Download Physics 169

Physics 169
Kitt Peak National Observatory
Monday, January 30, 17
Luis anchordoqui
1
we see that like charges repel. Conversely, a rubbed balloon and your hair,
which do not have the same kind of charge, are attracted to one another.
Thus, unlike charges attract.
1.1 Electric Charge
(a)
n across
the balloon
charged and
Two
he other
.
6
(b)
(a) If you rub a balloon across your hair on a dry day
the balloon and your hair become charged and attract each other
(b) Two charged balloons, on the other hand, repel each other
.
The two balloons must have the same kind of charge
because each became charged in the same way
Because two charged balloons repel one another we see that like charges repel
Conversely ☛ a rubbed balloon and your hair
which do not have the same kind of charge
are attracted to one another ☛ unlike charges attract
Monday, January 30, 17
2
are applied.
Another important aspect of electricity that arises from experimental obs
Electric charge is conserved
tions is that electric charge is always conserved in an isolated system. Th
when one object is rubbed against another, charge is not created in the process.
Charge is electrified
conserved
quantized
stateand
is due
to a transfer of charge from one object to the other.
object gains some amount of negative charge while the other gains an equal am
of a
positive
example,
when
When
glass charge.
rod is For
rubbed
with
silka glass rod is rubbed with silk, as in Fi
23.2, the silk obtains a negative charge that is equal in magnitude to the
–
electrons
are transferred from the glass to the silk
+
–
+
tive
charge
on the glass rod. We now know from our understanding of atomic s
– +++
Because
of
conservation
of charge
–
ture that electrons are transferred
from the glass to the silk in the rubbing pro
+
–
eachSimilarly,
electron
adds
negative
charge
to electrons
the silk are transferred from the f
when
rubber
is rubbed
with fur,
–
rubber,positive
giving the
rubberisa left
net negative
and the
an equal
charge
behind charge
on theand
rodthe fur a net po
This process is consistent with the fact that neutral, uncharged m
Also charge.
☛
because
charges are transferred in discrete bundles
contains as many positive charges (protons within atomic nuclei) as neg
charges
on (electrons).
the two objects are
charges
In 1909, Robert Millikan (1868–1953) discovered that electric charge a
Figure 23.2 When a glass rod is
occurs as some integral multiple of a fundamental amount of charge e (see Se
rubbed with silk, electrons are
25.7). In modern terms, the electric charge q is said to be quantized, where q i
transferred from the glass to the
standard symbol used for charge as a variable. That is, electric charge exis
silk. Because of conservation of
left
charge, each electron adds negadiscrete “packets,” and we can write q ! Ne, where N is some integer. Other ex
tive charge to the silk, and an equal
A negatively
rubber
rod " e and the pr
ments in the same period
showed that charged
the electron
has a charge
positive charge is left behind on
has a charge of equalsuspended
magnitude but
# e. Some particles, suc
by aopposite
threadsign
is attracted
the rod. Also, because the charges
the neutron,Rubber
have no charge.
to a positively charged glass rod
are transferred in discrete bundles,
Rubber
From
our
discussion
thus far, we conclude that electric charge has the followin
the charges on the two objects are
Fportant properties:
$ e, or $ 2e, or $ 3e, and so on.
±e, ±2e, ±3e, · · ·
Fields
–– –
F
– –
F
+
+
+ + +
+ +
–– –
Glass
–
(a)
– –
– –– –
– –
Rubber
F
(b)
Figure 23.1 (a) A negatively charged rubber rod suspended by a thread is attracted
to a positively charged glass rod. (b) A negatively charged rubber rod is repelled by
another negatively charged rubber rod.
