gauss's law in pictures

Gauss’s Law in Pictures

Branislav K. NikolićDepartment of Physics and Astronomy, University of Delaware, U.S.A.

PHYS 208 Honors: Fundamentals of Physics IIhttp://www.physics.udel.edu/~bnikolic/teaching/phys208/phys208.html

PHYS 208 Honors: Gauss’s Law

Symmetry Operations

A charge distribution is symmetric if there is a set of geometrical transformation that do not cause any physical change.


Symmetry of Charge Distribution vs. Symmetry of Electric Field

The symmetry of the electric field must match the symmetry of the charge distribution


Symmetry of Charge Distribution vs. Symmetry of Electric Field

The symmetry of the electric field must match the

symmetry of the charge distribution


Electric Field Pattern of Cylindrically Symmetric Charge Distribution


Electric Field Pattern of Charge Distributions with Basic Symmetries


Example: Problem 27.1

The electric field of a cylindrically symmetric charge distribution cannot have a component parallel to the cylinder axis. Also, the

electric field of a cylindrically symmetric charge distribution

cannot have a component tangent to the circular cross section. The only shape for the electric field that

matches the symmetry of the charge distribution with respect to (i) translation

parallel to the cylinder axis, (ii) rotation by an angle

about the cylinder axis, and (iii) reflections in any plane containing or perpendicular to the cylinder axis is the one shown in the figure.


The Concept of Flux of Electric Field


Gaussian surface which does not match symmetry of charge is not useful!

The electric field pattern through the surface is particularly simple if the closed surface matches the symmetry of the charge distribution inside.


Calculus for the Flux of Vector Fields

Motivation for the definition: Flux of a fluid flow

Flux of uniform electric field passing through a surface:


Flux of a Nonuniform Electric Field

Simplified situations where integration goes away:

1 1

( )N N

e i ii i

E Aδ δ= =

Φ = Φ = ⋅∑ ∑

,N A dA

e

surface

E dA

δ→∞ →

Φ = ⋅∫


Gauss’s Law For Point Charge

22

0 0

14

4e

q qr

rπ

πε εΦ = =

24e SphereE dA EA E rπΦ = ⋅ = =∫

Electric flux is

independent of surface shape:

20 0

cos

1cos

4 4

ed EdA

q qdA d

r

α

απε πε

Φ = =

= = ΩdΩ


Gauss’s Law For Charge Distribution = First Maxwell Equation

Unlike Coulomb’ law for static point charges, Gauss’s law is valid for moving charges and fields that change with time.

1 2

0 0 0

enclosede

Qq qE dA

ε ε εΦ = ⋅ = + + =∫

…


Applications: Charged Sphere

1. Determine the symmetry of the electric field.

2. Select a Gauss surface (as imaginary surface in the space surrounding the charge) to match the symmetry of the field.

3. If 1. and 2. are possible, flux integral should become algebraic product electric field x Gauss surface area.

0

22

0 0

ˆ44

e outside sphere

outside outside

QE dA E A

Q qE r E r

r

ε

πε πε

Φ = ⋅ = =

= ⇒ =

∫

0

3 3

233

0 0 0

4 43 3 ˆ4

4 43

insidee sphere

inside inside

QE dA EA

r r Q QE r E rr

RR

ε

πρ ππ

ε ε πεπ

Φ = ⋅ = =

= = ⇒ =

∫


Applications: Charged Wire

0 0e top bottom w all cylinderE dA EAΦ = ⋅ = Φ +Φ + Φ = + +∫

0 0 0 0

22 2

inw ire

QQ L LE rL E

r r

λ λπε ε πε πε

= = ⇒ = =


Applications: Charged Plane

2e E d A E AΦ = ⋅ =∫

0 0 02in

plane

Q AEA E

η ηε ε ε

= = ⇒ =


Conductors in Electrostatic Equilibrium

BASIC FACT: Electric field is zero at all points within the conductor – otherwise charges will flow, thereby violating electrostatic equilibrium.

0

0insidee inside

QE dA Q

εΦ = ⋅ = ⇒ ≡∫

Any excess charge resides entirely on the exterior surface.

Any component of the electric field tangent to the conductor surface would cause surface charges to move thereby violating the assumuptionthat all charges are at rest.


Gauss’s Law Determines Electric Field Outside of Metal

0 0 0

ˆine surface surface

Q AE A E n

η ηε ε ε

Φ = = = ⇒ =


Electric Field of Conducting Shells

+ ++

- -- 0closed path

qE ds⋅ ≠∫


Applications: Faraday’s Cage


Example: Electric Force Acting on the Surface of Metallic Conductor

0E =

AδAEδ

AEδ

02AEδηε

=

restE

restE

0 A restE E Eδ= ⇒ =

22

surfacesurface A rest rest rest

EE E E E Eδ= + = ⇒ =

surfaceE

1

2 2surface A

A surface

E FF A f E

Aδ

δ ηδ ηδ

= ⇔ = =

2 200

1ˆ

2 2surface conductor surfacef E n F E dAεε= ⇒ = ∫

NOTE: Regardless of the sign of surface charge density and corresponding direction of the electric field , the force is always directed outside of the conductor tending to stretch it.

ηsurfaceE

surfaceF

gauss's law in pictures

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