Total Charge: Line, Surface and Volume Integrals

Lecture 4
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
LECTURE 4 slide 1

Line Elements – 1
metric increment due to the differential increment of a coordinate
a y
dya
a
y
dl b
dxa
dl = dxa + dya
x
ax
x y
dφ
aρ
a dρa φ ρ
a b
dl
ρdφaφ dl = dρa ρ + ρdφaφ
LECTURE 4 slide 2

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 dz are linear by default
CCS: dρ and dz are linear, dφ is not
SCS: dr is linear, dθ and dφ are not
example: angular increment dφ corresponds to linear
increment ρdφ
LECTURE 4 slide 3

CCS:
dl = dρa + ρdφa + dza
ρ φ
SCS:
dl = dra + rdθa + rsinθdφa r
θ φ
x
Line Elements – 3
z
z
θ
dθ
φ dφ
r sinθ
dl
rdθ
dr
a
r
ar
a
LECTURE 4 slide 4
θ
φ
rsinθdφ y

Line Integration: Charge on Lines – 1
B
Q= ∫ ρldl
A
the direction of integration does
not matter: charge is scalar
SPECIAL CASES: CHARGE ON COORDINATE GRID LINES
straight line: choose RCS xB
axis along charged line Q= ∫ ρl
( x) dx
φ-line in CCS: circular
φB
charges in the x-0-y plane Q= ∫ l ( ) ⋅ 0d
θ-line and φ-line in SCS
x
φ
A
A
ρ φ ρ φ
θB
φB
∫ ρl ( θ) 0 θ = ∫
l ( ) ⋅ 0sin 0
θ
φ
Q= ⋅r
d
Q ρ φ r θ dφ
A
A
LECTURE 4 slide 5

x
Line Integration: Charge on Lines – 2
GENERAL CASE curved line: need line equation in parametric form
r( u) = x( u) ax+ y( u) ay + z( u)
az ⇒ dl= dr = dxax+ dyay + dzaz
2 2 2 2
⇒ dl = dx + dy + dz
dl
⇒
⎛ ⎞
⎜ ⎟
⎝du ⎠
dx
=
⎛ ⎞
⎜ ⎟
⎝du ⎠
⎛ dy
+
⎞
⎜ ⎟
⎝du ⎠
⎛ dz
+
⎞
⎜ ⎟
⎝du ⎠
2
⎛ dl ⎞ dr dr
⇒ ⎜ ⎟ = ⋅
⎝du ⎠ du du
z
dl
⇒
⎛ ⎞
⎜ ⎟=
⎝du ⎠
dr dr
⋅
du du
⇒ dl =
dr dr
⋅ du
du du
u
B
dr dr
LAB = ∫ ⋅ du
du du
uA
r(
u)
A
dr
y
B
2 2 2 2
B
dr dr
Q= ∫ ρl
( u) ⋅ du
du du
LECTURE 4 slide 6
u
u
A

Surface Elements – 1
The surface element is defined by two line elements:
ds= dl × dl
1 2
surface elements in the RCS principal planes
z = const plane: ds=± dxdya
y = const plane: ds=± dxdza
x = const plane: ds=± dydza
dl2 ds
z
y
x
dz
dl
dx
LECTURE 4 slide 7
1
x = a /2
dy
y = −a
/ 2
dy
z
dz
0 y dx
x
2 /
x = − a /2
z = −a
/2
a
z = a
/2
y = a

Surface Elements – 2
surface elements in the CCS principal planes
ρ = const plane: ds=± ρdφdza φ = const plane: ds=± dρdzaφ z = const plane: ds=± ρdρdφaz ρ
dz
LECTURE 4 slide 8
z
z = consta
z
a d ρ
ρdφ aφ
dz aρ
ρdφ dρ
const
φ =
ρ =
const

Surface Elements – 3
surface elements in the SCS principal planes
r = const surface (sphere):
2 ds= r sinθdθdφar θ = const surface (cone):
ds= rsinθdrdφaθ φ = const surface (circular plane):
ds= rdrdθaφ LECTURE 4 slide 9

Surface Integration: Charge on Surfaces – 1
We will limit ourselves to SPECIAL CASES
charges on principal coordinate planes
charge on a plane
y x
2 2
Q= ∫∫ρs(
x, y) dxdy
y x
1 1
charge on a circular disk or ring
ρ φ
2 2
Q= ∫∫ρs(
ρφρ , ) dφdρ ρ φ
1 1
y2
y1
LECTURE 4 slide 10
y
Q= ∫∫
ρsds
x
ρ1
S
x1 x2
φ2
ρ2 1 φ

x
Surface Integration: Charge on Surfaces – 2
charge on a cylinder
z
2
2
Q= ∫∫ρs
(, z φρ ) 0dφdz
z
φ
φ
1 1
charge on a sphere
z
0
θ1
r
0
θ2
φ1
φ2
y
φ θ
2 2
1 1
ρ0
2
0
LECTURE 4 slide 11
z2 1 φ φ2
z1
Q= ∫∫ρs
( θφ , ) r sinθdθdφ
φ θ

Surface Integration: Charge on Surfaces – 3
charge on a cone
φ2
r2
= ∫∫ρs (, φ)sinθ0 φ r
φ
r2
1
Q r r drd
1 1
NOTE: If you set ρ s = 1 in the above formulas, you can compute
the area of the respective surfaces.
φ
r1
θ0
LECTURE 4 slide 12
φ2

Volume Elements – 1
The volume element is defined by three line elements
RCS: dv = dxdydz
dv = ( dl × dl ) ⋅dl
1 2 3
dl2
dl 3 dl
LECTURE 4 slide 13
1

CCS: dv = ρd ρdφdz Volume Elements – 2
LECTURE 4 slide 14

2
SCS: dv = r sinθ
drdθdφ Volume Elements – 3
LECTURE 4 slide 15

parallelogram
Volume Integration: Volume Charges
z y x
2 2 2
Q = ∫∫∫ρv(
x, y, z) dxdydz
z y x
1 1 1
cylindrical volume
z
1 2 2
Q= ∫∫∫ρ
v ( ρφ , , z) ρdρdφdz z
φ ρ
φ ρ
1 1 1
spherical volume
φ θ
r
2 2 2
Q= ∫∫∫ρv(,
r θφ , ) r sinθdrdθdφ
φ θ
r
1 1 1
2
Q= ∫∫∫ ρvdv
NOTE: You can find the volume of the element by setting ρ v = 1.
LECTURE 4 slide 16
V

Volume Integration: Example
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?
Work in SCS. Assume the z-axis is along the axis of the
symmetrical conical light beam. The sine of the half the
subtended angle of the beam is
d /2
sinα = = 0.2
2
⇒ cosα = 1− sin α =
0.9798
a
2π
α a
V = ∫∫∫ dv= ∫ ∫ ∫
2 r sin drd d
V φ= 0θ= 0r= 0
θ θ φ
3 3
a
5
V = × 2 π× (1− cos α) = × 2π× 0.0202 = 0.21, m
3 3
LECTURE 4 slide 17
3

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
LECTURE 4 slide 18