Monday, January 30, 17
right
A negatively charged rubber rod
is repelled by another negatively charged
rubber rod
3
2.1
Electric Force
1.2 Electric Force
q1 and
Electric
force force
between
two two
charges
The electric
between
charges
q1 and byq2☛ Coulomb’s
can be described
by
described
Law
Coulomb’s Law.
q2
F~12 = Force on q1 exerted by q2
1
q1 q2
· 2 · r̂12
r12q2
0
F~12 = F orce on q14⇡✏
exerted
by
~r12
☛ unit vector which locates
1 relative to particle 2
=
qparticle
q2
1
1
~
|~r12 |
F12 = 4º≤0 · r2 · r̂12
r12 = ~r1 12~r2
i.e. ~
F~12 =
r̂12
~r12
where r̂12 =
is the unit vector which locates particle 1 relative to particle 2.
• q1 , q2 are|~relectrical
charges in units of Coulomb (C)
|
12
19
• Charge is quantized ☛i.e.
electron
carries
1.602
⇥
10
C
~r12 = ~r1 ° ~r2
• Permittivity of free space ✏0 = 8.85 ⇥ 10 12 C2 /Nm2
• q1 , q2 are electrical charges in units of Coulomb(C)
• Charge is quantized
Monday, January 30, 17
Recall 1 electron carries 1.602 £ 10°19 C
4
°12
2
2
COULOMB’S LAW:
(1) q1 , q2 can be either positive or negative
(2) If q1 , q2 are of same sign
force experienced by q2 is in direction away from q1 i.e. ☛ repulsive
(3) Force on q2 exerted by q1 :
F~21 =
BUT
1
q2 q1
· 2 · r̂21
4⇡✏0
r21
r12 = r21 = distance between q1 , q2
r̂21
~r21
~r2 ~r1
~r12
=
=
=
=
r21
r21
r12
F~21 =
Monday, January 30, 17
r̂12
F~12 Newton’s 3rd Law
5
SYSTEM WITH MANY CHARGES:
2.2
SYSTEM WITH MANY CHARGES:
The total force experienced by charge
q1 is the vector sum of the forces on q1
exerted by other charges.
Total force experienced by charge q1
vector sum of forces on q1 exerted by other charges
~
F1 = Force experienced by q1
~
F1 = Force experienced
by q1
= F~1,2 + F~1,3 + F~1,4 + · · · + F~1,N
= F~1,2 + F~1,3 + F~1,4 + · · · + F~1,N
PRINCIPLE OF SUPERPOSITION:
N
X
PRINCIPLE OF SUPERPOSITION
F~1 =
F~1,j
j=2
Monday, January 30, 17
The Electric Field
F~1 =
N
X
F~1,j
j=2
6
1.3 Electric Field
While we need two charges to quantify electric force
we define electric field for any single charge distribution
2.2. THE ELECTRIC FIELD
to describe its effect on other charges
Total force
F~ = F~1 + F~2 + · · · + F~N
Total force F~ = F~1 + F~2 + · · · + F~N
The electric field is defined as
Electric field is defined as F
~
~
F
~ = lim
E
q0 !0 q0
lim
q0 !0
q0
~
=E
(a) E-field due to a single charge qi :
Monday, January 30, 17
7
(a) E-field due to a single charge qi :
(i) E-field due to a single charge qi
From definitions of Coulomb’s Law
~ =
Recall E
where r̂0,i
experienced
at location
of qthe
0 (point P )
Fromforce
the definitions
of Coulomb’s
Law,
q0 qiP)
force experienced at location 1
of q0 (point
~
F0,i =
· 2 · r̂0,i
4⇡✏0 r0,i
r̂0,i ☛ unit vector along direction from charge qi to q0
~0,i = 1 · q0 qi · r̂0,i
F
~
F
2
4º≤
r
0
~
0,i
)
-field
due to qi at point P
lim
E
q0 !0
q0 along the direction from charge qi to q0 ,
is the unit
vector
r̂0,i =
=
1
qi
~
Ei =
Unit vector from
charge qi to ·point
P· r̂i
2
4⇡✏
r
r̂i (radical unit vector from0qi ) i
~ri ☛ vector pointing from qi to point P
r̂i ☛ unit vector pointing from qi to pointP
F~
~
Recall E = lim
q0 !0 q0
) E-field due to qi at point P:
Note:
~ -field is a vector
~ i = 1 · qi · r̂i
E
(1) E
4º≤0 ri2
~ -field depends on both position of
(2) Direction of E
where ~ri = Vector pointing from qi to point P,
thus r̂i = Unit vector pointing from qi to point P
Monday, January 30, 17
Note:
(1) E-field is a vector.
