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Unformatted text preview: Physics 212
Lecture 2
Today's Concept:
The Electric Field
Continuous Charge Distributions Physics 212 Lecture 2, Slide 1
Physics Music BB Who is the Artist?
A)
B)
C)
D)
E) Eric Clapton
Bill Frisell
Jimmy Page
Jeff Beck
Buddy Guy Why?
Theme: guitar players (Englilsh)
How about one of those guys listed last time?
Nice acoustic album from 8 years ago…
Physics 212 Lecture 2, Slide 2
Physics Your Comments
What's the difference between the green and blue completion bars on the website? What exactly
on
is the Smartscore, and does it have any effect on my Class score?
is
all of it. it was like crawling the first day to tour de france today
Some of the math magical calculations are a little confusing and some help on those
Some
some
would be appreciated.
That is really creepy  llogging all of our keystrokes and stuff! What else do you guys know?
ogging
That
know?
I would like to go over the infinite lines of charge as well as the equations and their meaning
would
for the physics presented. Especially lambda. Also, keeping rocking the good tunes!
for
50
40
30
20
10
0 Confused
04 Confident How do X (distance of segment from
origin) and L (length of line) relate? Isn't
dx (length of segment) more related to L
than dx is related to X? How is the
integration of dE over L worked out, step
by step? How do E and F relate
conceptually? What is an electric field?
How do charges exert this electric
field?More questions and comments are
to come.
Physics 212 Lecture 2, Slide 3
Physics Coulomb’s Law (from last time)
If there are more than two charges present, the total force
If
on any given charge is just the vector sum of the forces due
vector
of
to each of the other charges:
to
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 ﬂ direction reversed F2,1 q4 F1
F3,1 F4,1 MATH:
MATH: F1
33 kq1q3
kq1q2
kq1q4
ˆ
ˆ
ˆ
= 2 r12 + 2 r13 + 2 r14
r12
r13
r14 kq
F1 kq2
kq
ˆ
ˆ
ˆ
= 2 r12 + 23 r13 + 24 r14
q1 r12
r13
r14
Physics 212 Lecture 2, Slide 4
Physics Electric Field “What exactly does the electric field that we calculate mean/represent? “
esent?
“What is the essence of an electric field? “ 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
ˆ
E = ∑ k 2 ri
ri
i Q
ˆ
E=k 2 r
r
q2 E4
E2
E Field points toward negative and
Field
Away from positive charges. E3 q4 q3
08 Physics 212 Lecture 2, Slide 5
Physics Preflights
“I don't completely understand how to determine the direction of the electric field.”
don't
the 4) 70
60 B? 100
80 50
40
30 60
40 20
10
0 09 20
0
simulation Physics 212 Lecture 2, Slide 6
Physics Preflight
E E 80 “The upper left +Q only affects the x direction in
both and the lower right (+/)Q only affects the y
direction so in both, nothing cancels out, so they'll
have the same magnitude. ” 60 40 20 0 12 1 2 Same Physics 212 Lecture 2, Slide 7
Physics BB 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 yaxis.
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
There
related
related
Physics 212 Lecture 2, Slide 8
Physics  + + +
23  Physics 212 Lecture 2, Slide 9
Physics BB Preflight A B C INTERESTING: statement is correct,
but given in support of “to the left” !! D
“The force is proportional to the charge divided by
The
the square of the distance. Therefore, the force of
the 2Q charge is 1/2 as much as the force of the
Q charge. “
“Even though the charge on the right is larger,
Even
it is twice as far away, which makes the force
it exherts on the test charge half that as the
charge on the left, causing the charge to
move to the right.”
move 70
60
50
40
30
20
10
0 12 left right stay still The ratio between the R and Q on both sides is
The
1:1 meaning they will result in the same
magnitude of electric field acting in opposite
directions, causing q to remain still.
directions,
Physics 212 Lecture 2, Slide 10
Physics BB Example
“Show me more electric field examples, please!”
+q What is the direction of the electric
field at point P, the unoccupied corner
of the square? P d
q +q
d (A) Calculate E at point P. (D) know d (E) Need to know d & q Qi
ˆ
E = ∑ k 2 ri
ri
i 1 q Ex =
− d2
4πεo 1 q Ey =
− d2
4πεo 20 Need to (C) E = 0 (B) ( ( q
2d q
2d ) 2 ) 2 cos sin π 4 π 4 Physics 212 Lecture 2, Slide 11
Physics Continuous Charge Distributions
“I don't understand the whole dq thing and lambda.” Summation becomes an integral (be careful with vector nature) Qi
ˆ
E = ∑ k 2 ri
ri
i dq
ˆ
E = ∫k 2 r
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: λ = Q/L
dE pt for E
r
charges
25 dq = λdx
dq
Physics 212 Lecture 2, Slide 12
Physics BB Charge Density
“I would like to know more about the charge density.” • Linear (λ=Q/L)
Coulombs/meter
• Surface (σ = Q/A)
Coulombs/meter2
• Volume (ρ = Q/ V)
Coulombs/meter3 Some Geometry
Asphere = 4πR 2 Acylinder = 2πRL Vsphere = 4 πR 3 Vcylinder = πR 2 L 3 What has more net charge?.
