1. Physics 121 - Electricity and Magnetism Lecture 04 - Gauss’ Law Y&F Chapter 22 Sec. 1 – 5
Flux Definition (fluid flow example)
Gaussian Surfaces
Flux of an Electric Field
Flux Examples
Gauss’ Law
Gauss’ Law Near a Dipole
A Charged, Isolated Conductor
Spherical Symmetry: Conducting Shell with Charge Inside
Cylindrical Symmetry: Infinite Line of Charge
Field near an infinite Non-Conducting Sheet of Charge
Field near an infinite Conducting Sheet of Charge
Conducting and Non-conducting Plate Examples
Proof of Shell Theorem using Gauss Law
Examples
Summary

2. Evaluate integrand at all points on surface S
Gauss’ Law depends on the idea of Flux (symbol F): Basically vector field x area
In a fluid: flux measures volume or mass flow per second.
Definition: differential amount of flux dF of field v crossing vector area dA v
“unit normal”
outward and
perpendicular to
surface dA
“Phi”
Dimensions: Volume/unit time
Flux through a closed or open surface S: calculate “surface integral” of field over S
Evaluate integrand at all points on surface S
EXAMPLE : FLUID FLUX THROUGH A CLOSED, EMPTY, RECTANGULAR BOX IN A UNIFORM VELOCITY FIELD
No sources or sinks of fluid inside
DF from each side = 0 since v.n = 0, DF from ends cancels
TOTAL F = 0
Example also applies to gravitational or electric field v
What do field lines look like if a flux source or sink (or mass, or charge) is in the box? Can field be uniform? Can net flux be zero?

3. MORE ON FLUID FLUX: WATER FLOWING ALONG A STREAM
Assume:
fluid with constant mass density (incompressible)
constant flow velocity parallel to banks
no turbulence (smooth laminar flow)
Flux measures the flow (current):
flow means quantity/unit time crossing area
Two related fluid flow fields (currents/unit area):
Velocity field v is volume flow across area/unit time
Mass flow field J is mass flow across area/unit time
Flux interpretation = volume of fluid crossing an area per unit time
and
Chunk of
fluid moves distance DL
in time Dt: A J v t L m
lux f
mass V
Mass of chunk r o D = \
Continuity: Net flux (fluid flow) through a closed surface = 0
………unless a source or sink is inside

4. Flux is a SCALAR, Units: Nm2/C.
Applied to Electric Field:
Gaussian Surfaces are
closed 3D surfaces
Choose surface to match symmetry of the field or charge distribution where possible
Field lines cross a closed surface:
An odd number of times
for charges that are inside S
An even number of times
for charges outside of S
sphere
no end caps
not closed
not a GS
cylinder with end caps
closed box
The flux of electric field crossing a closed surface equals the net charge (source or sink) inside the surface (times a constant).
Example: Point charge at center of a spherical “Gaussian surface”
Principle holds for arbitrary charges and closed surfaces.
Flux is a SCALAR, Units: Nm2/C.

5. Electric Flux: Integrate electric field over a surface
Definition: flux dE crossing (vector) area dA E
“unit normal”
outward and
perpendicular to
surface dA
Divide up surface S into tiny chunks of dA each and consider one of them
Gauss’ Law: The flux through a closed surface S depends only on the net enclosed charge, not on the details of S or anything else.
Flux through any surface S (closed or open): Integrate on S
To do this: evaluate integrand at all points on surface S

6. DF depends on the angle between the field and chunks of area
E.n = 0 ZERO
tangent to surface
E.n < 0 NEGATIVE
from outside to inside of surface
E.n > 0 POSITIVE
from inside to outside of surface
FLUX F IS
WHEN E POINTS
DF=0

7. SUM Over Small chunks of surface DA
Can evaluate flux through open or closed surfaces
If field is constant across flat pieces of a surface, sum up
a finite number of flux terms instead of integrating:
SUM Over Small chunks of surface DA
Uniform E field everywhere
E along +x axis
Cube faces are normal to axes
Each side has area length2
Field lines cut through two surface areas and are tangent to the other four surface areas
For side 1, F = -E.l2
For side 2, F = +E.l2
For the other four sides, F = 0
Therefore, Ftotal = 0
EXAMPLE: Flux through a cube
Assume:
What if the cube is oriented obliquely??
How would flux differ if net charge is inside??

8. Flux through a cylinder in a uniform electric field
Closed Gaussian surface
UNIFORM E means zero
enclosed charge
Calculate flux directly
Symmetry axis along E
Break into areas a, b, c
Cap a:
Cap c:
Area b:
What if E is not parallel to cylinder axis:
Geometry is more complicated...but...
Qinside = 0 so F = 0 still !!