P and sign of qi
8
RIC FIELD
~ -field due to system of charges:
(ii) E
ole
11
Principle ofX
Superposition
1 X qi
~
~
Ei =
r̂ ~
2 i E
N charges
☛
total
-field due to all charges
4º≤0 i ri
~ -field due to individual charges
vector sum of E
X
X qi
1
~i =
~ =
E
r̂i
i.e. ☛ E
2
4⇡✏0
ri
i.e.
E=
In a system with
i
(iii) Electric Dipole
i
i
d opposite charges
tance d.
System of
equal2.1:
and An
opposite
by of
a distance d
Figure
electriccharges
dipole.separated
(Direction
d~ from negative to positive charge)
Electric Dipole Moment ☛ p
~ = q d~ = qddˆ
p = qd
Electric Dipole Moment
Monday, January 30, 17
p~ = q d~ = qddˆ
9
p = qd
~ due
mple: Example:
E
dipole along x-axis
~ to
E
due to dipole along x -axis
P at distance x along perpendicular axis of dipole p~
nsider point P at distance x along the perpendicular axis of the dipole p~ :
ice:
Consider point
~
E
~+ E
~E
~ =E
~+ + + E
~°
= E
"
"
(E-field
(E-field
due
to +q)
°q)due to q
-field
due to +q dueEto
E-field
~ + and E
~ ° cancel out.
Horizontal E-field components of E
Monday, January 30, 17
10
due to °q)
due to +q)
Notice:
~ + and E
~ ° cancel out.
Horizontal E-field components of E
~ cancel out
~ + and E
~-field components of E
Notice: Horizontal E
) Net E-field points along the axis opp
site to the dipole moment vector.
2.3. CONTINUOUS CHARGE DISTRIBUTION
12
~ -field points along axis opposite to dipole moment vector
) Net E
Magnitude of E-field = 2E+ cos µ
~ -field = 2E+ cos ✓
Magnitude of E
) E
But
= 2
!
1
q
· 2
cos ✓
4⇡✏0 r
| E{z magnitude!
}
or
EE+ or E °magnitude
+
r⇣ ⌘µ z 1 }| q{ ∂
2
) E =d2
cos µ
2· 2
r =
+
x
2 4º≤0 r
d/2
cos ✓ = But
r
r=
s
≥ d ¥2
+ x2
2
1
p
d/2 2 3
)E =
·
µ 2=+ d ] 2
4⇡✏0 cos[x
r2
(p = qd)
Monday, January 30, 17
)E =
1
p
·
4º≤0 [x2 + ( d2 )2 ] 32
11
Special case ☛ When x
⇥
d
x +
2
2
• Binomial Approximation
d
2 ⇤ 32
= x
3
⇥
d
1+
2x
(1 + y)n ⇡ 1 + ny
~
E
• Compare with
if
y⌧1
1
1
p
field of dipole '
· 3 / 3
4⇡✏0 x
x
1 ~ -field for single charge
E
2
r
• Result also valid for point
Monday, January 30, 17
2 ⇤ 32
P along any axis with respect to dipole
12
2.3
Continuous Charge Distribution
1.4 Continuous Charge Distribution
E -field at point P due to dq
E-field ~at point1P duedq
to dq:
dE =
· 2 · r̂
4⇡✏0 r
1
dq
· 2 · r̂
) E -field due to charge
4º≤distribution
r
0
~ =
dE
~ =
E
Z
~ =
dE
Z
1
dq
· 2 · r̂
4⇡✏0 r
(1) Take advantage of symmetry of system to simplify integral
(2) To write down small charge element dq ☛
1
D
dq =
ds
= linear charge density
2
D
dq =
dA
= surface charge density
3
D
dq = ⇢ dV
Monday, January 30, 17
⇢ = volume charge density
ds = small length element
dA = small area element
dV = small volume element
13
1-D
2-D
3-D
dq = ∏ ds
dq = æ dA
dq = Ω dV
∏ = linear charge density
æ = surface charge density
Ω = volume charge density
ds = small length element
dA = small area element
dV = small volume element
Example 1:
Uniform line of charge
Uniform line of charge
Example
1
charge per
unit length
= ∏ per unit
charge
(1) Symmetry considered ☛E -field from +z and
length =
z directions cancel along
(1) Symmetry considered: The E-field from +z and °z directions cancel along
z-direction,
) Only horizontal
components
needcomponents
to be considered.
