The document provides examples of calculating electric scalar potential using the superposition principle. In the first example, the potential is calculated at a point due to two equal but opposite point charges. In the second example, the potential is found at a point due to a line charge with uniform density. The third example finds the potential at the center of a square due to four point charges, both when all charges have the same polarity and when opposite charges are adjacent. Further examples include calculating potential due to a square loop of line charge and circular ring and disc with uniform charge density. The last example determines the potential and electric field at two points due to a dipole.

1. Series: EMF Theory
Lecture: #1.33
Dr R S Rao
Professor, ECE
ELECTROSTATICS
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Electric scalar potential for a source distribution using superposition principle.

2. Electrostatics
Electric
Scalar
Potential
2
Example I
Two point charges of opposite polarity, each with charge equal to q C is separated by a
distance d m. Assuming medium has a permittivity of ε, find the potential at a point which is
at a distance of h m from the center of the line joining the two charges.
Solution:
The potential obeys superposition principle. The total potential at an arbitrary point due to
two point charges is equal to sum of the potentials
at that point due to individual point charges.
Figure Electric potential of two equal but
opposite polarity point charges.
Given that the charges −q and q are separated by a
distance of d and the observation point P is at a
distance of h above the charge distribution.
Arrange them as shown in Figure. Now, the
potentials due to −q and q are,
 
 
 
 
2 2
2 2
1 1
V & V
4 4
2 2
q q
V P V P
h d h d
 
 

 
 
Then, the potential due to both the charges would be
   
2 2
2 2
1 1
( ) ( ) ( ) 0 V
4 4
2 2
q q
V P V P V P
h d h d
 
 

    
 

3. Electrostatics
Electric
Scalar
Potential
3
Example II
A straight line charge of 2Lm long is carrying a charge with a uniform density of λC/m.
Assuming a medium permittivity of ε, find potential at point P which is at a distance of h m
from the center of the line charge.
Solution:
Place the given line charge along x-axis with its center right over the origin and extending from
–L to L. Now, the field point is on the z-axis at a height of h. Identify a differential length dx,
carrying a charge λdx over the line charge, at a distance of x from the origin. The potential at the
observation point is P can be found as
2 2 2 2
0
1 λ 2 λ
( )
4 4
L L
L
dx dx
V P
h x h x
 

 
 
 
 
2 2
2 2
0
λ λ
ln ln V
2 2
L
L h L
x h x
h
 
 
 
     
 
 

4. Electrostatics
Electric
Scalar
Potential
4
Example III
Assuming a medium permittivity of ε, find the potential at the center of a a m side square, due
to four point charges at its corners, each with q C, when (a) all charges are positive and
(b)adjacent charges are of opposite polarity,
Solution:
Both the charge distributions are illustrated in Figure. It can be noticed that all four charges are
at the same distance from the observation point P.
Each point charge q C gives rise to potential given by
1
( ) V
4 / 2
q
V P
a


In case (a), potentials due to all the four charges get added resulting in a total potential of
1 1 2
( ) 4 V
4 / 2
tot
q q
V P
a
a
 
  
In case (b), potential due to two positive charges gets cancelled by potentials due to two
negative charges, resulting in nil total potential.

5. Electrostatics
Electric
Scalar
Potential
5
Example IV
A line charge in the form of square loop with a side length of L m, is carrying a uniform charge
with density, λ C/m. Given permittivity of medium is ε, (a) find the potential at point, P which
is center of the loop. (b)Also find potential at P if adjacent sides carry opposite polarity charge.
Solution:
All the four charges are at the same distance from the observation point P. Each side gives rise
to a potential,
2 2
2 ( 2) ( 2)
λ
( ) ln
2 ( 2)
L L L
V P
L

 
 
 

 
 
λ 0.44λ
ln(1 2) V
2 
  
In case (a), potentials due to all the four charges get added resulting in a total potential of
λ
( ) 4 ln(1 2)
2
tot
V P

  
2λ 1.76λ
ln(1 2) V
 
  
In case (b), potentials due to two positive charged sides gets cancelled by potentials due to two
negative charged sides, resulting in nil total potential.

6. Electrostatics
Electric
Scalar
Potential
6
Example V
Assuming the medium permittivity as ε, find the potential at a point which is at a distance of h
m from the center of (a) a circular ring with radius a m carrying a uniform charge with a density
of λC/m and (b) a charged circular disc with radius a m carrying a uniform surface charge with
a density of σ C/m2
. Extend the result to find the potential at their centers.
Solution:
Given the observation point, P is at a distance of h above the charge distributions, each with a
radius of a m. Place them over xy-plane with their centers over the origin
(a). The potential V at point P, using expression for the differential charge, dQ = λad is
2 2
2 2 2 2 2 2
0 0
1 λ 1 λ λ
( )
4 4 2
ad ad a
V P
a h a h a h
 
 
  
  
  
 
The ring center is h=0, and hence, potential there is,
0 2 2
0
λ λ
2
2
h
h
a
V
a h 



 


7. Electrostatics
Electric
Scalar
Potential
7
Example V
(b).To find the potential V at point P, using the expression for the differential charge, dQ =
σρdρd ,
2
2 2 2 2
0 0 0
1 σ 1 σ
( )
4 2
a a
d d d
V P
h h

    
 
 
 
 
  
2 2
0
σ
2
a
h


   
2 2
σ
2
a h h

  
At the disc center, h=0, and hence, potential there becomes,
 
2 2
0
0
σ σ
2 2
h
h
a
V a h h
 


   

8. Electrostatics
Electric
Scalar
Potential
8
Example VI
At points, (a)(3,4,5)m and (b)(2m,π/4,3m), determine the potential, V and field, E due to a
dipole, with +3mC at z=‒1mm and ‒3mC at z=+1mm.
Solution:
The given dipole is shown in Figure.
3 3 6 6
ˆ ˆ
3 10 2 10 ( 6 10 6 10 C-m
q qd
   
            
p d z) z
(a). At point (3,4,5)m: Radial distance to and polar angle of field point, respectively, are
2 2 2 1 2
(3 4 5 ) 7.07m
r    
= 1 2 2 1 2
1
5
tan (3 4 ) 0.785rad

 
 
 
Figure Dipole and field point.
The potential, V and field, E, respectively, are
6
2 2 9 2
ˆ cos 36 6 10 0.707
763.79 V
4 4 4 10 7.07
o o
qd
V
r r
 
  


   
    
 
p r
 
3
ˆ
ˆ
2cos sin
4 o
qd
r
 

 
r
 
 
6
9 3
36 6 10 ˆ
ˆ
2 0.707 0.707
4 10 7.07




  
  
 
r   
ˆ
ˆ
152.80 1.414 0.707 V/m
  
r 
1 2 2
tan x y z
  
 
 
 

9. Electrostatics
Electric
Scalar
Potential
9
Example VI
(b). At point (2m,π/4,3m): Radial distance to and polar angle of field point, respectively, are,
2 2
2 3 3.60m
   & 1 1
tan ( ) tan (2 3) 0.588rad
z
 
 
  
The potential, V and field, E, respectively, are
6
2 2 9 2
ˆ cos 36 6 10 0.832
3.47 kV
4 4 4 10 3.60
o o
qd
V
r r
 
  


   
    
 
p r
6
3 9 3
36 6 10
ˆ ˆ
ˆ ˆ
(2cos sin ) (2 0.832 0.555 )
4 4 10 3.60
o
qd
r

 
 


  
    
 
r r
  
ˆ
ˆ
1.157(1.664 0.555 )kV/m
  
r 
2 2
r z


 

10. ENOUGH
FOR
TODAY
ENOUGH
FOR
TODAY
ENOUGH
FOR
TODAY
ENOUGH
FOR
TODAY
ENOUGH
FOR
TODAY
10