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EEE 23
DC 1-3: Electric Fields II
Continuous Charge Distributions
Streamlines

1

Electric Field Computations
Coulomb’s Law
For continuous
charge distributions


E

dQ
  2
40 rP  rQ

1

Define the coordinates of a generic point
charge P where you want to evaluate E

P(x, y, z)

2

Define the coordinates of a generic point
charge Q on the charge distribution

Q(x’, y’, z’)

3

Divide the charge distribution into
infinitesimal point charges*

4
5
6

Find the distance from Q to P

 
rP  rQ
 
rP  rQ

dQ can be ρV dV, ρS dS, or ρL dL
|P – Q|

Find the unit vector from Q to P

__P – Q__
|P – Q|

Setup the integral to “add” the
contributions of all infinitesimal charges

Integrate!
2

*The form of dQ will depend on the coordinate system used in defining point Q
...


Streamline Sketches

(Left) http://numbercrunch
...
no/2010/09/25/using-mayavi-to-visualize-electric-fields/

4

Problem 1
Circular Ring of Charge
a
...

b
...
5 nC/m
...


Derive the electric field at any point on the z-axis due to a flat
circular disc of charge with uniform surface charge density ρs
lying on the x-y plane centered at the origin
...


b
...

c
...

6

Problem 3 Streamlines
Given the field E = 15x2 y ax +5x3ay , find:
a
...
a unit vector aE specifying the
direction of E at P
...
The
semi-circular ring has a uniform charge

density of ρL C/m
...

8

Homework 1–2
Streamlines of a Sheet Charge
Sketch the E-lines in the Cartesian
coordinate system given a uniform sheet of

charge with ρs =10 nC/m2 on the x-z plane
and a -500 pC point charge at point (0, 2, 3)
m
...

9

Homework 1–3
Flat Circular Discs of Charge
• A flat holed disc on the x-y plane and with center at the origin
has a total charge of 20 nC
...
5 m
...


• A flat disc on the x-y plane with its center at the origin carries

both uniformly distributed positive and negative charges
...
5 (in meters) is 10nC
...
5< r <1 (in

meters) is 30nC
...
Derive the electric field at any point on the z-axis
due to circular ring of charge lying on the x-y
plane of radius a centered at the origin with a
uniform charge density ρL
...
Then, find E and F experienced by a -2nC charge at
P (0,0,4) m if a = 2m and ρL = 3
...


12

Problem 1
Circular Ring of Charge
We can define the needed expressions as follows:

1

Define the coordinates of a generic point
charge P where you want to evaluate E

P(0,0, z )

2

Define the coordinates of a generic point
charge Q on the charge distribution

Q(a , ,0)

3

Divide the charge distribution into
infinitesimal point charges

z
P(0,0, z)

dQ   Lad
y
x

Q(a, φ, 0)

13

Problem 1
Circular Ring of Charge
4
5

Find the distance from Q to P

Find the unit vector from Q to P*

r  a2  z 2
  a cos ,a sin  , z
r
a2  z 2

xyz

z
P(0,0, z)

y
x

Q(a, φ, 0)

14

Problem 1
Circular Ring of Charge
Setup the integral to “add” the contributions
of all infinitesimal charges

6

dE 


dQ
ˆ
aPQ
2
4 0rPQ

 Lad  a cos ,a sin  , z

4 0
a2  z 2 3/2



E

xyz

 Lad  a cos
 ad  a sin 
 ad
z
ˆ
ˆ
ˆ
ax  L
ay  L
az
2
2 3/ 2
2
2 3/ 2
2
2 3/ 2
4 0   z 
4 0   z 
4 0   z 
a
a
a

2

2
2
 Lad  a cos
 Lad  a sin 
 ad
z
ˆ
ˆ
ˆ
ax  
ay   L
0 4 0 a2  z 2 3/2  0 4 0 a2  z 2 3/2  0 4 0 a2  z 2 3/2 az


15

Problem 1
Circular Ring of Charge
Evaluating the integral, we have

E



2

 Lad  a cos 0 2  Lad  a sin  0 2  Lad
z
ˆ
ˆ
ˆ
ax  
ay  
0 4 0 a2  z 2 3/2  0 4 0 a2  z 2 3/2  0 4 0 a2  z 2 3/2 az




az L

2 0 a  z
2



2 3/ 2

ˆ
az V / m

16

Problem 1
Circular Ring of Charge
Solving for E and F at P(0,0,4) when Q = –2nC, a = 2m and
ρL = 3
...
5  109 a
ˆ




2 0 22  4



2 3/ 2

z

ˆ
 17
...
678az   2  109  35
...


Derive the electric field at any point on the z-axis due to a flat
circular disc of charge with uniform surface charge density ρs
lying on the x-y plane centered at the origin
...


b
...

c
...

18

Problem 2
Flat Circular Disk of Charge
We can define the needed expressions as follows:

1

Define the coordinates of a generic point
charge P where you want to evaluate E

P(0,0, z )

2

Define the coordinates of a generic point
charge Q on the charge distribution

Q(  , ,0)

3

Divide the charge distribution into
infinitesimal point charges

z
P(0,0, z)

dQ   S dd
Q(ρ,φ,z)

19

Problem 2
Flat Circular Disk of Charge
4
5

Find the distance from Q to P

Find the unit vector from Q to P*

r   2  z2

   cos ,  sin  , z
r
 2  z2

xyz

z
P(0,0, z)

Q(ρ,φ,z)

20

Problem 2
Flat Circular Disk of Charge
6

Setup the integral to “add” the contributions
of all infinitesimal charges


dE 


dQ
ˆ
aPQ
2
4 0rPQ



 S dd   cos ,  sin  , z
4 0
 2  z 2 3/2



 S dd   cos
 dd   sin 
 dd
z
ˆ
ˆ
ˆ
ax  S
ay  S
az
2
2 3/ 2
2
2 3/ 2
2
2 3/ 2
4 0   z 
4 0   z 
4 0   z 


E

2

b

 
 

0  a

  S  2 cosdd



4 0   z
2



2 3/ 2

ˆ
ax 

2

xyz

b

 
 

0  a

  S  2 sin dd



4 0   z
2



2 3/ 2

ˆ
ay 

2

b

 
 

0  a

 S zdd



4 0   z
2



2 3/ 2

ˆ
az

21

Problem 2
Flat Circular Disk of Charge
Evaluating the integral, we have

E

2

b

 
 

0  a



  S  2 cosdd



4 0   z
2



2 3/ 2

0

ˆ
ax 

0
2

b

 
 

0  a

  S  2 sin dd



4 0   z
2



2 3/ 2

ˆ
ay 

2

b

 
 

0  a

 S zdd



4 0   z
2



2 3/ 2

ˆ
az


z S 
1
1



2
2
2 0  a2  z 2
b  z 

EXERCISE: The x- and y-components will cancel out due to symmetry
...


22

Problem 2
Flat Circular Disk of Charge
As we take the limit as b approaches infinity, we get E for an infinite
sheet with a circular hole with radius a as follows,


z S 
S z
1
1
lim b E  lim b




2 0  a2  z 2
b2  z 2  2 0 a2  z 2 

As we take the limit as a approaches zero of the above expression,
we get E for an infinite sheet as follows,

lim a0

S z
2 0 a2  z 2



S
2 0 

23

Problem 3 Streamlines
Given the field E = 15x2 y ax +5x3ay , find:
a
...
a unit vector aE specifying the
direction of E at P
...


25

Problem 3 Streamlines
You can solve for the unit vector at P by solving for E at P then
divide by the magnitude
...
976ax  0
...
523

Hence, the unit vector is given by

ˆ
ˆ
ˆ
ˆ
ˆ
aP  cosax  sin a y   0
...
217a y 

27



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