional area A.
For a portion l of the solenoid, we know from §8.1,
L = µ0 n2 lA
∴
Energy stored in inductor:
UB =
1 2 1
Li = µ0 n2 i2|{z}
lA
2
2
Volume of
solenoid
∴
Energy density (= Energy stored per unit volume) inside inductor:
uB =
UB
1
= µ0 n2 i2
lA
2
Recall magnetic field inside solenoid (Chap7)
B = µ0 ni
∴
uB =
B2
2µ0
This is a general result of the energy stored in a magnetic field.
9.4. LC CIRCUIT (ELECTROMAGNETIC OSCILLATOR)
9.4
113
LC Circuit (Electromagnetic Oscillator)
Initial charge on capacitor = Q
Initial current = 0
No battery.
Assume current i to be in the direction that increases charge on the positive
capacitor plate.
dQ
⇒
i=
(9.1)
dt
By Lenz Law, we also know the ”poles” of the inductor.
Loop rule:
VC + VL = 0
Q
di
− −L
= 0
C
dt
(9.2)
Combining equations (9.1) and (9.2), we get
d2 Q
1
+
Q=0
dt2
LC
This is similar to the equation of motion
of a simple harmonic oscillator:
d2 x
k
+ x=0
2
dt
m
Another approach (conservation of energy)
Total energy stored in circuit:
U = UE
↓
Q2
U =
2C
+
UB
↓
1 2
+
Li
2
Since the resistance in the circuit is zero, no energy is dissipated in the circuit.
Energy contained in the circuit is conserved.
∴
∴
dU
=0
dt
Q dQ¶
di
⇒
· ¶ + L¢i
=0
C ¶dt
dt
(∵ i =
dQ
)
dt
9.4. LC CIRCUIT (ELECTROMAGNETIC OSCILLATOR)
114
di Q
+ =0
dt C
d2 Q
1
+
Q=0
2
dt
LC
⇒ L
⇒
The solution to this differential equation is in the form
Q(t) = Q0 cos(ωt + φ)
dQ
= −ωQ0 sin(ωt + φ)
∴
dt
d2 Q
= −ω 2 Q0 cos(ωt + φ)
2
dt
= −ω 2 Q
d2 Q
+ ω2Q = 0
2
dt
∴
∴
ω2 =
1
LC
Angular frequency
of the LC oscillator
Also, Q0 , φ are constants derived from the
¯ initial conditions. (Two initial condidQ ¯
tions, e.g. Q(t = 0), and i(t = 0) = dt ¯
are required.)
t=0
Energy stored in C =
Q2
2C
=
Energy stored in L =
1 2
Li =
2
1
C
=
∵ Lω 2 =
∴
Total energy stored =
=
Q20
cos2 (ωt + φ)
2C
1
Lω 2 Q20 sin2 (ωt + φ)
2
Q20
sin2 (ωt + φ)
2C
Q20
2C
Initial energy stored in capacitor
9.5. RLC CIRCUIT (DAMPED OSCILLATOR)
9.5
115
RLC Circuit (Damped Oscillator)
In real life circuit, there’s always resistance.
In this case, energy stored in the LC oscillator is
NOT conserved,
dU
= Power dissipated in the resistor = −i2 R
dt
Negative sign shows that energy U is decreasing.
and
(Joule’s heating)
i
z}|{
∴
Li
di Q dQ
+ ·
= −i2 R
dt C dt
d2 Q R dQ
1
+
·
+
Q=0
dt2
L dt
LC
This is similar to the equation of motion of a damped harmonic oscillator (e.g.
if a mass-spring system faces a frictional force F~ = −b~v ).
Solution to the equation is in the form Q(t) = eλt
If damping is not too big (i.e. R not too big), solution would become
⇒
R
Q(t) = Q0 |e−{z2L }t cos(ω1 t + φ)
|
{z
}
exponential oscillating
decay term term
where
³ R ´2
1
−
LC
2L
³ R ´2
ω12 = ω 2 −
2L
ω12 =
Damped oscillator always oscillates
at a lower frequency than the
natural frequency of the oscillator.
(Refer to Halliday, Vol1, Chap17 for
more details.)
Check this at home: What is UE (t) + UB (t) for the case when damping is small?
