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CHAPTER-23

Gauss’ Law

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CHAPTER-23 Gauss’ Law

Topics to be coveredThe flux (symbol Φ ) of the electric fieldGauss’ lawApplication of Gauss’ law to determine the electric field generated by:An infinite, uniformly charged insulating planeAn infinite, uniformly charged insulating rod A uniformly charged spherical shellA uniform spherical charge distributionApplication of Gauss’ law to determine the electric field inside and outside charged conductors.

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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

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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

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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

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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

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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

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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

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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= Esurf dA = EA=qenc/ 0

For a sphere A= 4R2

Then E= qenc/ A0 = qenc /4 0 R2

E= qenc /4 0 R2=k qenc/R2

n̂

dA

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A Charged ConductorIf 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

Ch 23-6 A Charged Isolated Conductor

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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

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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)

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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

ObserverRotational symmetry

(23-10)

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.

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Featureless cylinder

Rotation axis

Observer

Rotational symmetry 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)

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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)

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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: enc

0

q

(23-13)

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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= 2rh Then EA= E 2rh= qenc/0

E=(qenc/2rh 0)= 1/2r0*(qenc /h) E=(1/20) /r =2k /r; = qenc/h

Ch 23-7 Applying Gauus’ Law: Cylindrical Symmetry

S1

1̂n

2n̂S2

S3

3n̂

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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=/20 (Sheet of charge )

1̂n

2n̂

3n̂ S1

S2S3

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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

1

0

2iE

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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(-)

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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

1̂n iE

2n̂ 0E

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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)

Suggested problems Chapter 23

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