1. Electric Field
Define electric field, which is independent of the test charge, q, and depends only on position in space:
dipole
One is > 0, the other < 0
electric dipole of dipole moment: -q - q +

2. Dipole in uniform electric fields
No net force. The electrostatic forces on the constituent point charges are of the same magnitude but along opposite directions. So, there is no net force on the dipole and thus its center of mass should not accelerate.
Net torque! There is clearly a net
torque acting on the dipole with respect to its center of mass, since the forces are not aligned.

3. Electric Field from Coulomb’s Law+ -
Bunch of Charges
Continuous Charge Distribution dq P P k
(volume charge)
(surface charge)
(line charge)
Summation over discrete charges
Integral over continuous charge distribution

4. Reading Quiz 1 Which one of the following statements is incorrect ?
A) Electric fields leave positive charges and end on negative charges
B) Electric field lines can intersect at some points in space.
C) Electric field field lines from a dipole fall off faster than 1/r2.
D) Electric fields describe a conservative force field.

5. SUMMARY: FIND THE ELECTRIC FIELD
GIVEN THE CHARGES
1) GEOMETRY FOR qi or dqi
2) DISCRETE CHARGES qi
3) CONTINUOUS CHARGES dqi
line charge density λ (x)
surface charge density σ (x.y)
volume charge density ρ (x,y,z)
Geometry may suggest other coordinate systems, R,θ,Φ or R,θ,Z

6. Continuous Charge Distribution 1 Charged Line
At a point P on axis:
Ex = k λ (1/r1 – 1/r2 ) ( Eqn 22-2a)
Ex = k λ (1/( XP – L/2 ) - 1/( XP + L/2 )
Ex = k ( Q/L) L [ XP2 – (L/2)2 ] -1
For XP2 >> (L/2) Ex = k Q / XP2
For XP = Ex = 0

7. Again: Continuous Charge Distribution 1: Charged Line
At a point P on perpendicular axis:  x

8. Physics 241 –Warm-up quiz 2
The rod is uniformly charged with a positive charge density . What is the direction of the electric field at a point P on a line perpendicular to the rod? Note that the line and the rod are in the same plane.
to the right
to the left
up
down
lower right
 p

9. General location of P: Charged Line
At a point P off axis:

10. Continuous Charge Distribution 2: Charged Ring
At point P on axis of ring: ds
Use symmetry! 
Ex = k Q x ( x2 + a2 )-3

11. Continuous Charge Distribution 3: Charged Disk
At a point P on axis:
Use the ring with radius a EX value
dEx = k dq x ( x2 + a2 ) -3
Integrate rings from 0 to R
Ex = -2πσkx ( 1/( x2 + R2 )1/2 – 1/x )
Superposition of rings!
E = σ/2εo
<= Independent of x

12. Continuous Charge Distribution 4: Charged Sheets
E=const in each region
Superposition!
Capacitor geometry

13. MULTIPLE CHARGE SHEET EXAMPLE DOCCAM 2

14. Gauss’s Law: Qualitative Statement
Form any closed surface around charges
Count the number of electric field lines coming through the surface, those outward as positive and inward as negative.
Then the net number of lines is proportional to the net charges enclosed in the surface.

15. General definition of electric flux:
To state Gauss’s Law in a quantitative form, we first need to define Electric Flux.
# of field lines N
= density of field lines
x “area”
where “area” = A2 x cos  
Sum over surface
General definition of electric flux:
(must specify sense, i.e., which way)

16. Electric Flux through Closed Surface
The integral is over a CLOSED surface.
Since is a scalar product, the electric flux is a SCALAR quantity
The integration element is a vector normal to the surface and points OUTWARD from the surface. Out is +, In is -
E proportional to # field lines coming through outward

17. Why are we interested in electric flux?
is closely related to the charge(s) which cause it.
Consider Point charge Q
If we now turn to our previous discussion and use the analogy to the number of field lines, then the flux should be the same even when the surface is deformed. Thus should only depend on Q enclosed.

18. Gauss’s Law: Quantitative Statement
The net electric flux through any closed surface equals the net charge enclosed by that surface divided by 0.
How do we use this equation??
The above equation is TRUE always but it doesn’t
look easy to use.
BUT - It is very useful in finding E when the physical situation exhibits a lot of SYMMETRY.

19. Physics 241 – 10:30 Quiz 3
The left half of a rod is uniformly charged with a positive charge density , whereas the right half is uniformly charged with a charge density of . What is the direction of the electric field at a point on the perpendicular bisector and above the rod as shown?
to the right
to the left
up
down
E is zero.



20. Physics 241 – 11:30 Quiz 3
The upper half of a ring is uniformly charged with a positive charge density , whereas the lower half is uniformly charged with a charge density of . What is the direction of the electric field at a point on the perpendicular axis and to the left of the ring as shown?

to the right
to the left
up
down
E is zero.


22. Gauss’ Law - Examples
Shell Theorem
Outside shell: E is as if Q at center
Inside: E is zero

23. Proof of the Shell Theorem
Electric Field Outside
By symmetry, the electric field
must only depend on r and is
along a radial line everywhere.
Apply Gauss’s law to the blue surface , we get

24. Uniformly charged thin shell: Inside
E = 0 inside
By symmetry, the electric field
must only depend on r and is
along a radial line everywhere.
Apply Gauss’s law to the blue surface , we get E = 0.
Equal and opposite contributions from charges on diagonally opposite surface elements.
Discontinuity in E

25. Electric Field of a Uniformly Charged Sphere
Apply Gauss’s Law directly or use superposition of the shell results

26. Physics 241 –Quiz 2a
Two identical point charges are each placed inside a large cube. One is at the center while the other is close to the surface. Which statement about the net electric flux through the surface of the cube is true?
The flux is larger when the charge is at the center.
The flux is the same (and not zero).
The flux is larger when the charge is near the surface.
Not enough information to tell.
The flux is zero in both cases. +Q +Q

27. Physics 241 –Quiz 2b
Two identical point charges are each placed inside a large sphere. One is at the center while the other is close to the surface. Which statement about the net electric flux through the surface of the sphere is true?
The flux is larger when the charge is at the center.
The flux is larger when the charge is near the surface.
The magnitude of the flux is the same (and not zero).
Not enough information to tell.
The flux is zero in both cases. +Q +Q

28. Physics 241 –Quiz 2c
Two identical point charges are placed, at the center of a large sphere in one case, and outside an identical sphere in the other case. Which statement about the net electric flux through the surface of the sphere is true?
The flux is larger when the charge is inside.
The flux is larger when the charge is outside.
The flux is the same (and not zero).
Not enough information to tell.
The flux is zero in both cases.
Q>0
Q>0