1. Electric Fields Electric Fluxa R a 2R

2. But how would we calculate this??R x y +
Consider a circular ring with total charge +Q. The charge is spread uniformly around the ring, as shown, so there is λ = Q/2pR charge per unit length.
The electric field at the origin is
(a) zero
(b)
(c)
But how would we calculate this??

3. Electric Fields from Continuous Charge Distributions
Examples:
line of charge
charged plates
electron cloud in atoms, … r
E(r) = ?
Principles (Coulomb’s Law + Law of Superposition)
remain the same.
Only change:

4. a) up b) down c) left d) right e) up and left f) up and right
Preflight 3: B A L
2) A finite line of positive charge is arranged as shown. What is the direction of the Electric field at point A?
a) up b) down c) left d) right
e) up and left f) up and right
3) What is the direction of the Electric field at point B?
a) up b) down c) left d) right
e) up and left f) up and right

5. small pieces of charge dq
Charge Densities
How do we represent the charge “Q” on an extended object?
total charge Q
small pieces of charge dq
Line of charge:
l = charge per unit length
dq = l dx
Surface of charge:
s = charge per unit area
dq = s dA
Volume of Charge:
r = charge per unit volume
dq = r dV

6. How We Calculate (Uniform) Charge Densities:
Take total charge, divide by “size”
Examples:
10 coulombs distributed over a 2-meter rod.
14 pC (pico = 10-12) distributed over a shell of radius 1 μm.
14 pC distributed over a sphere of radius 1 mm.

7. Electric field from an infinite line charge r
E(r) = ?
Approach:
“Add up the electric field contribution from each bit of
charge, using superposition of the results to get the final field.”
In practice:
Use Coulomb’s Law to find the E-field per segment of charge
Plan to integrate along the line…
x: from -¥ to +¥ OR q : from -p/2 to +p/2
Any symmetries ? This may help for easy cancellations q x

8. Infinite Line of Charge x y dx r' r q dE
Charge density = l
We need to add up the E-field contributions from all segments dx along the line.

9. Infinite Line of Charge x y dx r' r q dE
To find the total field E, we must integrate over all charges along the line. If we integrate over q, we must write r’ and dq in terms of q and dq .
The electric field due to dq is:
Solution: After the appropriate change of variables, we integrate and find:
*the calculation is shown in the appendix

10. Infinite Line of Charge x y dx r' r q dE
Conclusion:
The Electric Field produced by an
infinite line of charge is:
- everywhere perpendicular to the line
- is proportional to the charge density
- decreases as
- next lecture: Gauss’ Law makes this trivial!!

11. Summary Electric Field Lines Electric Field Patterns
Dipole
~ 1/R3
Coming up:
Electric field Flux
and
Gauss’ Law
Point Charge
~ 1/R2
~ 1/R
Infinite
Line of Charge

12. The Story Thus Far
Two types of electric charge: opposite charges attract, like charges repel
Coulomb’s Law:
Electric Fields
Charges respond to electric fields:
Charges produce electric fields:

13. The Story Thus Far
We want to be able to calculate the electric fields from various charge arrangements. Two ways:
1. Brute Force: Add up / integrate contribution from each charge.
Often this is pretty difficult.
Ex: electron cloud around nucleus
Ack!
2. Gauss’ Law: The net electric flux through any closed surface is proportional to the charge enclosed by that surface.
In cases of symmetry, this will be MUCH EASIER than the brute force method.

14. Examine the electric field lines produced by the charges
in this figure.
Which statement is true?
(a) q1 and q2 have the same sign
(b) q1 and q2 have the opposite signs and q1 > q2
(c) q1 and q2 have the opposite signs and q1 < q2 q1 q2

15. Examine the electric field lines produced by the charges
in this figure.
Which statement is true?
(a) q1 and q2 have the same sign
(b) q1 and q2 have the opposite signs and q1 > q2
(c) q1 and q2 have the opposite signs and q1 < q2 q1 q2
Field lines start from q2 and terminate on q1.
This means q2 is positive; q1 is negative; so, … not (a)
Now, which one is bigger?
Notice along a line of symmetry between the two, that the E-field
still has a positive y component. If they were equal, it would be zero;
This indicates that q2 is greater than q1

16. Electric Dipole: Lines of Force
Consider imaginary spheres centered on : c
c) midpoint (yellow) a
a) +q (blue) b
b) -q (red)
All lines leave a)
All lines enter b)
Equal amounts of leaving and entering lines for c)

17. Electric Flux
Flux:
Let’s quantify previous discussion about field-line “counting”
Define: electric flux FE through the closed surface S
“S” is surface
of the box

18. Flux How much of something is passing through some surface
Ex: How many hairs passing through your scalp.
Two ways to define
Number per unit area (e.g., 10 hairs/mm2)
This is NOT what we use here.
Number passing through an area of interest
e.g., 48,788 hairs passing through my scalp.
This is what we are using here.

