Summary of last class:
 Charge distributions:
(1) point
(2) Line
(3) Surface
(4) Volume
 Coordinate Systems:
(1) Cylindrical and
(2) Spherical
𝑄
𝜌
𝑑𝑣
z
P(r,, z)
x

y
 Electric Conduction Currents:
 Today’s Class
(1) Electric Field
(2) Electric Force
1
1
Electric field intensity 𝐸: a single charge q source
field
An isolated electric charge q produces
an electric vector field E at any point P
in space located at a distance R from
the charge. 𝐸 is given by
source
𝐸
𝑅
𝑅


(unit: volt/meter) with 𝑅
V/m
𝑅/R
ε is the electrical permittivity:
𝜀
𝜀 𝜀 (F/m)
 It is based on Coulomb’s experience
 The magnitude of E
• is proportional to the magnitude of q εo is the permittivity of free
• is inversely proportional to R2
space (air)
• acts along the unit vector
εr is the relative permittivity
connecting q () to the field point P
or dielectric constant
 Electric Field Contributions are directed away from
positive charges, & toward negative charges
2
2
Set #1 - part 2
1
Calculation of 𝑅
An important step in calculating the
electric field 𝐸 is to calculate the vector 𝑅
𝐸

𝑅

𝑅
Example: Given two points source @ P(1,2,3) and field @ Q (2,-2,1)
What is RPQ?
Source
General expression for 𝑅 calculation:
𝑅
𝑟⃗
𝑟⃗
Field
i.e., vector from P to () Q
𝑅
𝑟⃗
𝑟⃗
2, 2,1
2 1, 2 2,1 3
1,2,3
1, 4, 2
𝚤̂
4𝚥̂
2𝑘
3
3
Field due to multiple point sources: Superposition
For several Sources
(charges q1, q2, …), the
total filed is given by:
𝑬
𝑬𝟏
𝑬𝟐
Field (Force )
point
Source 1
point
⋯
Source 2 point

ε is the electrical permittivity: 𝜀
𝜀 𝜀
with:
Free space (air)
 For all examples and problems the medium is assumed
to be free space unless otherwise stated:
 Calculate the R vector for each charge,
 Apply Calculation of all R vectors:
4
4
Set #1 - part 2
2
Example – E due to multiple of charges
• Start with
• Write coordinates of all points
i.e. ,P , Q1, Q2, Q3, Q4
• Calculate all R vectors
𝑅
𝑟⃗
𝑟⃗
1,1,0
0,0,5
1, 1,5
𝚤̂ 𝚥̂ 5𝑘
𝑅
1 1 25
27 |𝑅|
Repeat for R2, R3 & R4
• Sub and calculate total E vector
𝐸
800/ 9 3
𝑘 kV/m
Note: from symmetry around the zaxis, x and y components cancel each 5
5
Electric flux density vector D:
Definition:
The flux density vector D is expressed in terms of Electric filed:


D   E (C/m2)
• D is independent of the medium
• It is a vector with units C/m2
• Electric flux begins on positive
charges and terminates on
negative charges
• Flux is in the same direction as E.
•
D
Material space with
Permittivity ε
ε is the electrical permittivity of the medium:
   r o
with:
and
is the permittivity of free space (air)
εr is the relative permittivity or dielectric constant
More later on εr
6
6
Set #1 - part 2
3
Examples of field lines (Flux)
Equal + charges
+ + + + + + + + + + + + + +
2-parallel plate
capacitor
7
             
7
Coulomb’s Law:
Electric force F on a test charge placed at a point P:
The force F (unit newtons N) exerted by q (source charge)
on q’ (test charge) is given by:
𝐹⃗
with k 
q’ at P
𝑞′𝑞
𝑞′𝑞
𝑘
𝑅 𝑘
𝑅
𝑅
𝑅
𝑞′𝐸 N
1
4