) Only E-field
z -direction,
horizontal
need
E-field
to be considered
(2) For each element of length dz, charge dq = ∏dz
(2) For each
dzdz, charge
) Horizontal
E-field element
at point P of
due length
to element
=
dq
=∏dz dz
1
~ cos µ =
E|
·
cos µ
to element
) Horizontal E-field at point P |ddue
4º≤0 r2dz is
|
{z
}
1
dz dEdz
~
dz = |dE| cos
✓ = to entire
· 2 cos
✓
) E-field due
charge at point P
4⇡✏0 rline
| {z }
Monday, January 30, 17
dEdz
ˆL/2
E =
°L/2
1
∏dz
· 2 cos µ
4º≤0 r
14
⌥ cos ⇥ =
|dE|
4⌅
⇤
0
⇥
·
dEdz
r2
) E -field due to entire line charge at point P
E-field due to entire line charge a
Z L/2
1
dz
E =
· L/2
cos ✓
2
rˆ
L/2 4⇡✏0
1
⇤dz
E =
· 2 cos
Z L/2
4⌅ 0 r
dz
= 2
· L/22 cos ✓
4⇡✏0 rL/2
0
ˆ
⇤
dz
= 2
· 2 cos
4⌅ 0 r
To calculate this integral ☛
0
• First, notice that x is fixed, but z, r, ✓ all varies
• Change of variable (from z to ✓ )
Monday, January 30, 17
15
(1)
z = x tan ✓
x = r cos ✓
z = 0
) dz = x sec2 ✓ d✓
) r2 = x2 sec2 ✓
✓ = 0
L/2
✓ = ✓0 where tan ✓0 =
(2) When z = L/2
Z ✓0
x
2
x sec ✓ d✓
E = 2·
· cos ✓
2
2
4⇡✏0 0
x sec ✓
Z ✓0
1
= 2·
· cos ✓ d✓
4⇡✏0 0 x
✓0
1
= 2·
· · (sin ✓)
4⇡✏0 x
0
1
= 2·
· · sin ✓0
4⇡✏0 x
1
L/2
r
= 2·
· ·
⇣ ⌘2
4⇡✏0 x
x2 + L2
Monday, January 30, 17
16
1
E =
·
4⇡✏0
x
r
L
x2 +
Important limiting cases
(1)
But
(2)
x
L:
⇣ ⌘2
along x-direction
L
2
1
L
E +
· 2
4⇡✏0 x
L = Total charge on rod ) System behave like a point charge
L
x:
1
L
E +
·
4⇡✏0 x · L2
Ex =
2⇡✏0 x
ELECTRIC FIELD DUE TO INFINITELY LONG LINE OF CHARGE
Monday, January 30, 17
17
2.3. CONTINUOUS CHARGE DISTRIBUTION
Example 2:
15
Ring of Charge
Example 2
Ring of Charge
z above
at a height
at a height
z abovea aring
ring of
of charge of radius R
E-field E-field
charge of radius R
(1) Symmetry considered: For every charge element dq considered, there exists
(1) Symmetry considered
☛ For every charge element dq considered,
⌃ field components
dq⇤ where the horizontal E
cancel.
0
~ field components cancel
⇥there
Overall
E-fielddq
lies along
z-direction.
exists
where
horizontal E
(2) For each element of length dz, charge
Monday, January 30, 17
18
(1) Symmetry considered: For every charge element dq considered, there exists
⌃ field components cancel.
dq⇤ where the horizontal E
⇥ Overall E-field lies along z-direction.
(2) For each element of length dz , charge
(2) For each element of length dz, charge
dq =
dq
=
⇤
·
ds
·
ds
Linear
Circular
Linear
Circular
charge
density
element
charge
density length
length
element
dqdq
= ⇤=· R d⇧,
· R d where ⇧ is the angle
measured on the ring plane
where
is angle measured on ring plane
Net
along
z-axis
duezto
dq: due to dq
) E-field
Net E
-field
along
-axis
1
dq
dE =
· 2 · cos ⇥
4⌅ 0 r
1
dq
dE =
· 2 · cos ✓
4⇡✏0 r
Monday, January 30, 17
19
Total E -field
=
=
Z
Z
0
dE
2⇡
1
·
4⇡✏0
Rd
r2
· cos ✓
Note: Here in this case, ✓, R and r are fixed as
BUT we want to convert r, ✓ to
z
(cos ✓ = )
r
varies!
R, z
1
Rz
E =
· 3
4⇡✏0
r
Z
2⇡
d
0
1
(2⇡R)z
E =
· 2
4⇡✏0 (z + R2 )3/2
BUT ☛
along z -axis
(2⇡R) = total charge on ring
Monday, January 30, 17
20
ample 3:
4⌅
BUT:
0
(z + R )
⇤(2⌅R) = total charge on the ring
Example
3 a disk E-field
fromdensity
a disk
of surface charge density
E-field from
of surface charge
⇧
We find E-field of a disk by integrating concentric rings
of charges
2.3. CONTINUOUS CHARGE DISTRIBUTION
Total
We find the E-field of a disk by
charge
of ring
integrating
concentric rings of
charges.
dq =
Total charge of ring
17
view from top
· (2⇡r
| {zdr})
Area of ring
dq = ⇤ · ( 2⇥r⌥ dr⌦ )
Area of the ring
Recall from Example 2:
E-field from ring: dE =
Monday, January 30, 17
1
E =
4⇥ 0
ˆ
0
ˆ
R
1
dq z
· 2
4⇥ 0 (z + r2 )3/2
2⇥⇤r dr · z
(z 2 + r2 )3/2
21
Recall from Example 2
1
dq z
· 2
4⇡✏0 (z + r2 )3/2
Z R
1
2⇡ r dr · z
) E =
4⇡✏0 0 (z 2 + r2 )3/2
E-field from ring ☛ dE =
1
=
4⇡✏0
Z
R
0
r dr
2⇡ z 2
(z + r2 )3/2
• Change of variable:
u = z 2 + r2 )
) du = 2r dr )
Monday, January 30, 17
(z 2 + r2 )3/2 = u3/2
1
r dr = du
2
22
• Change of integration limit:
⇢
BUT Z
u = z2
u = z 2 + R2
Z z2 +R2
1
1
) E=
· 2⇡ z
u
4⇡✏0
2
z2
u
Monday, January 30, 17
3/2
r = 0
r = R
du =
u
1/2
=
2u
3/2
du
1/2
1/2
z 2 +R2
1
) E =
z ( u 1/2 )
2✏0
z2
!
