1. 21. Gauss’s Law Electric Field Lines Electric Flux & Field Gauss’s Law
Using Gauss’s Law
Fields of Arbitrary Charge Distributions
Gauss’s Law & Conductors

2. Huge sparks jump to the operator’s cage, but the operator is unharmed
Huge sparks jump to the operator’s cage, but the operator is unharmed. Why?
Ans: Gauss’ law  E = 0 inside cage

3. Electric Field Lines
Vector gives E at point
Field line gives direction of E
Spacing gives magnitude of E
Electric field lines = Continuous lines whose tangent is everywhere // E.
They begin at + charges & end at  charges or .
Their density is  field strength or charge magnitude.

4. Field Lines of Electric Dipole
Field is strong where lines are dense.
Direction of net field tangent to field line

5. Field Lines

6. Electric Flux & Field
8 lines out of surfaces 1, 2, & 3. But 88 = 0 out of 4.
16 lines out of surfaces 1, 2, & 3. But 0 out of 4.
8 lines out of surfaces 1, 2, & 3. But 0 out of 4.
8 lines out of surfaces 1 & lines out of surface out of 4.
 8 lines out of surface 1.
8 lines out of surface 2.
88 = 0 lines out of surface 3.
1: 4
2: 8
3: 4
4: 0
Count these.
Number of field lines out of a closed surface  net charge enclosed.

7. Electric Flux A flat surface is represented by a vector
where A = area of surface and
Electric flux through flat surface A :
[  ]  N m2 / C.
E, 
Open surface: can get from 1 side to the other w/o crossing surface.
Direction of A ambiguous.
A, 
Closed surface: can’t get from 1 side to the other w/o crossing surface.
A defined to point outward.

8. GOT IT?
The figure shows a cube of side s in a uniform electric field E.
What is the flux through each of the cube faces A, B, and C with the cube oriented as in (a) ?
Repeat for the orientation in (b), with the cube rotated 45 °.

9. Gauss’s Law
Gauss’s law: The electric flux through any closed surface is proportional to the net charges enclosed.
 depends on units.
For point charge enclosed by a sphere centered on it:
SI units 
= vacuum permittivity
Field of point charge:
Gauss’s law:

10. Gauss & Coulomb For a point charge: E  r 2 A  r 2   indep of r.
Outer sphere has 4 times area.
But E is 4 times weaker.
So  is the same
For a point charge:
E  r 2
A  r 2
  indep of r.
Principle of superposition  argument holds for all charge distributions
Gauss’ & Colomb’s laws are both expression of the inverse square law.
For a given set of field lines going out of / into a point charge,
inverse square law  density of field lines  E in 3-D.

11. GOT IT?
A spherical surface surrounds an isolated positive charge, as shown.
If a 2nd charge is placed outside the surface,
which of the following will be true of the total flux though the surface?
it doesn’t change;
it increases;
it decreases;
it increases or decreases depending on the sign of the second charge.
Repeat for the electric field on the surface at the point between the charges.
Opposite charges
Same charges

12. Using Gauss’s Law
Useful only for symmetric charge distributions.
Spherical symmetry:
( point of symmetry at origin ) 

13. Example 21.1. Uniformily Charged Sphere
A charge Q is spreaded uniformily throughout a sphere of radius R.
Find the electric field at all points, first inside and then outside the sphere.  
True for arbitrary spherical (r).

14. Example 21.2. Hollow Spherical Shell
A thin, hollow spherical shell of radius R contains a total charge of Q. distributed uniformly over its surface.
Find the electric field both inside and outside the sphere. 
Contributions from A & B cancel.

15. GOT IT?
A spherical shell carries charge Q uniformly distributed over its surface.
If the charge on the shell doubles, what happens to the electric field
inside and
outside
the shell?
Stays 0.
Doubles.

16. Example 21.3. Point Charge Within a Shell
A positive point charge +q is at the center of a spherical shell of radius R carrying charge 2q, distributed uniformly over its surface.
Find the field strength both inside and outside the shell. 