(i.e. R ¿ ω)
Chapter 10
AC Circuits
10.1
Alternating Current (AC) Voltage
Recall that an AC generator described in Chapter 9 generates a sinusoidal emf.
i.e. E = Em sin(ωt + δ)
Note :
This circuit is the RLC circuit with one
additional element : the time varying AC
power supply. This is similar to a driven
(damped) oscillator.
d2 Q
dQ
1
+R
+ Q = Em sin(ωt + δ)
2
dt
dt
C
The general solution consists of two parts:
L
transient : rapidly dies away in a few cycles (not interesting)
steady state : Q(t), i(t) varies sinusoidally with the same frequency as input
³ R ´2
1
Note : Current does NOT vary at frequency ω12 =
−
LC
2L
Since we only concern about the steady state solution, therefore we can take any
time as starting reference time = 0
For convenience, we can write
E = Em sin ωt
And we can write
i = im sin(ωt − φ)
where im is current amplitude, φ is phase constant.
Our goal is to determine im and φ.
10.2. PHASE RELATION BETWEEN I, V FOR R,L AND C
10.2
117
Phase Relation Between i, V for R,L and C
(A) Resistive Element
∴
∆VR = VA − VB = iR
∆VR = im R sin(ωt − φ)
∆VR and i are in phase, i.e. what’s
inside the ”sine bracket” (phase) is the
same for ∆VR and i.
Graphically, we introduce phasor diagrams properties of phasors:
(1) Length of a phasor is proportional to the maximum value.
(2) Projection of a phasor onto the vertical axis gives instantaneous value.
(3) Convention: Phasors rotate anti-clockwise in a uniform circular motion with angular velocity.
∴
∆VR = (∆VR )m sin(ωt − φ)
(∆VR )m = im R
”Ohm’s Law like” relation for AC resistor
10.2. PHASE RELATION BETWEEN I, V FOR R,L AND C
118
(B) The Inductive Element
Potential drop across inductor
∆VL = VA − VB = −EL = L
∴
∆VL = Lim ω cos(ωt − φ)
π
)
2
π
= im XL sin(ωt − φ + )
2
= Lim ω sin(ωt − φ +
[∵ cos θ = sin(θ +
XL = Inductive Reactance
XL = ωL
As
i ↑, VA > VB
i ↓, VA < VB
∴ ∆VL > 0
∴ ∆VL < 0
∆VL
leads
i
by
i
lags
∆VL
by
(C) Capacitive Element
∆VC = VA − VB =
π
)]
2
”Ohm’s Law like” relation for AC inductor
(∆VL )m = im XL
where
di
dt
Q
C
π
2
π
2
10.3. SINGLE LOOP RLC AC CIRCUIT
119
where Q = charge on the positive plate of the capacitor.
ˆ
dQ
∴
i=
⇒ Q =
i dt
dt
ˆ
=
im sin(ωt − φ) dt
= −
∴
im
cos(ωt − φ)
ωC
π
= im XC sin(ωt − φ − )
2
∆VC = −
∴
where
10.3
im
cos(ωt − φ)
ω
XC =
[∵ − cos θ = sin(θ −
”Ohm’s Law like” relation for AC capacitor
(∆VC )m = im XC
1
= Capacitive Reactance
ωC
∆VC
lags
i
by
i
leads
∆VC
by
π
2
π
2
Single Loop RLC AC Circuit
Given that E = Em sin ωt, we want to
find im and φ so that we can write i =
im sin(ωt − φ)
Loop rule:
E − ∆VR − ∆VL − ∆VC = 0
⇒ E = ∆VR + ∆VL + ∆VC
π
)]
2
10.3. SINGLE LOOP RLC AC CIRCUIT
120
Using results from the previous section, we can write
Em sin ωt =
im R sin(ωt − φ)
+im XL cos(ωt − φ) − im XC cos(ωt − φ)
h
i
Em sin ωt = im R sin(ωt − φ) + (XL − XC ) cos(ωt − φ)
Answer :
1. Take
tan φ =
2. Define Z =
XL − XC
R
q
R2 + (XL − XC )2
as the impedance of the circuit.
3. Then
im =
Em
Z
or Em = im Z
”Ohm’s Law like” relation
for AC RLC circuits
Check :
i
XL − XC
cos(ωt − φ)
Z
hZ
i
= im Z cos φ sin(ωt − φ) + sin φ cos(ωt − φ)
R.H.S. = im Z
hR
sin(ωt − φ) +


Use the relation:


 sin(A + B) = sin A cos B + cos A sin B 
Here: A = ωt − φ, B = φ
= im Z sin(ωt − φ + φ)
= im z sin ωt
= L.H.S. if Em = im Z
Phasor Approach :
QED.
10.4. RESONANCE
10.4
121
Resonance
Em
im =
is at maximum for an AC circuit of fixed input frequency ω when Z
Z
is at minimum.
s
q
Z=
R2 + (XL − XC )2 =
³
R2 + ωL −
1 ´2
ωC
is at a minimum for a fixed ω when
XL − XC = ωL −
⇒
⇒
1
= 0
ωC
1
ωL =
ωC
1
ω2 =
LC
same as that for
a RLC circuit
In Hong Kong, the AC power input is 50Hz.
