1. CHAPTER-23
Gauss’ Law

2. CHAPTER-23 Gauss’ Law Gauss’ law Topics to be covered
The flux (Φ ) of an electric field
Gauss’ law
Application of Gauss’ law to :
A uniformly charged insulating plane
An infinite long , uniformly charged insulating rod
A uniformly charged spherical shell
A uniform spherical charge distribution
Application of Gauss’ law to determine the electric field inside and outside charged conductors.

3. Ch 23-1 Gauss’ Law
Gaussian surface: a hypothetical (imaginary) surface, enclosing a charge distribution. The surface mimics the symmetry of charge distribution .
For spherical charges distribution : spherical surface; charged wires/ cylinders: cylindrical surfaces
Gauss’ law: relates E-field at points on a closed Gaussian surface to the net charge qencl enclosed by that surface or vice versa.
For calculation of charge enclosed by the Gaussian surface, we need to know how much E-field intercepted by the Gaussian Surface : FLUX of E-field

4. Ch 23-2 Flux
Flux of velocity field: Volume flow rate  of air through a small square loop with area A depends upon the angle between air velocity v and plane of loop A
=vA (v A) ; =0 (vA)
A is area vector , with magnitude equal to loop area and direction is normal to plane of loop
 = v cos A =v.A
 is angle between v and A

5. Ch 23-3 Flux of an Electric Field
Electric Flux through a Gaussian surface immersed in a non-uniform electric field
Gaussian surface is divided into element of area A with corresponding electric field E at that location then electric flux 
 =  E. A
If A becomes smaller then
 = surf E.dA
The electric flux  through a Gaussian surface is proportional to net number of electric field line passing through that surface

6. Ch 23-3 Flux of an Electric Field
 = surf E.dA
Net fux through the cylindrical surface
= -EdA+0+EdA=0

7. Ch 23-4 Gauss’ Law
Gauss Law: relates the net flux  of an electric field through a closed Gaussian surface and the net charge qenc that is enclosed by the surface
0  = qenc
 = qenc/0 = surf E.dA

8. Ch 23-4 Gauss’ Law Point charges enclosed in the surface
Surface S1: lines of E field parallel to area vector dA; E.dA is positive; qenc is positive
Surface S2: lines of E field anti-parallel to area vector dA; E.dA is negative; qenc is negative
Surface S3: lines of E field anti-parallel to area vector dA in upper half and parallel in lower half; E.dA is zero; qenc is zero
Surface S4: Similar to S3 and net charge in the surface is Zero

9. Ch 23-5 Gauss’ Law and Coulomb Law
Gauss Law: surf E.dA= qenc/ 0
If E is constant at the surface then
surf E.dA= Esurf dA
= EA=qenc/ 0
For a sphere A= 4R2
Then E= qenc/ A0
= qenc /4 0 R2
E= qenc /4 0 R2
=k qenc/R2 dA

10. Ch 23-6 A Charged Isolated Conductor
A Charged Conductor
If an excess charge is placed on an isolated conductor, that amount of charge will move entirely to the surface of the conductor.
An Isolated Conductor with a Cavity
There is no net charge on the cavity wall of a charged conductor

11. Induced Charges
A charged spherical shell with charge -100e enclosing another point charge -50e
Charge on shell inner surface: +50e
Charge on shell outer surface: - 150e

12. Ch 23-6 A Charged Isolated Conductor
Electric Field outside the Surface of a Conductor
Imagine a tiny Gaussian cylindrical surface embedded in the conductor with one end cap inside the conductor. E-field field through this section is zero.
Evaluate  =surf E.dA= qenc/ 0 on the surface
For the cylinder the net flux is only through the end cap lying outside the conductor, where E A . Then
= EA=qenc/0 =A/0
E=/0
where  is surface charge density
( charge per unit area)

13. Example of Spherical Symmetry
Symmetry. We say that an object is symmetric under a particular mathematical operation (e.g., rotation, translation, …) if to an observer the object looks the same before and after the operation Note: Symmetry is a primitive notion and as such is very powerful.
Example of Spherical Symmetry
Consider a featureless beach ball that can be rotated about a vertical axis that passes through its center. The observer closes his eyes and we rotate the sphere. When the observer opens his eyes, he cannot tell whether the sphere has been rotated or not. We conclude that the sphere has rotational symmetry about the rotation axis.
(23-10)

14. A Second Example of Rotational Symmetry
Consider a featureless cylinder that can rotate about its central axis as shown in the figure. The observer closes his eyes and we rotate the cylinder. When he opens his eyes, he cannot tell whether the cylinder has been rotated or not. We conclude that the cylinder has rotational symmetry about the rotation axis.
(23-11)

15. Example of Translational Symmetry:
Consider an infinite featureless plane. An observer takes a trip on a magic carpet that flies above the plane. The observer closes his eyes and we move the carpet around. When he opens his eyes the observer cannot tell whether he has moved or not. We conclude that the plane has translational symmetry.
(23-12)

16. 1. Make a sketch of the charge distribution.
Recipe for Applying Gauss’ Law
1. Make a sketch of the charge distribution.
2. Identify the symmetry of the distribution and its effect on the electric field.
3. Gauss’ law is true for any closed surface S. Choose one that makes the calculation of the flux  as easy as possible.
4. Use Gauss’ law to determine the electric field vector:

17. Ch 23-7 Applying Gauus’ Law: Cylindrical Symmetry
Electric field E at a distance r from the axis of a infinitely long plastic rod with uniform positive linear charge density .
Imagine a tiny Gaussian cylindrical surface coaxial with the rod with two end caps of the cylindrical surface.
Flux of E-field through the end caps is zero where E  A. Net flux of E-field through the circumference where E A.
= surf E.dA= E surf dA= qenc/0
For circumference surf dA= 2rh
Then EA= E 2rh= qenc/0
E=(qenc/2rh 0)= 1/2r0*(qenc /h)
E=(1/20)  /r =2k  /r; = qenc/h

18. Ch 23-8 Applying Gauss’ Law: Planar Symmetry
Non conducting Sheet
A thin, infinite, non conducting sheet with a uniform positive charge density  on one side only.
E at a distance r in front of the sheet. Choose a Gaussian cylinder with end caps passing through the surface
E field  to cylinder circumference net flux zero through it
E field  to end caps. Net flux through caps
= surf E.dA= qenc/0=A/0
surf E.dA=EA+EA= 2EA=A/0
E=/20 (Sheet of charge )

19. Ch 23-8 Applying Gauss’ Law: Planar Symmetry
Two conducting plates with charge density 1
All charges on the two faces of the plates
For two oppositely charged plates placed near each other, E field outer side of the plates is zero while inner side the E-field= 2 1/0

20. Ch 23-8 Applying Gauss’ Law: Planar Symmetry
Two non-conducting plates with charge density +
All charges on the one face of the plates
For two oppositely charged plates placed near each other, E field outer side of the plates is EL or ER with EL= ER= E(+)- E(+) and between the plates the field  EB = E(+)+ E(-)

21. Ch 23-9 Applying Gauss’ Law: Spherical Symmetry
Shell Theorem:
A shell of uniform charge attract or repel a charged particle that is outside the shell as if all the shell charge were concentrated at the center of the shell.
Eo=kq/r2 (r>R)
If a charged particle is located inside a shell of uniform charge, there is no electrostatic force on the particle from the shell.
Ei=0 (r<R)

22. Ch 23-9 Applying Gauss’ Law: Spherical Symmetry
Electric field outside a uniform sphere of charge
E=kq/r2 ( r>R)
Electric field inside a uniform sphere of charge
E=kqr/R3 ( r>R)

23. Thank you