1
1
1
p
=
z
+
2
2
2✏0
z
z +R
"
#
z
p
E =
1
2✏0
z 2 + R2
23
VERY IMPORTANT LIMITING CASE:
VERY IMPORTANT LIMITING CASE
If R ⇤ z, that is if we have an infinite sheet of charge with charge density ⇥:
z , that is if we have an infinite sheet of charge with charge
If R
⇤
⌅
"⇥
z #
density
E =
1 ⇧z 2
2
2
z
+
R
0
p ⇥
E =
1
z2 + R2
2✏0 ⇥
z
⌅
1
"2 0
#R
'
2✏0
1
z
R
E ⇡
2✏0 ⇥
E⇥
20
E-field is normal to the charged surface
E -field is normal to charged surface Figure 2.2: E-field due to an infinite sheet of charge, charge density = ⇥
Monday, January 30, 17
24
1.5 Electric Field Lines
To visualize electric field
we can use a graphical tool called electric field lines
Conventions
1. Start on positive 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
proportional to number of E-field lines per unit area perpendicular to lines
Monday, January 30, 17
25
2.4. ELECTRIC FIELD LINES
Uniform E-field
Monday, January 30, 17
19
Non-uniform E-field
26
~P > E
~P
E
1
2
Monday, January 30, 17
~ =
E
+q
r̂
2
4⇡✏0 r
27
Infinite sheet of charge
E =
Monday, January 30, 17
2✏0
28
2.4. ELECTRIC FIELD LINES
Monday, January 30, 17
20
29
~ at point O = 0
E
Monday, January 30, 17
30
2.5
Point Charge in E-field
When we
place a chargein
q in E-field
an E-field E, the force experienced by the charge is
1.6 Point
Charge
~ma
E
= qE =
When we place a charge q in an EF-field
,force experienced by charge is
Applications:
~
~
F printer,
= q ETV=cathoderay
m~a tube.
Ink-jet
Applications
☛ Ink-jet printer, TV cathode ray tube
Example:
Example
Ink particle has mass m, charge q (q < 0 here)
Ink particle has mass m & charge q (q
< 0 here)
that mass
inkdrop is
is small,
what’s
thes deflection
y of the
Assume Assume
that mass
of of
inkdrop
small,
what’
deflection
of charge?
charge?
Solution:
Monday, January 30, 17
First, the charge carried by the inkdrop is negtive, i.e. q < 0.
31
st, the charge carried by the inkdrop is negtive, i.e. q < 0.
Solution ☛
Charge carried by inkdrop is negative ☛
q<0
~
~ points in qopposite
Note: Note:
direction
of E
qE
E points
in opposite
Horizontal motion ☛
Net force = 0
) L = vt
rizontal motion:
~
Vertical motion ☛ |q E|
Net force = 0
) Net force =
|q|E
)a=
m
|m~g |,
q is negative
|q|E = ma ☛ Newton’s 2nd Law
L = vt
1 2
Vertical distance travelled ☛ y =
at
2
Monday, January 30, 17
32
di
Review everything for next class BUT don’t forget
Monday, January 30, 17
33
Monday, January 30, 17
34
HOMEWORK
Vector
Review these slides
B4Algebra
watching superbowl
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)
Euclidean vector, a geometric entity endowed with magnitude and
direction as well as a positive-definite inner product; an element of a
Euclidean vector space!
In physics, Euclidean vectors are used to represent physical quantities
that have both magnitude and direction, such as force, in contrast to
scalar quantities, which have no direction
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(e.g. force, velocity, pressure)
A scalar consists of magnitude only.
(e.g. mass, charge, density)
1.2 Vector Algebra
1.2 Vector Algebra
~b +algebra
~aFigure
+ ~b1.1:=Vector
~a
~a + (~b + ~c) = (~a + ~b) + ~c
~a + ~b = ~b + ~a
~ = (~a + ~c) + d~
~a + (~c + d)
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1.3. COMPONENTS OF VECTORS
2
1.3 1.3
Components
of of
Vectors
Components
Vectors
1.3 Components of Vectors
Usually
vectors
are expressed
according
to coordinate
can
Usually
vectors
are expressed
according
to coordinatesystem.
system. Each
Each vector
vector can
be expressed
in terms
of components.
be expressed
in terms
of components.
