L03 IntScalar Post
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Transcript of L03 IntScalar Post
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7/30/2019 L03 IntScalar Post
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LECTURE 3 slide 1
Lecture 3
Total Charge:
Line, Surface and Volume Integrals
Sections: 1.8, 1.9, 2.3
Homework: D2.4; 1.23, 1.27, 2.14, 2.16
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LECTURE 3 slide 2
Line Elements 1
metric increment due to a differential change in position along a line
a
bdl
xdxa
ydya
x yd ddx y= +l aa
a
d
ad a
d a
d dd = +l aa
dlxa
ya
a b
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LECTURE 3 slide 3
Line Elements 2
line increment is a vector has direction: point of integration
moves from point a to point b
each of its components is a linear increment (in meters)
RCS: dx, dy, and dzare linear by default
CCS: d and dzare linear, d is not
SCS: dris linear, dand d are not
angular increment d corresponds to linear increment d
zd d d dz = + +l a a a
x y zd dx dy dz = + +l a a a
sinrd dr rd r d = + +l a a asin
d rdd r d
6
6
d d 6
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LECTURE 3 slide 4
Line Elements 3
SCS:
ra
a
a
dr
drrd
dl
sinr
sinr d
d
x
y
zsinrd dr rd r d = + +l a a a
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LECTURE 3 slide 5
the direction of integration does
not matter: charge is scalar
Line Integration: Charge on Lines 1
B
lQ dl=
straight line: choose RCS
axis along charged line
EASY SPECIAL CASES: CHARGE ON PRINCIPAL GRID LINES
( )Bx
l
x
Q x dx=
-line in CCS: circularcharges in thex-0-yplane 0( )
B
A
lQ d
=
-line and -line in SCS
0( )B
lQ r d
= 0 0( ) sinB
A
lQ r d
=
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LECTURE 3 slide 6
Line Integration: Charge on Lines 2
GENERAL CASE curved line: needs line equation in parametric form( ) ( ) ( ) ( )x y zu x u y u z u= + +r a a a x y zd d dx dy dz = = + +l r a a a
x
y
z
B
( )ur dr
2 2 2 2dl dx dy dz = + +2 2 2 2
dl dx dy dz
du du du du
= + +
2dl d d
du du du
=
r r dl d d
du du du
=
r r d ddl du
du du =
r r
( )Bu
l
u
d dQ u du
du du=
r r
B
A
u
AB
u
d dL du
du du=
r r
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LECTURE 3 slide 7
The surface element is defined by two line elements:
1 2d d d= s l l
surface elements in the RCS principal planes
plane:
plane:
plane:
z
y
x
z const d dxdy
y const d dxdz
x const d dydz
= = = =
= =
s a
s a
s a
Surface Elements 1
1dl2dlds
0
z
y
x
a
/ 2z a=
/ 2z a=
/ 2x a=
/ 2x a= / 2
ya
=/ 2
ya
=
dy
dz
dy
dx
dzdx
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LECTURE 3 slide 8
surface elements in the CCS principal planes
plane:
plane:
plane: z
const d d dz
const d d dz
z const d d d
= =
= =
= =
s a
s a
s a
Surface Elements 2
z
a
dz
d
a
dd
za
const=
z const=
const
=
d
dz a
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LECTURE 3 slide 9
surface elements in the SCS principal planes
Surface Elements 3
2
surface (sphere):
sin r
r const
d r d d
==s a
surface (cone):sin
constd r drd
==s a
surface (circular plane):constd rdrd ==s a
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LECTURE 3 slide 10
Surface Integration: Charge on Surfaces 1
We will limit ourselves to SPECIAL CASES
charges on principal coordinate planes
charge on a plane2 2
1 1
( , )
y x
s
y x
Q x y dxdy=
charge on a circular disk or ringx
y
1x 2x
1y
2y
2 2
1 1
( , )sQ d d
=
s
S
Q ds=
1
21
2
x
y
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LECTURE 3 slide 11
Surface Integration: Charge on Surfaces 2
charge on a cylinder
charge on a sphere
22
1 1
0( , )
z
s
z
Q z d dz
= 1z
2z1 2
0
2 2
1 1
20( , ) sinsQ r d d
= 0
x
y
z
0r
12
1
2
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LECTURE 3 slide 12
Surface Integration: Charge on Surfaces 3
charge on a cone2 2
1 1
0( , ) sin
r
s
r
Q r r drd
= 1 2
1r
2r
0
NOTE: If you sets = 1 in the above formulas, you can compute
the area of the respective surfaces.
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LECTURE 3 slide 13
1 2 3( )dv d d d = l l l
Volume Elements 1
The volume element is defined by three line elements
RCS: dv dxdydz= 2dl1
dl3dl
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LECTURE 3 slide 14
CCS: dv d d dz =
Volume Elements 2
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LECTURE 3 slide 15
2SCS: sindv r drd d =
Volume Elements 3
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LECTURE 3 slide 16
v
V
Q dv=
Volume Integration: Volume Charges
parallelogram22 2
1 1 1
( , , )
yz x
v
z y x
Q x y z dxdydz =
cylindrical volume
1 2 2
1 1 1
( , , )
z
v
z
Q z d d dz
=
spherical volume
2 2 2
1 1 1
2( , , ) sin
r
v
r
Q r r drd d
=
NOTE: You can find the volume of the element by settingv = 1.
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LECTURE 3 slide 17
Volume Integration: Example
Work in SCS. Assume thez-axis is along the axis of thesymmetrical conical light beam. The sine of half the subtended
angle of the beam is
2
2
0 0 0
sin
a
V r
V dv r drd d
= = =
= = 3 3
352 (1 cos ) 2 0.0202 0.21, m3 3
aV = = =
A light source shines onto a hemispherical dome of radius a = 5 m,and makes a round spot 2 m in diameter, d= 2 m. What is the
volume of the light cone from the light source to the dome?
/ 2sin 0.2
d
a = =2cos 1 sin 0.9798 = =
Homework: Find the area of the dome lit up by the beam. 23.173, mA =
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LECTURE 3 slide 18
You have learned how to
find the total charge along a straight line or any other curved line
find the total charge on any portion of the surface of a plane, disk,
cylinder, sphere, cone
find the total charge on any portion of a parallelogram, cylinder,
sphere, cone
use integration to find length, area and volume