EE340 Electromagnetics
Lecture #2
Electrostatic Fields
Coulomb’s Law:
F12 = Force on Q2 due to Q1
 Q1Q2 
R

a , where aR12  12 , R12 = Distance vector pointing from Q1 to Q2
2  R12
R12
 4  R12 
 = Permittivity of the material, in free Space o 
109
F
 8.54 1012  
36
M 
Notice that F21 = force on Q1 due to Q2 =  F12
The force F12 can expressed as follows,
 Q1Q2  R12  Q1Q2 
F12  

R
2 
3  12
 4  R12  R12  4  R12 
But since R12  r2  r1 , that is writing the distance vector in terms of position vectors r1 and r2 , then:
F12 


4  r2  r1
3
Q1Q2 r2  r1
Which gives the force in terms of the charges and the position vectors of the charges.
Notice that if:
1.
Q1 & Q2 have same sign  F12 // aR12 , F21 // - aR12  Q1 & Q2 repel each other.
2.
Q1 & Q2 have opposite sign  F12 // - aR12 , F21 // aR12  Q1 & Q2 attract each other.
Updated 05 Sep2019
Page 1/6
EE340 Notes: Dr. Mohammed Khan
Example:
Find the force on Q2  1n C  at P2 (1,1,0) due to charge Q1  3n C  at P1 (-2,3,0), if the two charges
exist in free space?
Solution:
r2  ax  a y  0az
r1  2ax  3a y  0az
 R12  r2  r1  3ax  2a y
Hence,
3 10
QQ R
F12  1 2 123 
4 o R12

27 109
 13 
3
9
 110  3a
4
9
9
10
36

94

x
 2a y

3
3ax  2a y 


 1.73 109 ax  1.15 109 a y  N 
Representation of the force F12 :
Updated 05 Sep2019
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EE340 Notes: Dr. Mohammed Khan
Electric Field Intensity
The Electric field intensity E is defined as the force on charge divided by the charge (i.e. force per unit
charge) E 
F
Q
V 
 m 
Consider now two charges q (a test charge) and Q1 as shown below:
The force F on q due to Q1 :
F
qQ1 R
4  R3
The electric field intensity E at point P, which is the observation point (or at the position of q) due to Q1
is therefore:
E


Q1 r  r1
qQ1 R
F


q 4 q  R3 4  r  r 3
1
Where r is the position vector of the observation point P
r1 is the position vector of Q1
Hence, the above equation gives the electric field intensity E at the observation point P due to the point
charge Q1 .
Updated 05 Sep2019
Page 3/6
EE340 Notes: Dr. Mohammed Khan
Example:
Calculate the Electric field intensity at (-1,1,0) due to Q1  5 nC  located at (1,0,0). Assume free space.
Solution:
E


Q1 r  r1


4  r  r1
3
  
5 109  ax  a y  ax 



3
 109 
4 
  4  1 
 36 
45 
 2ax  a y 
3 

5

 

 8.05ax  4.025a y
V 
 m 
Example 3:
Derive an expression for E due to a point charge Q located at the origin
Solution:
E

Q1 r  0

 Q 

a
2  r
4


r


4  r  0
3
Updated 05 Sep2019
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EE340 Notes: Dr. Mohammed Khan
The electric field of a point charge therefore is as shown below:
Note the following:
1.
E lines are radial with respect to Q
2.
E lines enter ( Q <0) and leave ( Q >0)
3.
E perpendicular to spherical surfaces centered at Q
4.
E is equal on any spherical surface centered at Q
Electric field due to a number of point charges:
Using the “superposition principle”
E  E1  E2  E 3  ...  E N


   Q  R   ......  Q  R 
Q1 R1P
2
4  R

3
1P
Q1 r  r1

4  r  r1
Or: E 
2P
4  R
N
3
2P
3


Q2 r  r2

Qk r  rk
k 1
4  r  rk
Updated 05 Sep2019
3
 ...... 

QN r  rN

4  r  rN
3

N

3
4  RNP

4  r  r2
NP
3
Page 5/6
EE340 Notes: Dr. Mohammed Khan
Example:
Point charges 5 nC  and 2  nC  are located at (2,0,4) and (-3,0,5) in free space. Determine the E at
(1, -3,7).
Solution:
E


5 109 ax  3a y  3az
 10
4 
 36
9
45
 
19
3
 a
x
3

  19 

  2 10  4a
9
 10
4 
 36
9

 3a y  3az 

18
29

3
 4a
x
x
 3a y  2az

3

  29 

 3a y  2az

 0.54ax  1.630a y  1.630az  0.461ax  0.346a y  0.231az
 1.004ax  1.284a y  1.399az
Updated 05 Sep2019
V / m
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EE340 Notes: Dr. Mohammed Khan