Basic of electromagnetics

1. Lecture-2
Introduction to Gauss’s Law
Unit-I
Basics of Electromagnetics

2. Summary of the previous class
Coulomb’s law gives the electrostatic force of interaction between stationary point charges.
Electric field strength (E) = F/q0
Electric field due to a point charge Q,
Electric field due to group of point charges E = E1 + E2+ E3 +………………..En

3. Outline of lecture
● Electric lines of force
● Electric flux
● Gauss’s law
● Coulomb’s law from Gauss’s law

4. After today's lecture, you will be able to understand:
● The properties of lines of force and electric flux
● Statement and importance of Gauss’s law
● Derivation of Coulomb’s law from Gauss’s law
Lecture Outcomes

5. 5
Electric lines of force
The concept of electric field lines introduced by Michael Faraday provide us with an easy way to visualize the electric
field.
The electric lines of force originate from a positive charge and ends at a negative charge.
Electric field lines always flow from higher potential to lower potential.
 Lines of force start / terminate are always perpendicular to the surface of the charged body or conductor.
 The field lines never intersect each other.
 The lines of force of like charges repel each other and of unlike charges attract each other.
The concentration of field lines represents the electric field strength.
The number of lines is proportional to the magnitude of the charge.
The tangent to the line of force gives the direction of electric field.
Electric field lines also gives us an indication
of the equipotential surface.

6. Electric flux : Electric Flux is a measure of the number of electric field lines crossing a surface.
Electric Flux: The electric flux is defined as the product of the magnitude of the electric field and the surface
area.
when Electric field is perpendicular to the surface.
If E is not perpendicular, then
For an arbitrary surface containing the charge, the electric flux of the small surface element dΦ of surface
area dA is
By integrating over entire surface, the net flux is,
= E A cosΘ

7. Electric flux through a cylindrical surface
A closed cylinder immersed in a uniform electric field

8. Gauss’s law
“The net electric flux through any hypothetical closed surface is equal to
one by epsilon not times the net electric charge within that closed
surface”.
Integral form:
Gauss's law may be expressed as:

9.  q is the net charge taking its algebraic sign into account.
 Charge outside the surface don't contribute to value of q.
 Gauss law can be used to evaluate E.
 If E is known for all points on a Gaussian surface, we can compute the charge inside.
 If E has an outward component on a closed surface,
will be positive.
 If E has an inward component on a closed surface,
will be negative.
-q
+3q
E
-5q
+3q
E
+11q

10. Steps to apply Gauss’s Law:
1. Select a Gaussian surface suitable in a specific symmetry.
2. Find out the electric flux through the Gaussian surface.
3. Apply Gauss’s law.

11. More information about Gauss’s Law:
Here , q-Total charge enclosed by the closed surface S
•The term q on the right side of Gauss’s law includes the sum of all charges enclosed by the closed
surface.
The charges may be located anywhere inside the closed surface.
•The total electric flux through a closed surface is zero if no charge is enclosed by the surface.
•Gauss’s law is true for any closed surface, regardless of its shape (or) size.
•The closed surface in Gauss’s law is imaginary known as Gaussian surface.
The surface that we choose for the application of Gauss’s law is called the Gaussian’s surface.
We choose Gaussian surface(either sphere or cylinder) in such a way that electric field E is either
parallel or perpendicular to its various faces.
•Gauss’s law and Coulomb’s law are equivalent.

12. •In case of symmetrical charge distribution, Gauss’s law is very helpful in calculating electric field.
•Gauss’s law is valid for any closed surface and for any distribution of charge.
If charge distribution is given and is symmetrical, we can find electric field or if we know the field,
charge distribution can be determined.

13. Importance of Gauss’s law:
Gauss’s law is useful in calculating the electric field in the problems in which it is possible to choose
a closed surface such that the electric field has a normal component which is either zero (or) has a
single fixed value at every point on the surface.
The use of Gauss’s law makes the calculations easier compared to Coulomb’s law.
Gauss’s law is very important especially in situations where calculation of electric field with the
application of Coulomb’s law (or) Principle of superposition becomes difficult.
In these cases a closed surface is selected which enclosed the source charge. Such a surface is
called Gaussian surface.
Gaussian surface: An imaginary closed surface enclosing a charge is called as Gaussian surface.

14. Derivation of Coulomb Law From Gauss Law
The flux passing through the area element , that is,
Hence, the total flux through the entire Gaussian sphere is obtained as
But ∫dA is the total surface area of the sphere and is equal to 4πr2.
Image Source

15. But according to Gauss’s law for electrostatics
Therefore,
If a second point charge is placed at the point at which the magnitude
of E is calculated ,then the magnitude of the force acting on the second
charge would be
Therefore,
F= 1/4πϵ0. q q0 /r2

16. ●‘The electric flux is defined as the product of the magnitude of the
electric field and the surface area’.
Φ=EA
● Gauss’s law: “The net electric flux through any hypothetical closed
surface is equal to one by epsilon times the net electric charge within
that closed surface”.
●We have derived the Coulomb's law from Gauss’s law.
Summary

17. Session Quiz
Important note: Session quiz marks will be considered for internal
Assessment

18. Thank you