Today in Physics 122: Gauss’s Law

Today in Physics 122: Gauss’s Law

Electric field flux

Gauss’s Law: an easy way to calculate E for symmetric charge distributions.

(Easier than using Coulomb’s law, anyway.)

11 September 2019 Physics 122, Fall 2019 1

Carl Friedrich Gauss (Wikimedia Commons)

http://en.wikipedia.org/wiki/File:Bendixen_-_Carl_Friedrich_Gau%C3%9F,_1828.jpg


Flux of E

First, the flux of rain.

Suppose rain is falling straight down at a constant rate: call the mass per unit area falling on the ground f.

You have a bucket sitting flat on the ground, with area A at the top.

The rate at which the bucket fills up is determined by how big or small the flux fA is. A

f

Flux of E (continued)

Now tip the bucket, so that its axis makes an angle θ with the direction of the raindrops.

Clearly it will capture water at a smaller rate than before, because its mouth presents a smaller area Acosθ to the rain. So the flux, which determines how fast the bucket fills, is more generally fAcosθ .


A

f θ

Acosθ


Now, E:

Electric field doesn’t flow, but the direction and density of lines of Efunction like the velocity and density of raindrops.

For uniform E and planar A, the flux of E is therefore


cosEA θΦ =

A

θ

Acosθ

E


Or, inventing a vector A which has magnitude equal to the area A and direction perpendicular to the area, we can write the flux in a vector shorthand that will turn out to be convenient:

Flux is the scalar (“dot”) product of the vectors E and A:


.Φ = E AA

θ

E


Compare fluxes

Which flux is larger?

1) Φ1>Φ2 2) Φ1<Φ2 3) Φ1=Φ2

1 2


Compare fluxes

1 2

Which flux is larger?1) Φ1>Φ2 2) Φ1<Φ2 3) Φ1=Φ2


Compare fluxes

1 2Which flux is larger?

1) Φ1>Φ2 2) Φ1<Φ2 3) Φ1=Φ2


Compare fluxes

1 2Which flux is larger?

1) Φ1>Φ2 2) Φ1<Φ2 3) Φ1=Φ2


Compare fluxes

1 2



Compare fluxes

1 2


Nonuniform field, irregular surface

Even if the field varies in strength with position, and the surface is irregular, one can always go to the location of each infinitesimal area element in the surface and

find the local value of E

define an area vector dAfor the area element.

Then the total flux through that surface is the sum of the fluxes through all the infinitesimal elements:


surface

.Φ = ∫ E dA Don’t worry, we’ll only be using simple surfaces that lead to do-able integrals

Closed surfaces and fields from charges

Use your intuition. Electric field lines from charges originate at + charges and terminate at – charges. So which of these spheres has a non-zero flux through it?

1) Left 2) Right 3) Both


Closed surfaces and fields from charges (continued)

This sphere contains an electric dipole, but the + (red) charge is closest to its inner surface. The flux through it is

1) positive

2) negative

3) zero



Evidently, the flux of E through a closed surface depends upon how much charge it contains, since lines of Ecan only start and finish on charges.

For such a flux to be positive (negative), it needs to contain an net positive (negative) charge.

If it contains no net charge, every flux line that enters the surface has to leave it too.


+

-


We can write this in the form of our integral for flux as

where the symbol means an integral over the closed surface for which dA is an infinitesimal area element.

The vector dA points outward from convex surfaces.


enclosedsome constant

Q

= ×

∫ E dA

∫


To find the constant, consider a point charge at the center of an imaginary sphere.

Since for spheres the surface is perpendicular to the radius, cosθ =1 for all elements.

Since all the points on the sphere lie a distance r away from the charge, E is the same for all elements.

Thus


( )20

220

0

14

1 44

Q dArQ rr

Q

πε

ππε

ε

=

=

=

∫ ∫E dA

r

Gauss’s Law

So now we know the constant, and can write

This result, called Gauss’s Law, is the first of the four fundamental equations, called the Maxwell equations, that form the most compact expression of the theory of electromagnetism.

It also represents a very powerful technique for the calculation of electric fields, which can be used whenever there are identifiable symmetries in the charge and electric field distribution.


enclosed 0Q ε=∫ E dA

Example (1) of a Gauss’s Law field calculation

Given an infinite line charge, density λ: what’s E a distance r away from the wire?

We already know the answer: see last lecture, page 16.

We know that the field must point perpendicularly away from the wire and be cylindrically symmetrical. (How could it be otherwise?)

So if we construct an imaginary cylinder of radius r and height hco-axial with the line, the flux through the circular ends is zero and is uniform on the cylinder walls.


E

r

λ

h

E

http://www.pas.rochester.edu/%7Edmw/phy122/Lectures/Lect_05b.pdf

Example 1 (continued)

Furthermore, as with the sphere, cosθ = 1 everywhere on the cylinder walls, so

The cylinder encloses a charge

So the hs cancel – it doesn’t matter how long the cylinder is – and


enclosed .Q hλ=

cyl. walls

2E dA rhEπ= =∫ ∫E dA

0 02 .

2hrhE E

rλ λπε πε

= ⇒ =

Same result as last time, but easier.

E

r

λ

h

How to use Gauss’s Law

The flux of E which emerges from a closed surface S depends upon how much charge the volume V bounded by this surface contains:

where dA is perpendicular to S and points away from V, and


encl.

0S

Qε

=∫ E dA

( )encl.V

Q dVρ ′= ∫ r

S

dA

E

( )ρ ′r

dV

How to use Gauss’s Law

Can use profitably when

one is given a distribution of electric charge, and

the symmetry and/or extent of the charge distribution makes it clear what the pattern (direction!) of E is.

To use,

draw imaginary closed surface (Gaussian surface) through point at which one wants to know the field.

The Gaussian surface needs to be drawn so as to make the unknown E come out of the flux integral that is, is uniform on S.

This usually means the symmetry of surface matches symmetry of the charge distribution.

Then calculate and solve for E.


;E dA=∫ ∫E dA

encl.Q dVρ= ∫

E = E

Example 2: infinite sheet of charge


x

y

z

z

σ

?=EWe did this via Coulomb’s law on Monday:

What is the electric field a distance z from an infinite plane which carries a uniform charge per unit area σ?

First, coordinates and symmetry:

Let the plane be x-y.

Because the plane is infinite and the charge is spread uniformly, E can’t depend upon x or y.

Nor can E point along a direction other than ±z.


Infinite sheet of charge (continued)

Nor can E be anything but symmetrical about the plane: the field at –z must be equal and opposite to that at +z.

So the Gaussian surface has to be symmetrical about the plane, and have all sides either parallel or perpendicular to the plane. Any shape with these properties will do.


σ


Let’s choose the Gaussian surface with rectangular sides: let its dimensions be a×a×2z, and let the charged plane bisect the a-2z sides.

Because E has to be ⊥ to the plane (cosθ = 0), the flux is zero through all of the vertical, a×z faces.

And because E is uniform on the a×a faces and points the same way as the area vectors (i.e. away from the charged plane; cosθ = 1), the flux through each of these faces is Ea2.


aa

aa

z

zσ

E, dA

E, dA


Now we’re ready to use the formula: the left-hand side is

and because the Gaussian surface intercepts an area a×a on the plane, the right-hand side is

So

(Note: independent of a and z.)


22a a

E dA Ea×

= =∫ ∫E dA

2encl.

0 0

Q aσε ε

=

22

0 02 .

2aEa Eσ σ

ε ε= ⇒ =

aa

aa

z

zσ

E, dA

E, dA


The complete answer is

the same answer we got (on 9 September) with Coulomb’s law, of course.

Note that the process requires more spatial/geometrical reasoning and intuition than the Coulomb’s-law method, but much less calculus.


0

0

ˆ, 0

2ˆ ˆ , 0

2

z

z

σεσε

>= − <

z

Ez z

σ

0ˆ

2σε

=E z

0ˆ

2σε

= −E z


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Today in Physics 122: Gauss’s Law