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Physics 212
Lecture 2
Today's Concept:
The Electric Field
Continuous Charge Distributions
Physics 212 Lecture 2, Slide 1
Music
Who is the Artist?
A)
B)
C)
D)
E)
Bill Frisell
Cindy Cashdollar
Daniel Lanois
Marc Ribot
Tony Rice
Great musician/album names.
All these people were at the Ellnora Guitar
Festival this year.
Physics 212 Lecture 2
Our Comments
1. If your TA doesn’t show up – go to undergrad office 231/3
(or home page for telephone numbers of discussion and lab
masters)
2. Please operate only your own clicker
3. Unfortunately we really can’t help with PreLecture,
Checkpoint or homework physics questions by email – too
many students: Office Hours, CARE (see link), your friends
–
–
–
Fridays 2:00 – 7:00 p.m. in 271 Loomis
Sundays 2:00 – 9:00 p.m. in 143 Loomis
Mondays 11:00 a.m. – 9:00 p.m. in 143 Loomis
4. Quite a bit of homework this week
5. Don’t panic!
Physics 212 Lecture 2, Slide 3
Coulomb’s Law (from last time)
If there are more than two charges present, the total force
on any given charge is just the vector sum of the forces due
to each of the other charges:
q2
q2
F4,1
F4,1
F1
q1
F2,1
F1
F3,1
q3
F2,1
q3
F3,1
F4,1
F3,1
q4
F1
F2,1
q1
+q1 -> -q1  direction reversed
F2,1
q4
F1
F3,1
F4,1
MATH:
F1
33
kq1q3
kq1q2
kq1q4
 2 rˆ12  2 rˆ13  2 rˆ14
r12
r13
r14
kq3
F1 kq2
kq4
ˆ
ˆ

r

r

E1  q r 2 12 r 2 13 r 2 rˆ14
1
12
13
14
Physics 212 Lecture 2, Slide 4
Electric Field
“Electric fields and also their use with lines of charge really confuse me! “
“How fields add together“
F
E
q
The electric field E at a point in space is simply
the force per unit charge at that point.
Electric field due to a point charged particle
Superposition
Qi
E   k 2 rˆi
ri
i
Q
E  k 2 rˆ
r
q2
E4
E2
Field points toward negative and
away from positive charges.
E
E3
q4
q3
08
Physics 212 Lecture 2, Slide 5
Checkpoint 1
“Field directions were confusing.”
Two equal, but opposite charges are place on the x axis. The positive charge is placed
to the left of the origin and the negative charge to the right, as shown in the figure above.
What is the direction of the electric field at point A?
A. Up B. Down C. Left D. Right E. Zero
09
What is the direction of the electric field at point B?
A. Up B. Down C. Left D. Right E. Zero
simulation
Physics 212 Lecture 2, Slide 6
Checkpoint 2
In which of the two cases below is the magnitude of the electric field at the point labeled A
the largest?
E
A Case 1
E
B Case 2
C Same
“the two positive charges would cancel eachother out in the x
direction since they both have fields that point away from them,
while the charges in case one have fields that both point in the
same direction.”
“In two there is a component of the field pointing away.”
“The coulomb forces exerted at point A are at right angles in
both cases, so that the magnitude stays the same.”
12
Physics 212 Lecture 2, Slide 7
Two Charges
Two charges q1 and q2 are fixed at points (-a,0) and (a,0) as
shown. Together they produce an electric field at point (0,d)
which is directed along the negative y-axis.
y
(0,d)
E
(-a,0)
q1
q2
(a,0)
x
Which of the following statements is true:
a)
b)
c)
d)
22
Both charges are negative
Both charges are positive
The charges are opposite
There is not enough information to tell how the charges are
related
Physics 212 Lecture 2, Slide 8
23
-
-
+
+
+
Physics 212 Lecture 2, Slide 9
Checkpoint 3
A positive test charge q is released from rest at a distance r away from a charge of +Q and a
distance 2r away from a charge of +2Q. How will the test charge move immediately after being
released?
A. To the left B. To the right C. Stay still D. Other
INTERESTING: statement is correct,
but given in support of “to the left” !!
“although the charge on the right is doubled, the
force on q by the charge is inversely proportional
to the square of the distance between the
charges, so the force is effectively halved
compared to the charge on the left. “
“radius gets squared so it moves to the right
even though the second charge is stronger.”
“The forces from each charge will have the
same force because of their relative distances.”
12
Physics 212 Lecture 2, Slide 10
Example
“Just some more examples would be cool”
+q
P
What is the direction of the electric
field at point P, the unoccupied corner
of the square?
d
-q
+q
d
(A)
(B)
Calculate E at point P.
(E)
Need to
know d & q
Qi
E   k 2 rˆi
ri
i

q
Ex  k  2 
d


20
Need to
(D) know d
(C) E  0

q
Ey  k  2 
d





cos
2
4 
2d




sin 
4 

q
q
2d


2
Physics 212 Lecture 2, Slide 11
Continuous Charge Distributions
“my head is spinning, what does the dQ mean?”
Summation becomes an integral (be careful with vector nature)
Qi
E   k 2 rˆi
ri
i
dq
E   k 2 rˆ
r
WHAT DOES THIS MEAN ??
Integrate over all charges (dq)
r is vector from dq to the point at which E is defined
Linear Example:
l = Q/L
dE
pt for E
r
charges
25
dq = ldx
Physics 212 Lecture 2, Slide 12
Charge Density
“What exactly is charge density, how is it calculated?”
• Linear (l=Q/L)
Coulombs/meter
• Surface (s  Q/A)
Coulombs/meter2
• Volume (r = Q/ V)
Coulombs/meter3
Some Geometry
Asphere  4R 2
Acylinder  2RL
Vsphere  4 R 3
Vcylinder  R 2 L
3
What has more net charge?.
A) A sphere w/ radius 2 meters and volume charge density r = 2 C/m3
B) A sphere w/ radius 2 meters and surface charge density s = 2 C/m2
C)
Both A) and B) have the same net charge.
Q A  r V  r 4 R 3
3
Q B  sA  s 4R 2
28
3
4
Q A r 3 R
1r


R
2
QB s 4R
3s
Physics 212 Lecture 2, Slide 13
Checkpoint 4
Two infinite lines of charge are shown below.
Both lines have identical charge densities +l C/m. Point A is equidistant from both lines and Point
B is located above the top line as shown. How does EA, the magnitude of the electric field at point
A compare to EB, the magnitude of the electric field at point B?
A. EA < EB B. EA = EB C. EA > EB
“cancel at A, sum at B“
“Both fields are parallel and both lines are
parallel and identical.”
“b is twice as far from the bottom one as the
other, therefore smaller E.”
29
Physics 212 Lecture 2, Slide 14
Calculation
“I really think it would be nice to go over how to
find the electric field with the infinite lines of
charge with the integrals”
y
P
r
Charge is uniformly distributed along
the x-axis from the origin to x = a.
The charge denisty is l C/m. What is
the x-componen t of the electric field
at point P: (x,y) = (a,h)?
h
dq=ldx
x
x
a
We know:
dq
E   k 2 rˆ
r
What is
(A)
33
dx
x
2
dq
r
2
?
dx
(B) 2
a  h2
(C)
ldx
2
a h
2
(D)
ldx
2
(a  x)  h
2
(E)
ldx
x2
Physics 212 Lecture 2, Slide 15
Calculation
dE
dE x
Charge is uniformly distributed along
the x-axis from the origin to x = a.
The charge denisty is l C/m. What is
the x-componen t of the electric field
at point P: (x,y) = (a,h)?
r
q1
We know:
dq
r2
h
q2
x
dq
E   k 2 rˆ
r
q2
P
y
x
a
dq=ldx

ldx
(a  x) 2  h 2
E x   dE x
What is dE x ?
(A) dE cos q1
33
(B) dE cos q 2
(C) dE sin q1
(D) dE sin q 2
Physics 212 Lecture 2, Slide 16
Calculation
dE
dE x
r
Charge is uniformly distributed along
the x-axis from the origin to x = a.
The charge denisty is l C/m. What is
the x-componen t of the electric field
at point P: (x,y) = (a,h)?
q1
x
a
dq=ldx
dq
r
h
q2
x
We know:
dq
E   k 2 rˆ
r
q2
P
“The concept I found most difficult was integrating y
to find electric field when the charge is distributed
over a distance.”
2

ldx
E x   dE x   dE cos q 2
(a  x) 2  h 2
What is E x ?

dx
(A) k l cos q 2 
( a  x )2  h 2

dx
2
2
(
a

x
)

h
0
(B) k l cos q 2 
(C) neither of the above
33
a
cosq2 DEPENDS ON x !!
Physics 212 Lecture 2, Slide 17
Calculation
dE
dE x
Charge is uniformly distributed along
the x-axis from the origin to x = a.
The charge denisty is l C/m. What is
the x-componen t of the electric field
at point P: (x,y) = (a,h)?
r
q1
We know:
dq
r2
h
q2
x
dq
E   k 2 rˆ
r
q2
P
y
x
a
dq=ldx

ldx
E x   dE x   dE cos q 2
(a  x) 2  h 2
What is cos q 2 ?
(A)
33
x
2
a h
2
(B)
ax
2
(a  x)  h
2
(C)
a
2
a h
2
(D)
a
(a  x) 2  h 2
Physics 212 Lecture 2, Slide 18
Calculation
dE
Charge is uniformly distributed along
the x-axis from the origin to x = a.
The charge denisty is l C/m. What is
the x-componen t of the electric field
at point P: (x,y) = (a,h)?
We know:
dq
r2

dq
E   k 2 rˆ
r
ldx
dE x
r
q1
h
q2
x
x
a
dq=ldx
E x   dE x   dE cos q 2
(a  x) 2  h 2
q2
P
y
cos q 2 
ax
(a  x) 2  h 2
What is E x (P ) ?
a
Ex ( P )  k l  dx
0
33
ax
( a  x )
2
h

2 3/ 2
Ex ( P ) 

kl 
h
1



2
2
h 
h a 
Physics 212 Lecture 2, Slide 19
Observation
dE
Charge is uniformly distributed along
the x-axis from the origin to x = a.
The charge denisty is l C/m. What is
the x-componen t of the electric field
at point P: (x,y) = (a,h)?
Note that our result can be
rewritten more simply in terms of q1.

kl 
h
Ex ( P ) 
1  2

2
h 
h a 
q2
P
y
dE x
r
q1
h
q2
x
x
a
dq=ldx
kl
Ex ( P ) 
1  sin q1 
h
Exercise for student:
Change variables: write x in terms of q
Result: obtain simple integral in q
33
kl
Ex ( P ) 
h
 /2
q dq cos q
1
Physics 212 Lecture 2, Slide 20
Notes
•
•
•
•
Preflight + Prelecture 3 due by 8:00 AM Tuesday Jan. 24
Homework 1 is due Tuesday Jan. 24
Labs start Monday Jan. 23
Discussion Quiz next week will be on Coulomb’s Law and E
Physics 212 Lecture 2, Slide 21