1. Gauss’s law and its applications
Lect. (3)
Gauss’s law and its applications

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.

3. Electric Flux
The amount of electric field passing through a surface area
The units of electric flux are N-m2/C

4. Gauss’s Law
The total flux passing through a closed surface is proportional to the charge enclosed within that surface.
Note: The area vector points outward

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

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

7. 1. Spherically Symmetric Charge Distribution
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:

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.

9. 1. Spherically Symmetric Charge Distribution
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.

10. Spherically Symmetric Charge Distribution

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.

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.

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.

14. For points inside the thin spherical shell, r < R:
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.

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

16. Cylindrically Symmetric Charge Distribution

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.

18. Cylindrically Symmetric Charge Distribution
The equation for the area of a cylinder is: A = 2··r·L
Applying Gauss’ law: