Transcript Concept Question: Electric Field

Physics 8.02T
http://web.mit.edu/8.02t/www
For now, please sit anywhere, 9 to a table
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Course Details
We will not go over the course details in
class, you can find them at
http://web.mit.edu/8.02t/www/
Or go directly to
http://web.mit.edu/8.02t/www/miscellaneous/8.02%20Introduction_v02.pdf
Are their any burning questions about course
details?
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W01D2: Outline
Introductions
Course Overview
Vector and Scalar Fields
Charge
Electric Force
Electric Field
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Introductions
Announcements
Math Review Tuesday Tues Feb 14 from 9-11 pm in
32-082
PS 1 due Tuesday Tues Feb 14 at 9 pm in boxes
outside 32-082 or 26-152
W01D3 Reading Assignment Course Notes: Chapter
1.1-1.7, 2.1-2.7
Bring Clickers to Monday/Tuesday Class
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8.02: Electricity and Magnetism
Also new way of thinking…
How do objects interact at a distance?
Fields We will learn about electric & magnetic
fields: how they are created & what they affect
Maxwell’s Equations
òò E × d A =
S
Qin
e0
òò B × d A = 0
S
ò E×d s = -
C
d
B×dA
òò
dt S
ò B × d s = m0 I enc + m0e 0
C
Lorentz Force Law
d
E×dA
òò
dt S
F = q(E + v ´ B)
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Scalar and Vector Fields
Review Vector Analysis
in
Online Course Notes
http://web.mit.edu/8.02t/www/coursedocs/current/guide.htm
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Scalar Fields
Temperature Scalar Field: every location
has an associated value (number with units)
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Scalar Fields - Contours
Colors represent surface temperature
Contour lines show constant temperatures
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Vector Fields
Vector (magnitude, direction) at every
point in space
Example: Velocity vector field - jet stream
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Coulomb’s Law, Electric Fields and
Discrete Charge Distributions
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Electric Charge
Two types of electric charge: positive and negative
Unit of charge is the coulomb [C]
Charge of electron (negative) or proton (positive) is
±e,
-19
e = 1.602 ´ 10 C
Charge is quantized
Q = ±Ne
Charge is conserved
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Electric Force
The electric force between charges q1 and q2 is
(a) repulsive if charges have same signs
(b) attractive if charges have opposite signs
Like charges repel and opposites attract !!
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Coulomb's Law
Coulomb’s Law: Force
on q2 due to interaction
between q1 and q2
F12 = ke
ke =
r̂12 :
1
4pe 0
q1q2
r
2
12
r̂12
= 8.9875 ´ 109 N m 2 /C2
unit vector from q1 to q2
r12 : vector from q1 to q2
r12
r̂12 =
r12
Þ F12 = ke
q1q2
r
3
12
r12
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Coulomb's Law: Example
q3 = 3 C
a=1m
F32 = ?
(
r32
r32 =
q2 = 3 C
r32 = 1m
q1 = 6 C
F32 = keq3q2
r32
r
3
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(
1
2
î -
3
2
)( )( )
= 9 ´ 109 N m 2 C2 3C 3C
(
)
81 ´ 109
=
î - 3ĵ
2
)
ĵ m
1
2
( î - 3ĵ) m
(1m )
3
N
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The Superposition Principle
Many Charges Present:
Net force on any charge is vector sum of forces
from other individual charges
Example:
F3 = F13 + F23
In general:
N
Fj = å Fij
i=1
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In Class Problem: Force on a
Charged Object
Three charged objects
are located at the
positions shown in the
figure. Find a vector
expression for the
force on the
negatively charged
object located at the
point P.
NOTE: Solutions will be posted within two days of class
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Charging
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How Do You Get Charged?
• Friction
• Transfer (touching)
• Induction
+q
-
Neutral
+
+
+
+
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Demonstrations:
Instruments for Charging
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Electric Field
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Electric Field
The electric field at a point P due to a charged
object (source) with charge qs is the force acting on
a test point-like charged object with charge qt at that
point P, divided by the charge qt :
Es (P) º
Fst
qt
Eq (P) = ke
qs
r̂
2 st
rst
Units: N/C
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Superposition Principle
The electric field due to a collection of N point
charges is the vector sum of the individual electric
fields due to each charge
N
E = E1 + E2 + ..... = å Ei
i=1
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Concept Question: 5
Equal Charges
Six equal positive charges q sit at the vertices of a
regular hexagon with sides of length R. We remove
the bottom charge. The electric field at the center of
the hexagon (point P) is:
2kq
1. E = 2 ĵ
R
kq
3. E = 2 ĵ
R
2kq
2. E = - 2 ĵ
R
kq
4. E = - 2 ĵ
R
5. E = 0
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Concept Question Answer: 5 Equal
Charges
Answer 4.
kq
E=- 2
R
ĵ
E fields of the side pairs cancel (symmetry)
E at center due only to top charge (R away)
Field points downward
Alternatively:
“Added negative charge” at bottom
R away, pulls field down
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Group Problem: Electric Field on
Axis (Symmetry)
P
ĵ
î
s
q
q
d
Consider two point charges of equal magnitude but opposite
signs, separated by a distance d. Point P lies along the
perpendicular bisector of the line joining the charges, a
distance s above that line. What is the E field at P?
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Electric Field Lines
1. Direction of field at any point is tangent to field
line at that point
2. Field lines point away from positive charges
and terminate on negative charges
3. Field lines never cross each other
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Concept Question: Field Lines
Electric field lines show:
1. Directions of forces that exist in space at all
times.
2. Directions in which positive charges on those
lines will accelerate.
3. Paths that charges will follow.
4. More than one of the above.
5. I don’t know.
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Concept Question Answer: Field
Lines
Answer: 2. Directions positive charges
accelerate.
NOTE: This is different than flow lines (3).
Particles do NOT move along field lines.
(1) is not correct because we don’t know
sign of charge so we don’t know direction
of force
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Force on a Charge Object in an
Electric Field
Force on a charged object with charge q at a point
P in an electric field E s due to a source is:
Fq (P) = qEs (P)
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Summary
SOURCE:
CREATE:
FEEL:
Charge qs (±)
E s = ke
qs
r
2
st
r̂st
FE = qE
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Concept Question: Electric Field
Two charged objects are placed on a line as shown below.
The magnitude of the negative charge on the right is greater
than the magnitude of the positive charge on the left, qR > qL .
Other than at infinity, where is the electric field zero?
1.
2.
3.
4.
5.
Between the two charged objects.
To the right of the charged object on the right.
To the left of the charged object on the left.
The electric field is nowhere zero.
Not enough info – need to know which is positive.
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Concept Question Answer: Electric
Field
Answer: 3. To the left of the positively charged object on the left
http://web.mit.edu/viz/EM/visualizations/electrostatics/ElectricFieldConfigurations/pcharges/pcharges.htm
Between: field points from positively charged object to
negatively charged object
On right: field is dominated by negatively charged object
On left: electric field due to close smaller positively charged
object will cancel electric field due to further larger negatively
charged object.
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