Transcript Lect03

a
R
a
2R
Today…
• More on Electric Field:
– Continuous Charge Distributions
• Electric Flux:
– Definition
– How to think about flux
1
Reminder: Lecture 2, ACT 3
•
•
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)
2p
4p 0 R
1
y
+ +++
++
+
+
+
+
+
++
R
++
+
+
+
+ x
+
++
(c)
But how would we calculate this??
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:



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
e) up and left
c) left
d) right
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
Charge Densities
• How do we represent the charge “Q” on an extended object?
total charge
Q
• Line of charge:
 = charge per
unit length
• Surface of charge:
s = charge per
unit area
• Volume of Charge:
r = charge per
unit volume
small pieces
of charge
dq
dq =  dx
dq = s dA
dq = r dV
How We Calculate (Uniform) Charge
Densities:
Take total charge, divide by “size”
Examples:
10 coulombs distributed over a 2-meter rod.
Q 10C
λ 
 5 C/m
L 2m
14 pC (pico = 10-12) distributed over a shell of radius 1 μm.
Q
14 1012 C 14
2
σ


C/m
4p R 2 4π(10-6 m)2 4π
14 pC distributed over a sphere of radius 1 mm.
Q
14 1012 C (3) 14 3
3
ρ 4 3  4


10
C/m
-3
3
p
R
π(10
m)
4π
3
3
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
q
+++++++++++++++++++++++++++++
x
Any symmetries ? This may help for easy cancellations
Infinite Line of Charge
dE
y
Charge density = 
q
r
r'
++++++++++++++++ x
dx
We need to add up the E-field
contributions from all
segments dx along the line.
Infinite Line of Charge
• 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 .
dE
•
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
y
q
r'
r
++++++++++++++++ x
dx
Ex  0
Ey 
1 2
4p 0 r
Infinite Line of Charge
dE
y
q
r
Conclusion:
r'
++++++++++++++++ x
dx
• The Electric Field produced by an
infinite line of charge is:
- everywhere perpendicular to the line
- is proportional to the charge density
1
- decreases as r
- next lecture: Gauss’ Law makes this
trivial!!
Summary
Electric Field Lines
Electric Field Patterns
Dipole
Point Charge
Infinite
Line of Charge
~ 1/R3
~ 1/R2
~ 1/R
Coming up:
Electric field Flux
and
Gauss’ Law
The Story Thus Far
Two types of electric charge: opposite charges attract,
like charges repel
Coulomb’s Law:

F12 
1 Q1Q2
4p o
2
r12
Electric Fields
•
•
Charges respond to electric fields:
Charges produce electric fields:
rˆ12
F  qE
q
Ek 2
r
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.
2
Lecture 3, ACT 1
•Examine the electric field
lines produced by the charges
in this figure.
•Which statement is true?
q1
q2
(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
Lecture 3, ACT 1
•Examine the electric field
lines produced by the charges
in this figure.
•Which statement is true?
q1
q2
(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
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
Electric Dipole: Lines of Force
Consider imaginary
spheres centered on :
a) +q
(blue)
c
b) -q
(red)
a
c) midpoint (yellow)
• All lines leave a)
• All lines enter b)
• Equal amounts of
leaving and entering
lines for c)
b
Electric Flux
• Flux:
Let’s quantify previous discussion about fieldline “counting”
Define: electric flux FE through the closed
surface S
“S” is surface
of the box
Flux
• How much of something is passing
through some surface
Ex: How many hairs passing through your
scalp.
• Two ways to define
1. Number per unit area (e.g., 10 hairs/mm2)
This is NOT what we use here.
2. Number passing through an area of interest
e.g., 48,788 hairs passing through my scalp.
This is what we are using here.
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 “-”
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
w
z
“S” is surface
of the box
y
x
– consider each side separately
surface area vector:

S  Area  yˆ
 w 2 yˆ
• Let E-field point in y-direction

   2

S
– then E and are parallel and E  S  E w
• Look at this from on top
– down the z-axis
How to think about flux
• Consider flux through two
surfaces that “intercept
different numbers of field
lines”
case 1
– first surface is side of box from
previous slide
– Second surface rotated by an
angle q
Flux:
E-field surface area
case 1
case 2

E  Eo yˆ
w2

E  Eo yˆ
2
w
case 2
E S
Eo w 2
Case 2 is
smaller!
Eo w2  cos q
q
The Sign Problem
• For an open surface we
can choose the direction of
S-vector two different ways
left
right
– to the left or to the right
– what we call flux would be
different these two ways
– different by a minus sign
• For a closed surface we
can choose the direction
of S-vector two different
ways
A differential surface
element, with its vector
– pointing “in” or “out”
– define “out” to be correct
– Integral of EdS over a closed
surface gives net flux “out,”
but can be + or -
2
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.
a) Ф1 > Ф2
b) Ф1 = Ф2
c) Ф1 < Ф2
E
2
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
Lecture 3, ACT 2
2A
•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) FE = 0
2B
(b) FE  2a2
(c) FE  6a2
• 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
a
a
Lecture 3, ACT 2
2A
•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) FE = 0
(b) FE  2a2
a
a
(c) FE  6a2
• The electric flux through the surface is defined by:
•
•
is ZERO on the four sides that are parallel to the electric field.
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:
Lecture 3, ACT 2
2B
• 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
• 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!
Summary
• Electric Fields of continuous charge distributions
r
E(r) =
++++++++++++++++++++++++++
• Electric Flux:
– How to think about flux: number of field lines
intercepting a surface, perpendicular to that surface
• Next Time: Gauss’ Law
Appendix
Infinite Line of Charge
We use Coulomb’s Law to find dE:
dE
y
q
r
What is dq in terms of dx?
++++++++++++++++ x
dx
Therefore,
What is r’ in terms of r ?
r'
Infinite Line of Charge
We still have x and q variables.
dE
y
q
r
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.
r'
++++++++++++++++ x
dx
x and q are not independent!
x = r tan q
dx = r sec2 q dq
Infinite Line of Charge
• Components:
Ey
dE
y
q
Ex
q
r
r'
++++++++++++++++ x
dx
• Integrate:
Infinite Line of Charge
• Now
p / 2
p

p
dE
y
q
Ex
sin q dq  0
 /2
p / 2
Ey
cos q dq  2
 /2
• The final result:
q
r
r'
++++++++++++++++ x
dx