(In US, as mentioned in Halliday, is 60Hz.)
∴
10.5
ω = 2πf = 314.2s−1
Power in AC Circuits
Consider the Power dissipated by R in an AC circuit:
P = i2 R = i2m R sin2 (ωt − φ)
The average power dissipated in each cycle:
´ 2π/ω
P dt
2π
Pave = 0
(
is period of each cycle)
2π/ω
ω
ˆ
ˆ
2π/ω
P dt =
0
i2m R
ˆ
= i2m R
2π/ω
sin2 (ωt − φ) dt
0
2π/ω
0
i
1h
1 − cos 2(ωt − φ) dt
2
©¯
©
sin2 (ωt
−©
φ) i¯¯2π/ω
©
=
·
−
©
¯
2 ©© 4ω
0
1
2π
= i2m R · ·
2 ω
i2m R
ht
10.5. POWER IN AC CIRCUITS
∴
122
i2
Pave = m R = i2rms R
2
where irms = root-mean-square current
im
irms = √
2
Symbol :
∵ Current is a
sinusoidal func.
hP i = Pave = Average of P over time
For sine and cosine functions of time:
Average :
hsin ωti = hcos ωti = 0
Amplitude : Peak value, e.g. Em , im , (∆VR )m , · · ·
Root-Mean-Square(RMS) : It’s a measure of the ”time-averaged” deviation
from zero.
q
xrms = hx2 i
For sines and cosines, for whatever quantity x:
xm
xrms = √
2
(xm is amplitude)
For an AC resistor circuit:
2
Erms
R
hP i = i2rms R =
Laws for DC circuits can be used to describe AC circuits if we use rms values
for i and E.
For general AC circuits:
z
E
}|
{ z
i
}|
{
P = Ei = Em sin ωt · im sin(ωt − φ)
= Em im sin ωt [sin ωt cos φ − cos ωt sin φ]
2
P = Em im [ sin
| {zωt} cos φ − sin
| ωt{zcos ωt} sin φ ]
1
2
hP i =
Em im
cos φ
2
hP i = Erms irms cos φ
| {z }
power factor
0
(check this!)
10.6. THE TRANSFORMER
Recall
123
XL − XC
R
R
cos φ =
Z
tan φ =
∴
Maximum power dissipated in circuit when
cos φ = 1
Two possibilities:
(1) XL = XC = 0
(2) XL − XC = 0
⇒
XL = XC
⇒
ωL =
(Resonance Condition)
10.6
1
ωC
⇒
ω2 =
1
LC
The Transformer
Power dissipated in resistor
hP i = i2rms R
∴
For power transmission, we’d like to keep irms at minimum.
⇒ HIGH potential difference across transmission wires. (So that total power
transmitted P = irms Erms is constant.)
However, for home safety, we would like LOW emf supply.
Solution : Transformers
Primary : Number of winding = NP
10.6. THE TRANSFORMER
124
Secondary : Number of winding = NS
In primary circuit, RP ≈ CP ≈ 0
Pure inductive
∴
R
≈0
Z
No power delivered from emf to transformer.
Power factor :
∴
cos φ =
The varying current (∵ AC!) in the primary produces an induced emf in the
secondary coils. Assuming perfect magnetic flux linkage:
emf per turn in primary
= emf per turn in secondary
dΦB
= −
dt
∆VP
NP
∆VS
emf per turn in secondary =
NS
(∆VP is P.D.
across primary)
emf per turn in primary =
⇒
∆VP
NP
=
∆VS
NS
If NP > NS , then ∆VP > ∆VS Step-Down
If NP < NS , then ∆VP < ∆VS
Step-Up
Consider power in circuit:
iP ∆VP = iS ∆VS
In the secondary, we have
∆VS = iS R
Combining the 3 equations, we have
∆VP =
³ N ´2
P
NS
R · iP
”Equivalence Resistor” =
³ N ´2
P
NS
R
Chapter 11
Displacement Current and
Maxwell’s Equations
11.1
Displacement Current
We saw in Chap.7 that we can use
Ampère’s law to calculate magnetic
fields due to currents.
¸
~ · d~s
We know that the integral C B
around any close loop C is equal to
µ0 iincl , where iincl = current passing an
area bounded by the closed curve C.
e.g.
=
Flat surface bounded by loop C
=
Curved surface bounded by loop C
If Ampère’s law is true all the time, then the iincl determined should be independent of the surface chosen.
11.1. DISPLACEMENT CURRENT
126
Let’s consider a simple case: charging a
capacitor.
From Chap.5, we know there is a current
flowing i(t) = ER0 e−t/RC , which leads
~ With
to a magnetic ¸field observed B.
~
Ampère’s law, C B · d~s = µ0 iincl .
BUT WHAT IS iincl ?
If we look at
,
If we look at
iincl = i(t)
, iincl = 0
(∵ There is no charge flow between the
capacitor plates.)
∴ Ampère’s law is either WRONG or
INCOMPLETE.
Two observations:
1. While there is no current between the capacitor’s plates, there is a timevarying electric field between the plates of the capacitor.
2. We know Ampère’s law is mostly correct from measurements of B-field
around circuits.
⇓
Can we revise Ampère’s law to fix it?
σ
Q
=
, where Q = charge on
ε0
ε0 A
capacitor’s plates, A = Area of capacitor’s plates.
Electric field between capacitor’s plates: E =
∴
Q = ε0 E
· A} = ε0 ΦE
| {z
Electric flux
∴ We can define
dQ
dΦE
= ε0
= idisp
dt
dt
where idisp is called Displacement Current (first proposed by Maxwell).
Maxwell first proposed that this is the missing term for the Ampère’s law:
˛
~ · d~s = µ0 (iincl + ε0 dΦE ) Ampère-Maxwell law
B
dt
C
11.2. INDUCED MAGNETIC FIELD
127
Where iincl = current through any surface bounded by C,
´
~ ·d~a.
ΦE = electric flux through that same surface bounded by curve C, ΦE = S E
11.2
Induced Magnetic Field
We
learn
earlier
that
electric
field
can
be
generated
by
(
charges
.
changing magnetic flux
We
( see from Ampère-Maxwell law that a magnetic field can be generated by
moving charges (current)
.
changing electric flux
That is, a change in electric flux through a surface bounded by C can lead to an
induced magnetic field along the loop C.
Notes The induced magnetic field is along the same direction as caused by the
changing electric flux.
Example What is the magnetic field strength inside a circular plate capacitor
of radius R with a current I(t) charging it?
Answer Electric field of capacitor
E=
Q
Q
=
ε0 A
ε0 πR2
11.3. MAXWELL’S EQUATIONS
128
Electric flux inside capacitor through a
loop C of radius r:
ΦE = E · πr2 =
Qr2
ε0 R 2
Ampère-Maxwell Law inside capacitor:
˛
dΦE
~ · d~s = µ0 (©
© + ε0
)
B
iincl
dt
C
|
{z
}
~ induced kd~s
∵B
2πr
|{z} Binduced = µ0 ε0
Length of loop C
= µ0
d ³ Qr2 ´
dt ε0 R2
r2 dQ
R2 |{z}
dt
I(t)
∴ Binduced =
µ0 r
I(t)
2πR2
for r < R
Outside the capacitor plate:
Electric flux through loop C: ΦE = E ·
Q
πR2 =
ε0
˛
~ · d~s = µ0 (iincl + ε0 dΦE )
B
dt
C
2πrBinduced = µ0 ε0
∴ Binduced =
11.3
³1
ε0
·
dQ ´
dt
µ0 I(t)
2πr
Maxwell’s Equations
The four equations that completely describe the behaviors of electric and magnetic
fields.
11.3. MAXWELL’S EQUATIONS
˛
S
˛
129
~ · d~a = Qincl
E
ε0
~ · d~a = 0
B
S
˛
C
˛
~ · d~s = − d
E
dt
ˆ
~ · d~a
B
S
~ · d~s = µ0 iincl + µ0 ε0 d
B
dt
ˆ
~ · d~a
E
S
C
The one equation that describes how matter reacts to electric and magnetic fields.
~ + ~v × B)
~
F~ = q(E
Features of Maxwell’s equations:
(1) There is a high level of symmetry in the equations. That’s why the study
of electricity and magnetism is also called electromagnetism.
There are small asymmetries though:
i) There is NO point ”charge” of magnetism / NO magnetic monopole.
ii) Direction of induced E-field opposes to B-flux change.
Direction of induced B-filed enhances E-flux change.
(2) Maxwell’s equations predicted the existence of propagating waves of E-field
and B-field, known as electromagnetic waves (EM waves).
Examples of EM waves: visible light, radio, TV signals, mobile phone
signals, X-rays, UV, Infrared, gamma-ray, microwaves...
(3) Maxwell’s equations are entirely consistent with the special theory of relativity. This is not true for Newton’s laws!