Gauss’s Law

German Mathematician Born: 30 April 1777 Braunschweig Died: 23 February 1855 Göttingen

Johann Carl Friedrich Gauss

He contributed significantly to many fields: Number theory, algebra, statistics, analysis, differential geometry, geodesy, geophysics, mathematics, electrostatics, astronomy, matrix theory, and optics Princeps mathematicorum : "the foremost of mathematicians"

Ref: Wikipedia

Flux of a Vector: Electric Flux Flow rate of a quantity across a unit area is called flux, or you can say that flux is surface bombardment rate.

Solid angle is the 3-D equivalent of the 2-D angle.

! = lR ! = A

R2

24-1 (SJP, Phys 1120)

Electric flux, and Gauss' law

Finding the Electric field due to a bunch of charges is KEY! Once you know

E, you know the force on any charge you put down - you can predict (or

control) motion of electric charges! We're talking manipulation of anything

from DNA to electrons in circuits... But as you've seen, it's a pain to start from

Coulomb's law and add all those darn vectors.

Fortunately, there is a remarkable law, called Gauss' law, which is a universal

law of nature that describes electricity. It is more general than Coulomb's law,

but includes Coulomb's law as a special case. It is always true... and

sometimes VERY useful to figure out E fields! But to make sense of it, we

really need a new concept, Electric Flux (Called !). So first a "flux interlude":

Imagine an E field whose field lines "cut through" or "pierce" a loop.

Define " as the angle between E and the "normal"

or "perpendicular" direction to the loop.

We will now define a new quantity, the electric

flux through the loop, as

Flux, or ! = E# A = E A cos"

E# is the component of E perpendicular to the loop:

E# = E cos".

For convenience, people will often characterize the area of a small patch (like

the loop above) as a vector instead of just a number.

The magnitude of the area vector is just... the area! (What else?)

But the direction of the area vector is the normal to the loop.

That way, we can write .

(Can you see that this just gives the formula we had above?)

Flux is a useful concept, used for other quantities besides E, too. E.g. if you have solar

panels, you want the flux of sunlight through the panel to be large. House #2 has poorly

designed panels. Although the AREA of the

panels is the exact same, and the sunshine

brightness is the exact same, panel 2 is less

useful: fewer light rays "pierce" the panel, there

is less FLUX through that panel.

House Solar panel

(Lots of flux)

(Less flux)

Solar panel 2,

same area,

diffferent tilt.

1 2

Flux is a useful concept and can be used for other quantities besides E. For example if you have solar panels, you want the flux of sunlight through the panel to be large. House #2 has poorly designed panels. Although the AREA of the panels is exactly the same for house #1 and house #2, and the sunshine brightness is exactly the same, panel 2 is less useful: fewer light rays "pierce" the panel, therefore, there is less FLUX through that panel.

Flux

Electric Flux

Electric flux through an area is proportional to the total number of field lines crossing the area.

!E = E"A = EA" = EAcos# !E =!E "A"!

Flux through a closed surface !E =

!E "d!A"#

!E "

!Ei # $

!Ai

i=1

n

%

Uniform Electric Field

or

Electric Flux & Gauss’s Law Gauss’s law is equivalent to Coulomb’s law, but it has advantages:

!! Makes it easy to evaluate the electric fields of certain symmetries.

!! Provides clear insight into certain basic properties of the electric field of any charge distribution.

!E =!E "d!A"# = Qencl.

!0

“The total flux through a closed surface is equal to the total (net) electric charge inside, divided by .” !0

The flux through the planar surface below (positive unit normal to left)

1)! is positive.

2)! is negative.

3)! is zero.

Question (1)

Question (2)

The flux through the surfaces S1, S2, and S3 are:

1) !S1>!S2

>!S3.

2) !S3>!S2

>!S1.

3) !S3= !S2

= !S1.

4) is zero.

Question (3) Calculate the electric flux through the rectangle shown. The rectangle is a 10cm by 20cm, the electric field is uniform at 200 N/C, and the angle ! is 30o.

