Application of Gauss' Law for line, surface, volume charge densities

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 – Application of Gauss' Law
03-11-2021 Arpan Deyasi, EM Theory 1
Arpan Deyasi
Electromagnetic
Theory

2. i
l 0
q
(r) Lt C/m
l
 →

 =

Density of Charge Distribution
Line charge density (λ): charge is distributed over a line
Total charge for line charge distribution
line
Q (r)dl C
= 

03-11-2021 Arpan Deyasi, EM Theory 2
Arpan Deyasi
Electromagnetic
Theory

3. 03-11-2021 Arpan Deyasi, EM Theory 3
Application of Gauss’ law for line charge distribution
E
Consider the case of uniformly charged infinite cylinder n̂
n̂
r
Electric flux is zero for both the flat surfaces
Electric flux exists only for curved surface
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Electromagnetic
Theory

4. 03-11-2021 Arpan Deyasi, EM Theory 4
l
Application of Gauss’ law for line charge distribution
E
r
a
Case-I: r>a
enc
out
S
q
E .ds =


enc
out
q
2 rlE
 =

out
l
2 rlE

 =

out
E
2 r

=

‘r’ is the radius of the cylinder
‘a’ is the radius of Gaussian cylinder
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Electromagnetic
Theory

5. 03-11-2021 Arpan Deyasi, EM Theory 5
l
Application of Gauss’ law for line charge distribution
E
r
a
Case-I: r>a
out
r 2 r
 
= −
 
r
out dr
2 r


 = −


out ln(r)
2

 = −

Arpan Deyasi
Electromagnetic
Theory

6. 03-11-2021 Arpan Deyasi, EM Theory 6
l
Application of Gauss’ law for line charge distribution
E
r
a
Case-II: r<a
Let, ‘ρ’ be the volume charge density
2
enc
q a l
=  
2
l a l
 =  
2
a
 =  
2
a

 =

Arpan Deyasi
Electromagnetic
Theory

7. 03-11-2021 Arpan Deyasi, EM Theory 7
l
Application of Gauss’ law for line charge distribution
E
r
a
Let, ‘q0’ be the charge contained in the Gaussian cylinder
2
0
q r l
 =  
2
0 2
q r l
a

= 

2
0 2
r l
q
a

=
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Electromagnetic
Theory

8. 03-11-2021 Arpan Deyasi, EM Theory 8
0
in
S
q
E .ds =


0
in
q
2 rlE
 =

2
in 2
r l
2 rlE
a

 =
in 2
r
E
2 a

=

E
a
r
Application of Gauss’ law for line charge distribution
Arpan Deyasi
Electromagnetic
Theory

9. 03-11-2021 Arpan Deyasi, EM Theory 9
2
in 2
r 1
ln(a)
2 2 2a 2
 
 
 
 = − −
 
 
   
 
E
a
r
Application of Gauss’ law for line charge distribution
a r
in 2
a
r
dr dr
2 r 2 a

 
 = − −
 
 
a r
in out in
a
E dr E dr

 = − −
 
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Electromagnetic
Theory

10. 03-11-2021 Arpan Deyasi, EM Theory 10
Problem 1
A long charged cylinder of radius ‘a’ has volume charge density . Find electric
field inside And outside of the cylinder.
(r) r
 = 
Soln
l
r
r+dr
Amount of charge in cylindrical shell of radii ‘r’ and ‘r+dr’ is
dq (r). rl.dr
=  
r
0
q (r). rl.dr
 =  

r
0
q r. rl.dr
=  

3
lr
q
3

=
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Electromagnetic
Theory

11. 03-11-2021 Arpan Deyasi, EM Theory 11
By Gauss’ law,
electric field inside the cylinder
in
S
q
E .ds =


3
in
r
E . rl l.
3
 = 

2
in
r
E
3

=

2
in
r
ˆ
E r
3

=

Arpan Deyasi
Electromagnetic
Theory

12. 03-11-2021 Arpan Deyasi, EM Theory 12
l
r
r+dr
Amount of charge in cylindrical shell of radii ‘r’ and ‘r+dr’ is
dq (r). rl.dr
=  
a
0
q (r). rl.dr
 =  

a
0
q r. rl.dr
=  

3
la
q
3

=
Arpan Deyasi
Electromagnetic
Theory

13. 03-11-2021 Arpan Deyasi, EM Theory 13
By Gauss’ law,
electric field outside the cylinder
out
S
q
E .ds =


3
out
a
E . rl l.
3
 = 

3
out
a
E
3 r

=

3
out
a
ˆ
E r
3 r

=

Arpan Deyasi
Electromagnetic
Theory

14. Density of Charge Distribution
Surface charge density (σ): charge is distributed over the surface
2
i
l 0
q
(r) Lt C/m
s
 →

 =

Total charge for surface charge distribution
surface
Q (r)ds C
= 

03-11-2021 Arpan Deyasi, EM Theory 14
Arpan Deyasi
Electromagnetic
Theory

15. 03-11-2021 Arpan Deyasi, EM Theory 15
Application of Gauss’ law for surface charge distribution
Consider the case of
uniformly charged infinite plane
We consider an imaginary right circular cylinder
which is perpendicular to the flat surfaces
Electric flux exists for both the flat surfaces
E
n̂
E
n̂
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Electromagnetic
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16. 03-11-2021 Arpan Deyasi, EM Theory 16
Application of Gauss’ law for surface charge distribution
E
n̂
E
n̂
①
②
enc
1 2
q
E.ds E.ds
+ =

 
S
2ES

=

E
2

=

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Electromagnetic
Theory

17. 03-11-2021 Arpan Deyasi, EM Theory 17
Problem 2
r
R
z
P
A thin circular ring of radius ‘R’ carries a uniform surface charge
density σ. Calculate potential and electric field on axis at a point.
Soln
Let, ‘dR’ be the thickness of the ring
Therefore, charge contained by the ring
Q 2 RdR.
=  
Potential at ‘P’ due to ‘dq’
dq
d
4 r
 =

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Electromagnetic
Theory

18. 03-11-2021 Arpan Deyasi, EM Theory 18
Total potential
Q
4 r
 =

2 RdR.
4 r
 
 =

RdR.
2 r

 =
 ( )
1/ 2
2 2
RdR.
2 R z

 =
 +
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Electromagnetic
Theory

19. 03-11-2021 Arpan Deyasi, EM Theory 19
d
E
dz

= −
( )
1/ 2
2 2
d RdR.
E
dz 2 R z
 

 
= −
 
 +
 
( )
3/ 2
2 2
RdR. z
E
2 R z

=
 +
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Electromagnetic
Theory

20. 03-11-2021 Arpan Deyasi, EM Theory 20
Density of Charge Distribution
Volume charge density (ρ): charge is distributed over the volume
3
i
l 0
q
(r) Lt C/m
v
 →

 =

Total charge for volume charge distribution
volume
Q (r)dv C
= 

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Electromagnetic
Theory

21. 03-11-2021 Arpan Deyasi, EM Theory 21
Application of Gauss’ law for volume charge distribution
Consider the case of
uniformly charged infinite sphere
Electric flux is always along outward direction
E
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Theory

22. 03-11-2021 Arpan Deyasi, EM Theory 22
Application of Gauss’ law for volume charge distribution
r
R
Case-I: r>R
enc
out
S
q
E .ds =


‘R’ is the radius of the sphere
‘r’ is the radius of Gaussian sphere
3
2
out
4
R
3
E .4 r
 
 =

3
out 2
R
E
3r

=

Arpan Deyasi
Electromagnetic
Theory

23. 03-11-2021 Arpan Deyasi, EM Theory 23
r
R
Application of Gauss’ law for volume charge distribution
3
out
2
R
r 3r
 
= −
 
r 3
out 2
R
dr
3r


 = −


3
out
R
3r

 =

Arpan Deyasi
Electromagnetic
Theory

24. 03-11-2021 Arpan Deyasi, EM Theory 24
Application of Gauss’ law for volume charge distribution
r
R
Case-II: r<R
enc
in
S
q
E .ds =


3
2
in
4
r
3
E .4 r
 
 =

in
r
E
3

=

Arpan Deyasi
Electromagnetic
Theory

25. 03-11-2021 Arpan Deyasi, EM Theory 25
r
R
Application of Gauss’ law for volume charge distribution
R r
in out in
R
E .dr E .dr

 = − −
 
R r
3
in 2
R
R r
.dr .dr
3r 3

 
 = − −
 
 
( )
2 2
in 3R r
6

 = −

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Electromagnetic
Theory

26. 03-11-2021 Arpan Deyasi, EM Theory 26
Problem 3
A charge ‘q’ is distributed in a spherical volume of radius ‘a’ with volume charge density
(r) r
 =  . Determine potential and electric field inside and outside of the sphere.
Soln
a
r
Case-I: r>a
enc
out
S
q
E .ds =


3
2
out
4
a (r)
3
E .4 r
 
 =

3
2
out
4
a r
3
E .4 r
 
 =

3
out
a
E
3 r

=

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Electromagnetic
Theory

27. 03-11-2021 Arpan Deyasi, EM Theory 27
3
out a
r 3 r
 
= −
 
r 3
out
a
dr
3r


 = −


3
out
a
ln(r)
3

 = −

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Electromagnetic
Theory

28. 03-11-2021 Arpan Deyasi, EM Theory 28
Case-II: r<a
enc
in
S
q
E .ds =


3
2
in
4
r r
3
E .4 r
 
 =

2
in
r
E
3

=

r
a
Arpan Deyasi
Electromagnetic
Theory

29. 03-11-2021 Arpan Deyasi, EM Theory 29
a r
in out in
a
E .dr E .dr

 = − −
 
a r
3 2
in
a
a r
dr dr
3 r 3

 
 = − −
 
 
( )
3
3 3
in
a
ln(a) r a
3 9
 
 = − − −
 
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Electromagnetic
Theory