17. Tip: Symmetry Matters
Spherical charge distribution inside a spherical shell is zero  E = 0 inside shell
E  0 if either shell or distribution is not spherical.
Q = qq = 0
But E  0 on or inside surface

18. Line Symmetry Line symmetry:
r = perpendicular distance to the symm. axis.
Distribution is indpendent of r//  it must extend to infinity along symm. axis. 

19. Example 21.4. Infinite Line of Charge
Use Gauss’ law to find the electric field of an infinite line charge carrying charge density  in C/m.
(radial field)
No flux thru ends 
c.f. Eg. 20.7 
True outside arbitrary radial (r).

20. Example A Hollow Pipe
A thin-walled pipe 3.0 m long & 2.0 cm in radius carries a net charge q = 5.7 C distributed uniformly over its surface.
Fine the electric field both 1.0 cm & 3.0 cm from the pipe axis, far from either end. 
at r = 1.0 cm
at r = 3.0 cm

21. Plane Symmetry Plane symmetry:
r = perpendicular distance to the symm. plane.
Distribution is indpendent of r//  it must extend to infinity in symm. plane. 

22. Example 21.6. A Sheet of Charge
An infinite sheet of charge carries uniform surface charge density  in C/m2.
Find the resulting electric field.  
E > 0 if it points away from sheet.

23. Basic High Symmetry Fields
Dipole : E  r 3
Point charge : E  r 2
Line charge : E  r 1
Surface charge : E  const

24. 21.5. Fields of Arbitrary Charge Distributions
Field of charged disk: guesswork
infinite-plane-charge-like
point-charge-like

25. GOT IT? 21.4. A uniformly charged sheet measures 1m on each side,
and you are told the total charge Q.
What expression would you use to get approximate values for the field magnitude
1 cm from the sheet (but not near an edge) and
1 km from the sheet?
(a)
(b)

26. 21.6. Gauss’s Law & Conductors
Electrostatic Equilibrium
Conductor = material with free charges
E.g., free electrons in metals.
Neutral conductor
Uniform field
External E  Polarization
 Internal E
Total E = 0 : Electrostatic equilibrium
( All charges stationary )
Induced polarization cancels field inside
Microscopic view: replace above with averaged values.
Net field

27. Charged Conductors
Excess charges in conductor tend to keep away from each other
 they stay at the surface.
 = 0 thru this surface
More rigorously:
Gauss’ law with E = 0 inside conductor
 qenclosed = 0
 For a conductor in electrostatic equilibrium, all charges are on the surface.

28. Example 21.7. A Hollow Conductor
An irregularly shaped conductor has a hollow cavity.
The conductor itself carries a net charge of 1 C,
and there’s a 2 C point charge inside the cavity.
Find the net charge on the cavity wall & on the outer surface of the conductor, assuming electrostatic equilibrium.
E = 0 inside conductor
 = 0 through dotted surface
qenclosed = 0
Net charge on the cavity wall qin = 2 C
qout
qin
+2C
Net charge in conductor = 1 C = qout + qin
 charge on outer surface of the conductor qout = +3 C

29. GOT IT? 21.5. A conductor carries a net charge +Q.
There is a hollow cavity inside the conductor that contains a point charge Q.
In electrostatic equilibrium, is the charge on the outer surface of the conductor
2Q Q Q 2Q

30. Experimental Tests of Gauss’ Law
Measuring charge on ball is equivalent to testing the inverse square law.
The exponent 2 was found to be accurate to 1016 .

31. Application: Shielding & Lightning Safety
Coaxial cable
High frequency EM field easily blocked by metal (skin effect).

32. Field at a Conductor Surface
At static equilibrium,
E = 0 inside conductor,
E = E at surface of conductor.
Gauss’ law applied to pillbox surface: 
The local character of E ~  is incidental.
E always dependent on ALL the charges present.

33. Dilemma? E outside charged sheet of charge density  was found to be
E just outside conductor of surface charge density  is
What gives?
Resolution:
There’re 2 surfaces on the conductor plate.
The surface charge density on either surface is .
Each surface is a charge sheet giving E = /20.
Fields inside the conductor cancel, while those outside reinforce.
Hence,
outside the conductor

34. Plate Capacitor
inside
outside
Charge on inner surfaces only