9. Flux of an Electric Field
4-1: Which of the following figures correctly shows a positive electric flux out of a surface element? I.
II.
III.
IV.
I and III. q E DA q E DA
I II.
III IV. q E DA q E DA

10. See Divergence Theorem,
Gauss’ Law Statement
Qenc is the NET charge enclosed by a (closed) Gaussian surface S. The net flux F through the surface is Qenc/e0
See Divergence Theorem,
for the ambitious
F does not depend on the shape of the surface, assumed closed.
Ignore charge outside the surface S.
Surface integral yields 0 if E = 0 everywhere on surface
Example: Derive point charge formula (Coulomb’s Law) from Gauss’ Law
Point charge is at center of spherical Gaussian surface, radius r. r
Qenc
Coulomb’s Law

11. Shell Theorem: spherically symmetric shell of charge
What are the fields on Gaussian surfaces inside and outside shell?
Use Gauss’ Law and spherical symmetry
Spherical symmetry makes the flux integral trivial +Q
Points Outside
Above yields point charge field formula, so field of shell of charge mimics point charge field at locations outside. +Q
Points Inside:
Above implies E = 0 inside, since Qenc is zero. So field due to shell of charge vanishes everywhere inside the shell.

12. 4-2: What is the flux through the Gaussian surface below?
zero
-6 C./e0
-3 C./e0
+3 C./e0
not enough info

13. + - Example: Flux through surfaces near a dipole FLUX POSITIVE from S1
FLUX NEGATIVE
from S2
FLUX = 0
through S4
through S3 + - S1 S2 S3 S4
Equal positive (+Q) and negative (–Q) charges
Consider (closed) Gaussian surfaces S1...S4.
Surface S1 encloses only the positive charge. The field is everywhere outward. Positive flux. F = +Q/e0
Surface S2 encloses only the negative charge. The field is everywhere inward. Negative flux. F = -Q/e0
Surface S3 encloses no charges. Net flux through the surface is zero. The flux is negative at the upper part, and positive at the lower part, but these cancel. F = 0
Surface S4 encloses both charges. Zero net charge enclosed, so equal flux enters and leaves, zero net flux through the surface. F = 0

14. Where does net charge stay on an isolated conductor?
Charge flows until E = 0 at every
interior point (screening)
At equilibrium E = 0 everywhere
inside a conductor
Place a net charge q initially in the
interior of a conductor…where…
Charges can move, but can not leak off
Choose Gaussian surface S just below surface: E = 0 everywhere on it
Use Gauss’ Law:
The only place where un-screened charge can end up is on the outside surface of the conductor
E at the surface is everywhere normal to it; if E had a component parallel to the surface, charges would move to screen it out.
Just outside surface, E = s/e0 (E for infinitely large conducting plane)

15. All net charge on a conductor moves to the outer surface E=0
Gauss’s Law: net charge on an isolated conductor moves to the outside surface
1: SOLID CONDUCTOR WITH NET CHARGE ON IT S
All net charge on a conductor
moves to the outer surface
E=0
2. HOLLOW CONDUCTOR WITH A NET CHARGE, NO CHARGE IN CAVITY
There is zero net charge on the inner
surface: all net charge is on outer surface S’
Choose Gaussian surface S’ just outside the cavity
E=0 everywhere on S’, so the surface integral is zero
Gauss’ Law says:
E=0
Does this mean that the charge density is zero at every point on the inner surface?

16. Hollow conductor with cavity, charge inside
Electrically neutral, conducting shell
Arbitrary charge distribution +Q in cavity
Choose Gaussian surface S completely within
the conductor + Q -
Gaussian Surface S
E = 0 everywhere on S, so FS = 0  qenc= 0
Charge Qinner is induced on inner surface,
distributed so that E = 0 in the metal, hence…
The shell is neutral, so Qouter= -Qinner = +Q appears on outer surface
Logic above holds for non-spherical as well as spherical shells.
IF SHELL IS SPHERICAL, field outside is spherically symmetric:
Choose another spherical Gaussian surface S’ outside the spherical shell
Gauss Law:
Whatever the inside distribution may be, outside the shell it is shielded.
Qouter is uniform if the shell is spherical. The field outside then looks like that of a point charge at the center.