in air 𝑘
9 109
r
N  m2
C2
9
10 𝑁 ⋅ 𝑚 /𝐶


Note I: F   F
21
12
F
q’
R̂
q

F
R
Note II: The electric field E (at a given point in space) is the force per
unit charge that would be experienced by the test charge at that point. 8
8
Set #1 - part 2
4
Example:
Comparison of Strength of
Electromagnetic vs. Gravitational Force
FEM 
q1q2
4 0 r 2
Fg  G
1
m1m2
R122
mp=1.67x10-27 Kg
e =1.6x10-19 C
εo=8.84x10-12 C2/(Nm2)
G=6.67x10-11 (Nm2)/Kg2
FEM
 1.2 1036
Fg
Electromagnetism is much stronger!
9
9
Field/force due to multiple point sources: Superposition
For several Sources (charges Q1, Q2, …)
The force on a test charge q, located at R, by an
assembly of point charges Qi, is the vector sum
of the forces on q by the individual charges:
Field point (Force)
(x,y,z) q
𝐹⃗
𝐹⃗
…
𝑘
𝑘
𝑅
𝑄 𝑞 𝑅
𝑅
𝑅
Source 2
Q1
Source 1
𝑅
𝑅
𝑅
𝑅
(x1, y1, z1)
O
𝑄 𝑞 𝑅
(x2, y2, z2)
𝑅
𝑅
i.e., Consider force due
to 2 charges Q1 and Q2
on a 3rd charge q
𝐹⃗
Q2
⋯
in air
𝑘
9
10
𝑁⋅𝑚
𝐶
10
10
Set #1 - part 2
5
Example:
A charge q = 5 nC is at the origin. A charge Q1 = 2 nC is located
2 cm to the right on the x axis and and Q2 = -3 nC is located 4 cm
to the right on the x axis. What is the net force on q?
4cm
Steps:
 Geometry
 Magnitudes
 Directions
 Net Force
2cm
q = 5 nC
F1
x
Q1 = 2 nC
F2
Q2 = - 3 nC

Qq
F  k 2 Rˆ
R
• If the distance between Q2 and q is doubled,
what will be the change in F1 , F2 ?
• If the distance between Q1 and q is halved,
what will be the change in F1 , F2 ?
11
11
Example:
As shown, given 3
charges
q1 = 2 x 10-7 C @
q2 = -2 x 10-7 C @
q3 = 4 x 10-7 C @
Ä =4m
h=2m
Calculate force on q3
force at
+
Step 1 position vectors:
R1 = (-4, 0, 0) m
R2 = (4, 0, 0) m +
R3 = (0, 2, 0) m
R1
Step 2
distance vectors:
𝑟⃗
𝑟⃗
𝑟⃗
𝑟⃗
F23
F3
-
Origin
𝑟⃗
𝑟⃗
4𝚤̂ 2𝚥̂
4𝚤̂ 2𝚥̂
Step 3 individual force calculation:

qq
F13  k 1 2 3 rˆ13  32 .2iˆ  16 .1 ˆj N
r
13

qq
F23  k 2 2 3 rˆ23  32 .2iˆ  16 .1 ˆj  N
r
23
Finally add forces:
 

 F3  F13  F23  64.4 î  N
Note: from symmetry around the y-axis,
y components cancel each others. 12
12
Set #1 - part 2
6
Electric Field From Continuous
Distributions of Charge
Up to now we have only considered the electric field of point charges.
Now let’s look at continuous distributions of charge:
 lines
 surfaces
 volumes of charge and determine the resulting electric fields.
Surface distribution (Sheet)
Line distribution
(Ring)
L [C/m]
s [C/m2]
Volume distribution
(Sphere)
v [C/m3] 13
13
Electric Field Produced by a
Continuous Distribution of Charge
When charges are distributed within a region, then a summation or
integration must be carried out to calculate the total electric field.
Divide the distribution into small pieces,
then we sum the contribution:


For each dq: d E i ( r ) 
Then:
dq i
Rˆ '
4  R ' 2
R'
total E =  dEi
In the limit of very small pieces,
the sum is an integral
Rˆ ' dq
V ' 4R ' 2
 
E (R )  
Field dEi
 

E(r)   dE
  l dl

With dq    s dS
  dV
 v
dqi
Source
line distribution
surface distribution
volume distribution
14
14
Set #1 - part 2
7
Electric Field Due to Charge Distributions
For Continuous Distribution of Charges, we substitute the expression
of dq for each of the following distributions:
 Volume Charge Distribution
E
1
4𝜋𝜀
E
𝜌 𝑑𝑣′
𝑅′
𝑅′
Field
R' point
 Surface Charge Distribution
E
1
4𝜋𝜀
𝑅′
𝜌 𝑑𝑠′
𝑅′
ds'
 Line Charge Distribution
E
1
4𝜋𝜀
𝑅′
𝜌 𝑑𝑙′
𝑅′
15
15
Another reminder Review:
Cylindrical Coordinate System
Notation:
16
16
Set #1 - part 2
8
Another reminder Review: Cylindrical Coordinate System
17
17
Set #1 - part 2
9