Electric Field of a point Charge

F = Qq4!"0r

2

If another charge q is placed at r, it would experience a force

It is consistent with Coulomb’s law.

d! =!E "d!A

! =!E "d!A

sphere"# = E dA

sphere"# = Q

$0

! = E(4%r2 ) = Q$0

E = Q4%$0r

2 Electric field at radius r

!E !d!A

cyl ."" = Q

#0

=!E1 !d

!A1

1""

= 0#$% &%

+!E2 !d

!A2

2""

= 0# $% &%

+!E3 !d

!A3

3"" = Q

#0

!E !d!A

cyl ."" =

!E3 !d

!A3

3"" = Er ! r dA

cyl ."" = Q

#0

E dAcyl ."" = Q

#0

$ E 2%rl( ) = Q#0

E = 12%rl

Q#0

=Ql

2%#0

1r

& = Ql

!E = &

2%#0

1rr

Electric Field of Line Charge

Electric Field of a Uniform Charged Plane

!E = ±Ez

!E =!E "d!A

S"# = E dA

S"# = EAEndcaps

= E(2A) = Q$0

= % A$0

E = %2$0

!E = %

2$0

z upward& z downward

'()

Question (4) Two point charges, +q (in red) and –q (in blue), are arranged as shown. Through which closed surface(s) is the net electric flux equal to zero?

(a)! surface A (b)! Surface B (c)! Surface C (d)! Surface D (e)! both surface C and surface D

Thin Conductor Spherical Shell

The electric field inside a conductor must be zero, the net charge enclosed by the Gaussian surface shown in Figure is zero.

Example: Point Charge.

For an isolated point charge Q, any sphere surrounding the charge contains the same net

charge Q(r) = Q, hence eq. (8) reproduces the Coulomb Law,

E(r) =kQ

r2. (9)

Example: Thin Spherical Shell.

Now consider a thin spherical shell of radius R and uniform surface charge density

! =dQ

dA=

Qnet

4"R2. (10)

For this shell, a Gaussian sphere of radius r < R contains no charge at all, while a Gaussian

sphere of radius r > R contains the whole charge Qnet of the shell, thus

Q(r) =

!

0 for r < R,

Qnet for r > R,(11)

and therefore

E(r) =

"

#

$

0 for r < R,

kQnet

r2for r > R,

%

&

'

r

E(r)

(12)

In other words, inside the shell there is no electric field, but outside the shell the electric field

is the same as if the whole charge was at the center.

3

Example: Point Charge.

For an isolated point charge Q, any sphere surrounding the charge contains the same net

charge Q(r) = Q, hence eq. (8) reproduces the Coulomb Law,

E(r) =kQ

r2. (9)

Example: Thin Spherical Shell.

Now consider a thin spherical shell of radius R and uniform surface charge density

! =dQ

dA=

Qnet

4"R2. (10)

For this shell, a Gaussian sphere of radius r < R contains no charge at all, while a Gaussian

sphere of radius r > R contains the whole charge Qnet of the shell, thus

Q(r) =

!

0 for r < R,

Qnet for r > R,(11)

and therefore

E(r) =

"

#

$

0 for r < R,

kQnet

r2for r > R,

%

&

'

r

E(r)

(12)

In other words, inside the shell there is no electric field, but outside the shell the electric field

is the same as if the whole charge was at the center.

3

Solid Charged Spherical Ball

Conductor with a cavity

The electric field inside a conductor must be zero, the net charge enclosed by the Gaussian surface shown in Figure must be zero.

This implies that a charge –q must have been induced on the cavity surface. Since the conductor itself has a charge +Q, the amount of charge on the outer surface of the conductor must be Q + q.

Summary 1)! Look at Figure.

2)! Identify the symmetry.

3)! Determine the direction of the E-field.

4)! Divide the space into different regions.

5)! Choose Gaussian surface.

6)! Calculate electric flux.

7)! Calculate enclosed charge.

8)! Apply Gauss’s law to find E-field