17. Conducting spherical shell with charge inside
4-3: Place a charge inside a cavity in an isolated neutral conductor. Which statement is true?
E field is still zero in the cavity.
E field is not zero in the cavity, but it is zero in the conductor.
E field is zero outside the conducting sphere.
E field is the same everywhere as if the conductor were not there (i.e. radial outward everywhere).
E field is zero in the conductor, and negative (radially inward) outside the conducting sphere.
Spherical
cavity + Q
Positive
charge
Conducting
sphere
(neutral)

18. Gauss’ Law Example: Find the electric field formula at a distance r from an infinitely long (thin) line of charge z
Cylindrical symmetry around z-axis
Uniform charge per unit length l along line
Every point on the infinite line has the
identical surroundings, so….
E is radial, by symmetry (perpendicular to line charge)
E has the same magnitude everywhere on
any concentric cylindrical surface
Flux through end caps of surface = 0
as E is perpendicular to DA there
E on cylinder is parallel to unit vectors for DA
So…total flux is just cylinder area x a constant E
Trivial integration due to Gauss Law
Good approximation for finite line of
charge when r << L, far from the ends.
Cylinder area = 2prh

19. Field Lines and Conductors
4-4: The drawing shows cross-sections of three cylinders with different radii, each having the same total charge. Each cylindrical gaussian surface has the same radius (again, shown in cross-section). Rank the three arrangements according to the electric field at the Gaussian surface, greatest first.
I, II, III
III, II, I
All tie.
II, I, III
II, III, I
I II III.

20. Use symmetry arguments where applicable
4-5: Outside a sphere of charge the electric field mimics that of a point charge of
the same total charge, located at its center……
Outside of an infinitely long, uniformly charged conducting
cylinder, which statement below describes the electric field?
The charge per unit area on the cylinder surface is s.
E is like the field of a point charge at the center of the cylinder.
E is like the field of a circular ring of charge at its center.
E is like the field of an infinite line of charge along the cylinder axis.
Cannot tell from the information given.
The E field equals zero s

21. Use Gauss’ Law to find the Electric field near an infinite non-conducting sheet of charge
s is a uniform positive charge per unit area
Choose cylindrical gaussian surface penetrating
sheet (rectangular is OK too)
E remains constant as a chosen point moves
parallel to the surface
E points radially away from the sheet (both sides),
E is perpendicular to cylindrical tube of gaussian
surface.  Flux through it = 0
On both end caps E is parallel to A
 F = + EAcap on each
Uniform field – independent of distance from sheet
Same as near field result for charged disk but no integration needed
Good approximation for finite sheet when r << L, far from edges.

22. Electric field near an infinite conducting sheet of charge – using Gauss’ Law
F=EA
F= 0 r
Uniform charge per unit area s on one face of sheet
E points radially away from sheet outside metal
otherwise surface current would flow (!!)
Use cylindrical or rectangular Gaussian surface
End caps just outside and just inside (E = 0)
Flux through the cylindrical tube = 0 as
E is normal to surface
On left cap (inside conductor) E = 0 so F = 0
On right cap E is parallel to DA so F = EAcap
Field is twice that for a non-conducting sheet with same s
Same enclosed charge, same total flux now “squeezed” out
the right hand cap, not both
Otherwise like previous result: uniform, no r dependence, etc.

23. Example: Fields near parallel nonconducting sheets - 1
Two “large” nonconducting sheets of charge are near each other,
Use infinite sheet results
The charge cannot move, use superposition.
There is no screening, as there would be in a conductor.
Each sheet produces a uniform field of magnitude:
Oppositely Charged Plates, same |s| but opposite sign s -s - +
Left region: Field due to the positively charged sheet is canceled by the field due to the negatively charged sheet. Etot is zero.
Right region: Same argument. Etot is zero.
Between plates: Fields reinforce. Etot is twice E0 and to the right.

24. Example: Fields near parallel nonconducting sheets - 2
Positively Charged Plates, same |s| same sign +s +
Now, the fields reinforce to the left and to the right of both plates.
Between plates, the fields cancel.
Signs are reversed for a pair of negatively charged plates

25. Charge on a finite sized but thin conducting plate, isolatedor
E = 0 inside the conductor. Charge density s1 becomes the same on
both faces, o/w charges will move to make E = 0 inside
Cylindrical Gaussian surfaces S, S’  charges distribute on surfaces
Same field on opposite sides of plate, opposite directions
Same field if replace plate with 2 charged non-conducting sheets alone:
same s1
same cancellation inside conductor
reinforcement outside

26. Details: proof of the Shell Theorem using Gauss’ Law (depends on spherical symmetry)
Hollow shell, net charge Q
Spherically symmetric surface charge
density s, radius R
Two spherical Gaussian surfaces:
- S1 is just inside shell, r1 < R
- S2 is outside shell, r2 > R
qenc means charge enclosed by S1 or S2 S1 S2 R r1 r2
OUTSIDE:
(on S2)
Shell of charge acts like a point charge at the center of the sphere
INSIDE:
(on S1)
Shell of charge creates zero field inside (anywhere)
At point r inside a spherically symmetric volume charge distribution (radius R):
only shells of charge with radii smaller than r contribute to E as point charges.
shells with radius between r and R produce zero field inside.