19. Electric Flux What does this new quantity mean?
The integral is over a CLOSED SURFACE
Since is a SCALAR product, the electric flux is a SCALAR quantity
The integration vector is normal to the surface and points OUT of the surface is interpreted as the component of E which is NORMAL to the SURFACE
Therefore, the electric flux through a closed surface is the sum of the normal components of the electric field all over the surface.
The sign matters!!
Pay attention to the direction of the normal component as it penetrates the surface… is it “out of” or “into” the surface?
“Out of” is “+” “into” is “-”

20. How to think about flux
We will be interested in net flux in or out of a closed surface like this box
This is the sum of the flux through each side of the box
consider each side separately
Let E-field point in y-direction
then and are parallel and
Look at this from on top
down the z-axis w
“S” is surface
of the box x y z
surface area vector:

21. How to think about flux
Consider flux through two surfaces that “intercept different numbers of field lines”
first surface is side of box from previous slide
Second surface rotated by an angle q
case 1
case 2 q
case 1
case 2
E-field surface area E S
Case 2 is
smaller!
Flux:

22. The Sign Problem
For an open surface we can choose the direction of S-vector two different ways
to the left or to the right
what we call flux would be different these two ways
different by a minus sign
left
right
For a closed surface we can choose the direction of S-vector two different ways
pointing “in” or “out”
define “out” to be correct
Integral of EdS over a closed surface gives net flux “out,” but can be + or -
A differential surface
element, with its vector

23. a) Ф1 > Ф2 b) Ф1 = Ф2 c) Ф1 < Ф2 Preflight 3:
Wire loops (1) and (2) have the same length and width, but differences in depth. 1 5)
Wire loops (1) and (2) are placed in a uniform electric field as shown. Compare the flux through the two surfaces. E 2
a) Ф1 > Ф2
b) Ф1 = Ф2
c) Ф1 < Ф2

24. a) Фbottom < 0 b) Фbottom = 0 c) Фbottom > 0 Preflight 3:
6) A cube is placed in a uniform electric field. Find the flux through the bottom surface of the cube.
a) Фbottom < 0
b) Фbottom = 0
c) Фbottom > 0

25. Lecture 3, ACT 2 2A (a) FE = 0 (b) FE µ 2a2 (c) FE µ 6a2 2B
Imagine a cube of side a positioned in a region of constant electric field as shown
Which of the following statements about the net electric flux FE through the surface of this cube is true? a a
(a) FE = 0
(b) FE µ 2a2
(c) FE µ 6a2 2B R 2R
Consider 2 spheres (of radius R and 2R) drawn around a single charge as shown.
Which of the following statements about the net electric flux through the 2 surfaces (F2R and FR) is true?
(a) FR < F2R
(b) FR = F2R
(c) FR > F2R

26. 2A (a) FE = 0 (b) FE µ 2a2 (c) FE µ 6a2
Imagine a cube of side a positioned in a region of constant electric field as shown
Which of the following statements about the net electric flux FE through the surface of this cube is true? a a
(a) FE = 0
(b) FE µ 2a2
(c) FE µ 6a2
The electric flux through the surface is defined by:
on the bottom face is negative. (dS is out; E is in)
on the top face is positive. (dS is out; E is out)
Therefore, the total flux through the cube is:
is ZERO on the four sides that are parallel to the electric field.

27. 2B (a) FR < F2R (b) FR = F2R (c) FR > F2R
Consider 2 spheres (of radius R and 2R) drawn around a single charge as shown.
Which of the following statements about the net electric flux through the 2 surfaces (F2R and FR) is true? R 2R
(a) FR < F2R
(b) FR = F2R
(c) FR > F2R
Look at the lines going out through each circle -- each circle has the same number of lines.
The electric field is different at the two surfaces, because E is proportional to 1 / r 2, but the surface areas are also different. The surface area of a sphere is proportional to r 2.
Since flux = , the r 2 and 1/r 2 terms will cancel, and the two circles have the same flux!
There is an easier way. Gauss’ Law states the net flux is proportional to the NET enclosed charge. The NET charge is the SAME in both cases.
But, what is Gauss’ Law ??? --You’ll find out next lecture!

28. Summary Electric Fields of continuous charge distributions E(r) = r
Electric Flux:
How to think about flux: number of field lines intercepting a surface, perpendicular to that surface r
E(r) =
Next Time: Gauss’ Law

29. Infinite Line of Charge
Appendix
Infinite Line of Charge x y dx r' r q dE
We use Coulomb’s Law to find dE:
What is dq in terms of dx?
Therefore,
What is r’ in terms of r ?

30. Infinite Line of Charge
We still have x and q variables. x y dx r' r q dE
We are dealing with too many variables. We must write the integral in terms of only one variable (q or x). We will use q.
x and q are not independent!
x = r tan q
dx = r sec2 q dq

31. Infinite Line of ChargeEx x y dx r' r q dE Ey
• Components:
• Integrate:

32. Infinite Line of ChargeEx x y dx r' r q dE Ey
• Now
• The final result: