Solution of capacitive system using Laplace's equation

1. Course: Electromagnetic Theory
paper code: EI 503
Course Coordinator: Arpan Deyasi
Department of Electronics and Communication Engineering
RCC Institute of Information Technology
Kolkata, India
Topic: Electrostatics – Capacitor
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2. Types of Capacitor
[i] Parallel-plate capacitor
[ii] Cylindrical capacitor
[iii] Spherical capacitor
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3. Parallel-plate Capacitor
z=d
Laplace’s equation in Cartesian coordinate
2
2
0
d
dz

=
d
A
dz

=
Az B
 = +
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4. Parallel-plate Capacitor
z=d
1
0,
at z =  = 
2
,
at z d
=  = 
2 1
1
z
d
 −
 = + 
Electrostatic potential is obtained as
Boundary conditions are
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z
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5. Parallel-plate Capacitor
z=d
d
E
dz

= −
Electric field
1 2 12
E
d d
 − 
= =
Substituting the value of potential
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Theory

6. Parallel-plate Capacitor
z=d
Capacitance of the system
12
q
C =

E
C
Ed

= C
d

=
Let, σ be the uniform surface charge density
q
E

 
= =
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7. Problem 1
Distance between two plates of a parallel plate capacitor is ‘d’ having permittivity ε1. A
dielectric slab of thickness ‘x’ is introduced in the gap with permittivity ε2. Determine the
condition when capacitance of the system will be doubled.
Soln
Capacitance of a parallel plate capacitor of the undisturbed system
1 0
C
d
  
=
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Theory

8. Then capacitance per unit area
1 0 0
1
1
( / )
C
d d
  

= =
After insertion of dielectric with permittivity ε2
new capacitance per unit area
0
2
1 2
C
d x x

 
=
   
−
+
   
   
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Theory

9. According to the condition
0 0
1 2 1
2
d x x d
 
  
=
     
−
+
     
     
1 1 2
2
d d x x
  
 
     
−
= +
 
     
     
 
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10. 2 2 1
1 1 2
2
d x x
d   
  
− +
=
2
1 2 1 2 1 2 1
2 2 2
d d x x
      
= − +
2
1 2 1 1 2
2 2
x x d
    
− =
1
2
2 1
d x


 
= −
 
 
1
2
2 2
d x x


= −
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11. Cylindrical Capacitor
a
a b
l
1
0
d d
r
r dr dr

 
=
 
 
Laplace’s equation in Cylindrical coordinate
d
r A
dr

 
=
 
 
d A
dr r

=
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12. Cylindrical Capacitor
a
a b
l
ln( )
A r B
 = +
1
,
at r a
=  = 
2
,
at r b
=  = 
Boundary conditions are
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13. Cylindrical Capacitor
a
a b
l
Electrostatic potential is obtained as
1 2 1 2
ln( ) ln( )
ln( )
ln ln
b a
r
a b
b a
 −  −
 = +
   
   
   
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14. Cylindrical Capacitor
a
a b
l
d
E
dr

= −
1 2
1
ln
b
r
a
 −
 =
 
 
 
Electric field
Substituting the value of potential
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15. Cylindrical Capacitor
a
a b
l
Let, σ be the uniform surface charge density
E
 
=
r a
d
dr
 
=

= −
1 2
ln
b
a
a


 −
=
 
 
 
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16. Cylindrical Capacitor
a
a b
l
( )
2 a
r
 
=
1 2
2
ln
a
b
a
a

 
 −
=
 
 
 
12
2
ln
b
a
 

=
 
 
 
Also let, λ be the uniform line charge density
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17. Cylindrical Capacitor
a
a b
l
12
q
C =

ln
2
q
C
b
a


=
 
 
 
Capacitance of the system
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18. Cylindrical Capacitor
a
a b
l
ln
2
L
C
b
a



=
 
 
 
2
ln
L
C
b
a

=
 
 
 
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19. Problem 2
A capacitance is made of two coaxial metallic cylinders of radii ‘a’ and ‘b’ [a<b]
and length ‘l’ [l>>b]. The region between ‘a’ and ‘c’ [c=√(ab) ] is filled with a
medium of dielectric constant ε1 and remaining region is filled with a medium of
dielectric constant ε2. Find the capacitance of the system.
Soln
Let, E1 and E2 are the electric fields in the region ε1 and ε2 respectively
So, electric displacement
0 1 1 0 2 2
D E E
   
= =
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20. We assume a Gaussian cylinder of length ‘l’ and radius ‘r’ [a<r<b] having same axis
By Gauss’ law
2 rlD l
 
=
1
0 1
2
E
r

  
 =
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2
0 2
2
E
r

  
 =
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21. Potential difference between two cylinder is
1 2
a c
c b
E dr E dr
 = − −
 
0 1 0 2
2 2
a c
c b
dr dr
r r
 
     
 = − −
 
1 2
0 1 2
ln
2
b
a
 

  
 
+  
 =    
 
 
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22. Thus, capacitance of the system
q l
C

= =
 
1 2
0 1 2
ln
2
l
C
b
a

 

  
=
 
+  
   
 
 
0 1 2
1 2
2
ln
l
C
b
a
  
 
 
=  
+
   
 
 
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23. Spherical Capacitor
a
b
Laplace’s equation in Spherical coordinate
2
2
1
0
d d
r
r dr dr

 
=
 
 
2 d
r A
dr

 
=
 
 
2
d A
dr r

=
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24. Spherical Capacitor
a
b
2
,
at r b
=  = 
2
A
B
r
 = +
1
,
at r a
=  = 
Boundary conditions are
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Theory

25. Spherical Capacitor
a
b
1 2 2 1
b a
ab
b a r b a
 −  − 
 = +
− −
1 2
2
ab
E
b a r
 −
=
−
d
E
dr

= −
Electrostatic potential is obtained as
Electric field
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Theory

26. Spherical Capacitor
a
b
1 2
2
r a
ab
E
b a a
=
 −
 =
−
1 2
r a
b
E
b a a
=
 −
=
−
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27. Spherical Capacitor
a
b
Let, σ be the uniform surface charge density
r a
E
  =
=
1 2
b
b a a
 
 −
=
−
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28. Spherical Capacitor
a
b
Total charge on the outer surface of inner spherical shell
2
4
q a
 
=
2 1 2
4
b
q a
b a a
 
 −
=
−
12
4
ab
q
b a a


=
−
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Theory

29. Spherical Capacitor
a
b
Capacitance of the system
12
12
4
ab
b a a
C


 
 
  
−
  
=

4
ab
C
b a

 
=  
−
 
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Theory

30. Problem 3
A spherical conductor of radius ‘a’ is coated with a uniform thickness ‘d’ of shellac of
dielectric thickness ε. Show that capacity of the conductor is increased in the ratio
( )
( )
a d
a d


+
+
Soln
Let, E1 be the electric field outside the shellac
1 2
q
E
r
 =
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31. Let, E2 be the electric field inside the shellac
2 2
q
E
r

 =
Potential on the conductor
1 2
a d a
a d
E dr E dr
+
 +
 = − −
 
2 2
a d a
a d
q q
dr dr
r r

+
 +
 = − −
 
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Theory

32. ( )
( )
q a d
a a d


+
 =
+
Capacitance of the system
q
C =

( )
( )
q
C
q a d
a a d


=
+
+
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Theory

33. ( )
( )
a a d
C
a d


+
=
+
Increment of the system
0
'
C
C
C
=
0
( )
( )
a a d
a d
C
a


+
+
=
0
( )
( )
a d
C
a d


+
=
+
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Theory

34. Capacitance of isolated conductor
Charge of an isolated conductor ‘q’ can be written as
q C
= 
For single isolated conducting sphere of radius ‘r’
4
q
r

 =
4
C r

 =
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Theory

35. 16-11-2021 Arpan Deyasi, EM Theory 35
Problem 4
Calculate capacitance of Earth
Soln
0
4
Earth
C r

=
12 3
4 8.854 10 6400 10
Earth
C F
 −
=    
7
7117.2 10
Earth
C F
−
= 
711.72
Earth
C F

=
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