Usually vectors are expressed according to coordinate system
The
most
common
system:
The most
common
coordinate
system:
Cartesian
Each
vector
cancoordinate
be expressed
in Cartesian
terms of components!
The most common coordinate system
Cartesian
~a = ax + ay + az
~a = ~a + ~a + ~a
~a = ~ax x+ ~ayy+ ~azz
Magnitude of
Magnitude of ~a = |~aq
| = a,
~a = |~a| = a
Magnitude of ~a = |~a| = a, 2
q=
a
a +
a =q a2x + a2y +xa2z
a = a2x + a2y + a2z
a2y + a2z
~a = ax + ay
q
a = a2x + a2y
~a = ~ax + ~ay
axq = a cos
2
a2x~a+
~a a== ~ax +
y ay
aqya cos¡;
= a sin
a
=
x
a =
a2x + a2yay = a sin¡
ay
tan
= aayy /a
tan¡
=
ax = a cos¡;
= x
a sin¡
ax
Figure 1.2: ¡ measured anti-clockwise tan¡ =
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January 30,
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from position
x-axis
Figure 1.2: ¡ measured anti-clockwise
from position x-axis
ay
ax
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Unit vectors have magnitude of 1
~a
â =
= unit vector along ~a direction
|~a|
ı̂
|ˆ
k̂
x
y
z
are unit vectors along
directions
~a = ax ı̂ + ay |ˆ + az k̂
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1.3. COMPONENTS OF VECTORS
1. Polar Coordinates
1. Polar Coordinate:
!
~a = ar r̂ + a✓ ✓ˆ
~a = ar r̂ + aµ µ̂
Figure 1.3: Polar Coordinates
2. Cylindrical
Monday, January 30, 17Coordinates:
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Figure 1.3: Polar Coordinates
2. Cylindrical
Cylindrical
Coordinates:Coordinates
!
~a = ar r̂ + aµ µ̂ + az ẑ
~a = ar r̂ + a✓ ✓ˆ + az ẑ
r̂ originated from
r̂ originated from nearest point on
nearest point
on z-axis
(Point
z-axis
(Point
O’)O’)!
igure 1.4: Cylindrical Coordinates
pherical Coordinates:
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Figure 1.4: Cylindrical Coordinates
3. Spherical Coordinates
3. Spherical Coordinates:
!
~a = ar r̂ +~aa=✓ a✓ˆr r̂++aaµ µ̂ˆ+ a¡ ¡ˆ
r̂ originated
from Originfrom
O
r̂ originated
Origin O
Figure 1.5: Spherical Coordinates
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1.4 Multiplication of Vectors
1. Scalar multiplication
If
~b = m ~a ~b, ~a are vectors; m is a scalar
then b = m a (Relation between magnitude)
bx = m a x
by = m a y
i.e.
}
~a = ax
Components also follow relation
ı̂ +
ay
m~a = max ı̂ + may
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|ˆ +
az
k̂
|ˆ + maz
k̂
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2. Dot Product (Scalar Product) Cont’d
ı̂ · ı̂ = |ı̂| |ı̂| cos 0 = 1 · 1 · 1 = 1
ı̂ · |ˆ = |ı̂| |ˆ
|| cos 90 = 1 · 1 · 0 = 0
ı̂ · ı̂ = |ˆ · |ˆ = k̂ · k̂ = 1
ı̂ · |ˆ = |ˆ · k̂ = k̂ · ı̂ = 0
If
then
~a = ax ı̂ + ay |ˆ + az k̂
~b = bx ı̂ + by |ˆ + bz k̂
~a · ~b = ax bx + ay by + az bz
~a · ~a = |~a| · |~a| cos 0 = a · a = a2
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1.4. MULTIPLICATION OF VECTORS
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3. Cross Product (Vector Product):
2. Cross Product (Vector Product)
~c = ~a ⇥ ~b
If
If
~c = ~a £ ~b,
then c = |~c| = a b sin¡
c| = a b sin
then c = |~
~a £ ~b 6= ~b £ ~a !!!
