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From Last Time…
Forces between charges
Electric dipole
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Exam 1 Wed. Feb. 20, 5:30-7 pm
Covers Chap. 21.5-7, 22-23,25-26+ lecture, lab, discussion, HW
Review Group/Quiz (solutions on website).
Review lab question sheets.
Review sample exams on website.
HW4 (due Fri. next week) covers exam material.
Students with schedule class conflicts: stay after lecture today to arrange time
8 1/2 x 11 handwritten note sheet (both sides) allowed
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From Last Time…
Forces between charges
Electric dipole
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Unusual dipoles:Electrogenic fish
• Dipole + nearby conducting object
Some fish generate charge separation - electric dipole.
Dipole is induced in nearby (conducting) fish
Small changes detected by fish.
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The idea of electric fields• EM wave made up of
oscillating electric and magnetic fields.
• But what is an electric field?
• Electric field is a way to describe the force on a charged particle due to other charges around it.
• Force = charge electric field
• The direction of the force is the direction of the electric field.
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Electric field of a point charge
+
+
Force on this charge…
…due to this charge
+++
€
rF = k
Q1Q2
r2ˆ r
Q1
Q2
€
rE =
r F
r r ( ) /Q2
= Force/unit charge
Units? N/C
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QuestionWhich vector best represents the electric
field at the red dot?
-
A
B
C
-D
E
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Electric field• Electric field vector defined at every point in
space.
• Gives magnitude and direction of force on test particle
e.g. wind velocity (speed and direction) in different parts of the country.
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Relationship Between F and E
• Fe = qE This is valid for a test charge small
enough that it does not disturb the
source charge distribution
• If q is positive, F and E are in the same direction
+
r = 1x10-10 m
Qp=1.6x10-19 C
E
(to the right)
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
€
E =9 ×109 N ⋅M /C2
( ) 1.6 ×10−19C( )
10−10 m( )2 = 2.9 ×1011N /C
Electric field 1Å away from proton
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Electric Field Direction
• a) q is positive, F is directed away from q
• b) The direction of E is also away from the positive source charge
• c) q is negative, F is directed toward q
• d) E is also toward the negative source charge
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Superposition with Electric Fields
• At any point P, the total electric field due to a group of source charges equals the vector sum of electric fields of all the charges
Find the electric field due to q1, E1
Find the electric field due to q2, E2
E = E1 + E2Remember, the fields add as vectors
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Quick QuizWhich is the direction of the electric field at dot?
A. Left
B. Right
C. Up
D. Down
E. Zero
+ - x
y
Away from positive charge (right)
Net E field is to right.
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Quick QuizIn this electric dipole, what is the direction of the
electric field at point A?
A) Up
B) Down
C) Left
D) Right
E) Zero
+Q -Q
x=+ax=-a
A
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Electric field: summary
• Electric field -> will be a force on a charged particle.
• This force ( and electric field) can arise from electric charges (via Coulomb’s law)
• But once electric field is known, don’t need to know the charges that produce it.
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Calculating dipole electric field
On the y-axis
+
-€
s /2
€
−s /2
x
y
€
E = E yˆ y
Ey = E+ + E−
= kq1
y − s /2( )2 + −kq( )
1
y + s /2( )2
= kq2ys
y − s /2( )2
y + s /2( )2
€
≈k 2qs( )1
y 3 For
€
y >> s
Since points from - charge to + charge
€
rp
€
rE = k
2r p
r3 on y-axis of dipole only
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Question: electric dipoleA and B are the same (large) distance from dipole.
How do the magnitude of the electric fields at A and B compare?
A
B
A)
B)
C)
D)
€
rE A =
r E B
€
rE A <
r E B
€
rE A >
r E B
€
rE B = 0
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Pictorial representation of E: Electric Field Lines
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Electric field lines
• Local electric field tangent to field line
• Density of lines proportional to electric field strength
• Fields lines can only start on + charge
• Can only end on - charge.
• Electric field lines can never cross
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Point particles
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Electric field of a dipole
+
-
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Electric field of two + charges
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Question
How are the charges A and B related?
A) A+, B-,
B) A-, B+,
C) A+, B-,
D) A-, B+,
E) A+, B-,
€
A < B
€
A < B
€
A > B
€
A < B
€
A = B
A
B
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Hydrogen Atom Example
• The magnitude of the electric force between the electron and proton: Fe = ke|e| |-e| / r2 = 8.2 x 10-8 N, r ~ 0.53 x 10-10 m
• The gravitational force between the electron and the proton Fg = Gmemp / r2 = 3.6 x 10-47 N
me = 9.11 x 10-31 kg, mp = 1.67 x 10-27 kgG = 6.67 x 10-11 N m2/kg2
Fe/Fg ~ 2 x 1039
+ -
r = 0.53x10-10 m
Mp=1.67x10-27 kg Me = 9.11 x10-31 kg