A) A sphere w/ radius 2 meters and volume charge density ρ = 2 C/m3
B) A sphere w/ radius 2 meters and surface charge density σ = 2 C/m2
C) Both A) and B) have the same net charge. Q A = ρV = ρ 4 πR 3
3 QB = σA = σ 4πR 2
28 3
4
1ρ
Q A ρ 3 πR
=
=
R
2
QB σ 4πR
3σ
Physics 212 Lecture 2, Slide 13
Physics Preflight
10) 70
60
50
40 “At point A the two fields will cancel each other out
At
but at point B the fields will act in the sme direction”
but 30
20
10
0 29 Physics 212 Lecture 2, Slide 14
Physics BB Calculation
“How is the integration of dE over L
How
y worked out, step by step?”
worked
Charge is uniformly distributed along
the xaxis from the origin to x = a.
The charge denisty is λ C/m. What is
the xcomponen t of the electric field
at point P: (x,y) = (a,h)? P
r
h dq=λdx
x x
a We know: dq
ˆ
E = ∫k 2 r
r
What is
(A) 33 dx
x 2 dq
r 2 ?
dx (B) 2
a + h2 (C) λdx
2 a +h 2 (D) λdx
2 (a − x) + h 2 (E) λdx
x2 Physics 212 Lecture 2, Slide 15
Physics Calculation dE
dE x Charge is uniformly distributed along
the xaxis from the origin to x = a.
The charge denisty is λ C/m. What is
the xcomponen t of the electric field
at point P: (x,y) = (a,h)? r
θ1 We know: dq
r2 = λdx
(a − x) 2 + h 2 h θ2 x dq
ˆ
E = ∫k 2 r
r θ2 P y BB x a
dq=λdx E x = ∫ dE x What is dE x ?
(A) dE cos θ1 33 (B) dE cos θ 2 (C) dE sin θ1 (D) dE sin θ 2 Physics 212 Lecture 2, Slide 16
Physics Calculation dE
dE x Charge is uniformly distributed along
the xaxis from the origin to x = a.
The charge denisty is λ C/m. What is
the xcomponen t of the electric field
at point P: (x,y) = (a,h)? r
θ1 We know: dq
r2 = λdx
(a − x) 2 + h 2 h θ2 x dq
ˆ
E = ∫k 2 r
r θ2 P y BB x a
dq=λdx E x = ∫ dE x = ∫ dE cos θ 2 What is E x ?
dx
λ cos θ 2 ∞
(A)
∫
4πεo − ∞ (a − x) 2 + h 2 λ cos θ 2 a
dx
(B)
∫
4πεo 0 (a − x) 2 + h 2 (C) none of the above
33 cosθ2 DEPENDS ON x !!
Physics 212 Lecture 2, Slide 17
Physics Calculation dE
dE x Charge is uniformly distributed along
the xaxis from the origin to x = a.
The charge denisty is λ C/m. What is
the xcomponen t of the electric field
at point P: (x,y) = (a,h)? r
θ1 We know: dq
r2 = h θ2 x dq
ˆ
E = ∫k 2 r
r θ2 P y BB x a
dq=λdx λdx E x = ∫ dE x = ∫ dE cos θ 2 (a − x) 2 + h 2 What is cos θ 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
Physics Calculation dE Charge is uniformly distributed along
the xaxis from the origin to x = a.
The charge denisty is λ C/m. What is
the xcomponen t of the electric field
at point P: (x,y) = (a,h)? We know: dq
r2 = dq
ˆ
E = ∫k 2 r
r λdx dE x
r
θ1 h θ2 x x a
dq=λdx E x = ∫ dE x = ∫ dE cos θ 2 (a − x) 2 + h 2 θ2 P y cos θ 2 = a−x
(a − x) 2 + h 2 What is E x ( P) ?
a−x
λa
E x ( P) =
dx
∫
4πεo 0
(a − x) 2 + h 2 ( 33 ) 3/ 2 h
λ
1 −
E x ( P) =
4πεo h h2 + a 2 Physics 212 Lecture 2, Slide 19
Physics Observation dE Charge is uniformly distributed along
the xaxis from the origin to x = a.
The charge denisty is λ C/m. What is
the xcomponen t of the electric field
at point P: (x,y) = (a,h)? Note that our result can be
rewritten more simply in terms of θ1. h
λ
1 −
E x ( P) =
4πεo h h2 + a 2 θ2 P y dE x
r
θ1 h θ2 x x a
dq=λdx E x ( P) = λ
(1 − sin θ1 )
4πεo h Exercise for student:
Change variables: write x in terms of θ
Result: obtain simple integral in θ
33 λ π /2
E x ( P) =
∫ dθ cos θ
4πεo h θ
1 Physics 212 Lecture 2, Slide 20
Physics Notes
•
•
•
•
• Preflight + Prelecture 3 due by 8:00 AM Tuesday August 31
Homework 1 is due Tuesday August 31
Labs start Monday August 30
Discussion Quiz next week will be on Coulomb’s Law and E
Homework 2 is due Tuesday September 7 Physics 212 Lecture 2, Slide 21
Physics ...
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This note was uploaded on 02/09/2012 for the course PHYSICS 212 taught by Professor Mestre during the Spring '11 term at University of Illinois at Urbana–Champaign.
 Spring '11
 MESTRE
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