1. The document discusses electric flux and Gauss's law. Electric flux is defined as the product of the electric field and the perpendicular surface area. Gauss's law states that the total electric flux through a closed surface is proportional to the enclosed charge. 2. Examples are presented for applying Gauss's law to calculate electric fields produced by spherically symmetric charge distributions like a point charge or thin spherical shell, as well as a cylindrically symmetric charged rod. The calculations involve setting up Gaussian surfaces and relating the flux to the enclosed charge. 3. Key results are that the electric field of a point charge follows an inverse square law, a thin spherical shell produces no field inside but an inverse square field outside, and the

1. Lect. (3)
1

2. Electric Flux
• The product of the magnitude of the
electric field E and surface area A
perpendicular to the field is called the
electric flux E
• the SI units of electric flux (N.m2/C.)
• Electric flux is proportional to the
number of electric field
• lines penetrating some surface.
2

3. Electric Flux
• The amount of electric
field passing through a
surface area
• The units of electric flux
are N-m2/C
E E dA
   
A
 
3

4. Gauss’s Law
The total flux passing through a closed surface is
proportional to the charge enclosed within that
surface.
q
0 
 
E dA
E     
surface
Note: The area vector points outward
4

5. The Gaussian Surface
An imaginary closed surface created to enable the application of Gauss’s Law
5

6. Application of Gauss’s Law to Various
Charge Distributions
1. The Electric Field Due to a Point Charge
2. The Electric Field Due to a Thin Spherical Shell
3. A Cylindrically Symmetric Charge Distribution
6

7. 1. Spherically Symmetric Charge Distribution
1.The Electric Field Due to a Point Charge:
• The area vector is parallel to the electric field vector
at the surface of the spherical Gaussian surface.
• Applying Gauss’ law:
        
Φ E dA E dA E A cos θ
cos θ 1 sphereA 4 π r Φ E A E 4 π r
3
k Q r
Q r
3 3
in
k
4  π  k  q  E  4  π 
r
k q
2 2
4 π k q
2
in
2
in
2 2
R
R
r
r
  
4 π r
E
 


 


 

         
7

8. 1. Spherically Symmetric Charge Distribution
• As r increases to R, the magnitude of the electric
field will increase as more charge is included
within the volume of the spherical Gaussian
surface.
• The maximum electric field
will be obtained when r = R.
• For a spherical Gaussian
surface of radius r  R, we
can consider the charged
insulating sphere as a point
charge.
8

9. 1. Spherically Symmetric Charge Distribution
        
Φ E dA E dA E A cosθ
2 2
cosθ  1 sphereA  4  π  r Φ  E  4  π 
r
Φ  4  π  k  q 4  π  k  q  E  4  π 
r
in in
in
k Q
k q


2 2
4 π k q
2
in
2
r

r
  
4 π r

 

E
• Conclusion: the electric field E inside a uniformly charged solid insulating sphere varies
linearly with the radius until reaching the surface. The electric field E outside the sphere
varies inversely with 1/r2. The diagram on the next slide illustrates this relationship.
9

10. Spherically Symmetric Charge Distribution
10

11. 2. The Electric Field Due to a Thin Spherical Shell
A thin spherical shell of radius a has a total charge Q distributed uniformly over its surface .
Find the electric field at points
(A) outside and (B) inside the shell.
11

12. 2. The Electric Field Due to a Thin Spherical Shell
• Consider a thin spherical shell of
radius R with a total charge Q
distributed uniformly over its
surface.
• For points outside the shell, r  R:
the spherical Gaussian surface will
enclose the thin spherical shell, so
qin = Q.
12

13. 2. The Electric Field Due to a Thin Spherical Shell
• The area vector is parallel to the electric field vector at the surface of
the spherical Gaussian surface. The angle between the area vector
and the electric field vector is zero.
Φ E dA E dA E A cosθ
        
cosθ 1 sphereA 4 π r Φ E 4 π r
Φ  4  π  k  q 4  π  k  q  E  4  π 
r
in
2 2
k Q
k q
2 2
4 π k q
2
in
2
in in
r
r
  
4 π r




 

       
E
13

14. 2. The Electric Field Due to a Thin Spherical Shell
• For points inside the thin spherical
shell, r < R:
• The charge lies on the surface of the
thin spherical shell.
• No charge is found within the shell,
so qin = 0 and the electric field and
flux within the thin spherical shell is
also 0.
14

15. Cylindrically Symmetric Charge Distribution
• Consider a uniformly charged cylindrical rod of length L.
15

16. Cylindrically Symmetric Charge Distribution
16

17. Cylindrically Symmetric Charge Distribution
• Instead of the shortened Gaussian cylinder indicated in
the diagrams, I use a Gaussian cylinder that is the same
length as the charged cylindrical rod.
• The electric field lines are perpendicular to the charged
cylindrical rod and are directed outward in all directions.
• The electric field is perpendicular to the Gaussian
cylinder and is parallel to the area vector at those points
where it passes through the Gaussian cylinder. The angle
between the area vector and the electric field vector is
zero.
17

18. Cylindrically Symmetric Charge Distribution
• The equation for the area of a cylinder is: A = 2··r·L
• Applying Gauss’ law:
        
Φ E dA E dA E A cosθ
cosθ 1 cylinderA 2 π r L Φ E 2 π r L
Φ  4  π  k  q 4  π  k  q  E  2  π  r 
L
2 k λ
r
in in
2  k 
Q
r L
2  k 
q
r L
4  π  k 
q
2 π r L
E
in in
 





  

         
18