1 Lecture 3 Gauss’s Law Ch. 23 Physlet ch9_2_gauss/default.html Topics –Electric Flux...

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Lecture 3 Gauss’s Law Ch. 23• Physlet http://webphysics.davidson.edu/physletprob/ch9_problems/ch9_2_gauss/default.html• Topics

–Electric Flux–Gauss’ Law–Coulombs Law from Gauss’ Law–Isolated conductor and Electric field outside conductor–Application of Gauss’ Law–Conductor in a uniform field

• Demo–Faraday Ice pail: metal cup, charge ball, teflon rod, silk, electroscope, electrometer

• Elmo–Problems 29

• Polling and Clickers

Examples of Field Lines

Electric FluxElectric flux is the number of Electric field lines penetrating a surface or an area. Call it .

Electric Flux =Φ =(Ecosθ)A=rE ⋅

where rA =An̂

EA0cosEA ,So

=θ⇒rr

Electric Flux =Φ =(Ecosθ)A=rE ⋅

2. 0 A E =θ⇒rr

EA707.045cosEAThen, 45 Let=°=Φ

where rA =An̂

Gauss’s Law - Powerful relationship between charges and electric fields through closed

surfaces

• Gauss’s law makes it possible to find the electric field easily in highly symmetric situations.

• If there is no charge inside, then the flux is 0

• Let look at an example

Φnet = 0

rE—∫ ⋅d

Derivation of Gauss’ Law from a point charge and using Coulombs law:

Summary of steps

• Start with an isolated point charge.

• Choose a sphere around the charge.

• Calculate

• Show that

Φnet =

rE—∫ ⋅d

0εenc

Start with a point charge.

Note that the electric lines of flux are radial for a point chargeChoose a spherical surface to simplify the calculation of the flux

• Calculate the net flux through the closed surface.

EdA∫ ∫==Φ 2

)4( 222

dArkq π==Φ ∫

Φ=4πkq

10x85.8 where14NmCk −== ε

Φnet =qenc

Gauss’ Law

dr A = ˆ n dA

For a Point charge 2r

Φnet =

rE—∫ ⋅d

rANet Flux =

= E∫ cosθdA = EdA∫10cos =°

Valid even when charges are not at the center

Find the electric flux through a cylindrical surface in a uniform electric field E

Uniform electric fieldCylindrical surface choose because of symmetryNo charge insideFlux should be 0We want to calculate AdE

rr⋅=Φ ∫

Remember

ˆ n θ

00 =×==Φ AAEn

AEAEn ⋅==Φ θcos

En =Ecosθ

⋅=Φ ∫

Φ= E∫ cos180dA = − EdA∫ = −EπR2a.

20cos REEdAdAE π===Φ ∫∫

Φ= E∫ cos90dA = 0

Flux from a. + b. + c. 0 as expected

What would be the flux if the cylinder were vertical ?

Suppose it were any shape?

0εenc

This result can be extended to any shape surface

with any number of point charges inside and

outside the surface as long as we evaluate the

net flux through it.

Gauss’ Law

Δ⋅=Φ ∑

⋅=Φ ∫

Approximate Flux

Exact Flux

Circle means you integrate over a closed surface.

dr A = ˆ n dA

You should know these things

1 The electric field inside a conductor is 0.2 The total net charge inside a conductor is 0. It resides on the

surface.3 Find electric field just outside the surface of a conductor.4 Find electric field around two parallel flat conducting planes.5 Find electric field of a large non-conducting sheet of charge σ.6 Find electric field of an infinitely long uniformly line of charge λ.7 Find E inside and outside of a long non-conducting solid cylinder

of uniform charge density ρ.8 Find E for a thin cylindrical shell of surface charge density σ9 Find E inside and outside a solid non-conducting sphere of

uniform charge density ρ.

1. Electric field inside a conductor is 0. Why?

• Inside a conductor in electrostatic

equilibrium the electric field is always zero.

(averaged over many atomic volumes)

• The electrons in a conductor move

around so that they cancel out any

electric field inside the conductor

resulting from free charges anywhere including outside the conductor. This results in a net force of 0 on any particular charge inside the conductor.

2. The net charge inside a conductor is 0.

• Any net electric charge resides on the surface of the conductor within a few angstroms (10-10 m).

• Draw a Gaussian surface just inside the conductor. We know everywhere on this surface E=0.

• Hence, the net flux is zero. From Gauss’s Law the net charge inside is zero.

• Show Faraday ice pail demo.

3. Find electric field just outside the surface of a conductor

• The electric field just outside a conductor has magnitude and is directed perpendicular to the surface.

• Draw a small pill box that extends

into the conductor. Since there is

no field inside, all the flux comes

out through the top.

00 εσ

4. Find electric field around two parallel flat conducting planes.

E =σ1

ε0E =

E =σ1

6. Find the electric field for a very long wire of length L that is uniformly charged.

6. Find the electric field for a very long wire of length L that is uniformly charged. q

n̂EEn ⋅=r

sides EndA =E dA=E ⋅2πrh—∫—∫

rh2E 0 sideendcapsπ⋅+=

2πε0r090Cos

90n̂E

=°°=θ

hελ=

E =2kλr

EndA =

—∫ Gauss's Law

endcaps+sides

9. Electric field inside and outside a non-conducting solid uniformly charged sphere

• Often used as a model of the nucleus.• Electron scattering experiments have shown that the charge density is constant for some radius and then suddenly drops off at about

2 − 3×10−14 m.

For the nucleus,

ρ =10−26 C

=ρ R Charge density per unit volume

m10 14−×

Electric Field inside and outside a uniformly charged sphere

43 πR3

, r ≤ RQ= Total charge

= Z ×1.6 ×10-19C

Inside the sphere:To find the charge at a distance r<RDraw a gaussian surface of radius rBy symmetry E is radial and parallel to normal at the surface. By Gauss’s Law:

E ⋅4πr2 =q

3 πr3

E =ρr

EndA =

—∫

ρr3ε0

Electric Field inside and outside a uniformly charged sphere

43 πR3

, r ≤ RQ= Total charge

= Z ×1.6 ×10-19C

Outside the sphere:

E ⋅4πr2 =q

3 πR3

Eερ=

Same as a point charge q

Electric field vs. radius for a nonconducting solid sphere filled with a uniform charge

distribution

rE r ∝

2r r1E ∝rE

Problem 23 -4

A charge of - 5uC is placed inside a neutral conducting shell at R/2.What is the charge on the inner and outer surface of the shell?How is it distributed?What are the field lines inside and outside the shell?

Problem 23 -4

Chapter 23 Problem 29 Elmo

A long straight wire has a fixed negative charge with a linear charge density of magnitude 3.1 nC/m. The wire is to be enclosed by a coaxial, thin walled, nonconducting cylindrical shell of radius 1.8 cm. The shell is to have positive charge on its outside surface with a surface charge density that makes the net external electric field zero. Calculate the surface charge density.

1) Suppose you put a neutral ideal conducting solid sphere in a region of space in which there is, initially, a uniform electric field.a) Describe (as specifically as possible) the electric field

inside the conductor.b) Describe the distribution of charge in and on the conductor.c) Describe the electric field lines at the surface of the

conductor.

Here is a essay question you should be able to answer based on our ideas so far

External Electric field penetrates conductor

Free charges inside conductor move around onto the surface in such a way as to cancel out the field inside.

E=0 inside conducting sphere

Orientation of electric field is perpendicular to surface

If the electric field were not perpendicular, then there would be a component of electric field along the surface causing charges to accelerate on the surface and it would not be in electrostatic equilibriumcontrary to experience.

2) Suppose you put some charge on an initially-neutral, solid, perfectly-conducting sphere (where the sphere is not in a pre-existing electric field). – Describe the electric field inside the conductor, – Describe the electric field at the surface of the conductor,

and outside the conductor as a result of the unbalanced charge.

– Describe the distribution of the charge in and on the conductor.

3) Repeat questions 1-3 for the case of a hollow perfectly-conducting spherical shell (with the interior being vacuum).

4) How would your answers to questions 1-4 change if the conductor had some shape other than spherical?

1 Lecture 3 Gauss’s Law Ch. 23 Physlet ch9_2_gauss/default.html Topics –Electric Flux...

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