CHAPTER-23 Gauss’ Law

CHAPTER-23 Gauss’ Law Topics to be covered The flux (symbol Φ ) of the electric field Gauss’ law Application of Gauss’ law to determine the electric field generated by: An infinite, uniformly charged insulating plane An infinite, 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.

Ch 23-1 Gauss’ Law Gaussian surface: a hypothetical (imaginary) surface enclosing a charge distribution. Gauss’ law: relates the electric field at points on a closed Gaussian surface to the net charge enclosed by that surface

Ch 23-2 Flux Flux : Volume flow rate (Volume flux)  of air through a small square loop with area A  = v cos A =v.A * where v is air velocity, * A is area vector , with magnitude equal to loop area and direction is normal to plane of loop * and  is angle between v and A

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 Net fux through the cylindrical surface = -EdA+0+EdA=0

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 Net fux through the cylindrical surface = -EdA+0+EdA=0

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 but  = surf E.dA 0 surf E.dA = qenc  = surf E.dA=qenc/0

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

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

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

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

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 suface charge density ( charge per unit area)

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. Featureless sphere Rotation axis Observer Rotational symmetry 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)

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. Featureless cylinder Rotation axis Observer Rotational symmetry (23-11)

Example of Translational Symmetry: Observer Magic carpet Infinite featureless plane Translational symmetry 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)

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: (23-13)

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 S1 S2 S3

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 )

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

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(-)

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)

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)