~a ⇥ ~b 6= ~b ⇥ ~a
~a £ ~b = °~b £ ~a
~a ⇥ ~b =
!!!
~b ⇥ ~a
Figure 1.7: Note: How angle ¡ is measured
• Direction of cross product determined from right hand rule
• Direction of cross product determined from right hand rule.
• Also, ~
a ⇥ ~b ~is ? to ~a ~and ~b
i.e.
• Also, ~a £ b is ? to ~a and b, i.e.
~a · (~a ⇥ ~b) = 0
~b · (~a ⇥ ~b) = 0
~a · (~a £ ~b) = 0
~b · (~a £ ~b) = 0
• IMPORTANT:
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~a £ ~a = a · a sin0± = 0
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• IMPORTANT
~a ⇥ ~a = a · a sin 0 = 0
±
£ î| = |ı̂|î|⇥|î|
sin0
ı̂| = |ı̂| |ı̂| sin=0 1=· 1
1 ·· 0
1 ·=
0 0
=
±
£ ĵ| = |î| |ĵ| sin90 = 1 · 1 · 1 = 1
0
|ı̂ ⇥ |ˆ| = |ı̂| |ˆ
|| sin 90 = 1 · 1 · 1 = 1
⇥ îı̂ == ĵ|ˆ£⇥ĵ|ˆ==k̂k̂£⇥k̂k̂==0 0
îı̂ £
î £ ĵ = k̂; ĵ £ k̂ = î; k̂ £ î = ĵ
ı̂ ⇥ |ˆ = k̂; |ˆ ⇥ k̂ = ı̂; k̂ ⇥ ı̂ = |ˆ
~a ⇥ ~b =
Ø î
Ø
Ø
= Ø ax
Ø
bx
ı̂
ax
bx
|ˆ k̂
ay az
Ø
b y Ø bz
= (ay bz
az by ) ı̂ + (az bx
ax bz ) |ˆ + (ax by
ĵ k̂
Ø
(ay bz ° az by ) î
ay az ØØ =
+(az bx ° ax bz ) ĵ
by bz
ˆ
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ay bx ) k̂
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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)
~ (x, y, z) is a mathematical function
A vector field F
which has a vector output for a position input
(Scalar field U(x, y, z) )
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1.6
Other Topics
1.6 Analytic Geometry
Tangential Vector
Tangential Vector
igure 1.8: d~l is a vector that is always tangential to the curve C with infinitesimal
ength dl
urface Vector
Surface Vector
Figure 1.8: d~l is a vector that is always tangential to the curve C with infinitesimal
length dl
Surface Vector
igure 1.9:
d~a uncertainty!
is a vector that
to the surface S with
Some
( d~ais always
versus perpendicular
d~a )
nfinitesimal area da
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Figure 1.9: d~a is a vector that is always perpendicular to the surface S with
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Two conventions:
Some uncertainty!
(d~a versus ° d~a)
• Area formed from a closed curve
Two Two
conventions:
conventions:
• •Area
formed
from afrom
closedacurve
Area
formed
closed
curve
Figure 1.10: Direction of d~a determined from right-hand rule
• Closed surface enclosing a volume
Figure 1.10:
Direction
of d~a determined
from right-hand rule
• Closed
surface
enclosing
a volume
• Closed surface enclosing a volume
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30, 1.11:
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Figure
Direction of d~a going from inside to outside
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