Transcript W04D1_Equipotentials_mac_answers_v03_jwb

W04D1
Electric Potential and Gauss’ Law
Equipotential Lines
Today’s Reading Assignment
Course Notes: Sections 3.3-3.4, 4.4-4.6. 4.10.5
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Announcements
Exam One Thursday Feb 28 7:30-9:30 pm Room
Assignments (See Stellar webpage announcements)
Review Tuesday Feb 26 from 9-11 pm in 26-152
PS 3 due Tuesday Tues Feb 26 at 9 pm in boxes
outside 32-082 or 26-152
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Outline
Continuous Charge Distributions
Review E and V
Deriving E from V
Using Gauss’s Law to find V from E
Equipotential Surfaces
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Continuous Charge
Distributions
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Continuous Charge Distributions
Break distribution into
infinitesimal charged elements of
charge dq. Electric Potential
difference between infinity and P
due to dq.
dq
dV º Vdq (P) - Vdq (¥) = ke
r
Superposition Principle: V (P) - V (¥) = k
e
Reference Point:
dq
ò r
source
dq
V (¥) = 0 Þ V (P) = ke ò
r
source
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Group Problem
Consider a uniformly
charged ring with total
charge Q. Find the electric
potential difference between
infinity and a point P along
the symmetric axis a
distance z from the center of
the ring.
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Group Problem: Charged Ring
l = Q / 2p R
dq¢ = l ds¢ = l Rdq ¢
Choose
V (¥) = 0
r = z k̂
r¢ = Rr̂
r - r¢ = R2 + z 2
dq¢
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l Rdq ¢
dV =
=
4pe 0 r - r¢ 4pe 0 R 2 + z 2
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V (z) = kel ò
2p
Rdq ¢
=
ke l R
R +z
R +z
ke 2p Rl
k eQ
V (z) =
=
2
2
R +z
R2 + z 2
0
2
2
2
2
ò
b
a
dq ¢
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Electric Potential and Electric Field
Set of discrete charges:
V (r) -V (¥) = ke å
i
Continuous charges:
V (r) - V (¥) = ke
ò
source
qi
r - ri
d q¢
r - r¢
If you already know electric field (Gauss’ Law)
compute electric potential difference using
B
VB - V A = - ò E × d s
A
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Using Gauss’s Law to find
Electric Potential from Electric Field
If the charge distribution has a lot of symmetry, we
use Gauss’s Law to calculate the electric field and
then calculate the electric potential V using
B
VB  VA    E  d s
A
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Group Problem: Coaxial Cylinders
A very long thin uniformly charged cylindrical shell
(length h and radius a) carrying a positive charge
+Q is surrounded by a thin uniformly charged
cylindrical shell (length h and radius a ) with
negative charge -Q , as shown in the figure. You
may ignore edge effects. Find V(b) – V(a).
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Worked Example: Spherical Shells
These two spherical
shells have equal
but opposite charge.
Find
V (r) - V (¥)
for the regions
(i) b < r
(ii) a < r < b
(iii) 0 < r < a
Choose V (¥) = 0
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Electric Potential for Nested Shells
From Gauss’s Law
ì
Q
r̂,
a<r<b
ï
2
E = í 4pe 0 r
ï
0, elsewhere
ïî
r
B
ò
Use VB - V A = - E × d s
A
Region 1: r > b
r
V (r) - V (¥) = - ò 0 dr = 0
=0
¥
No field 
No change in V!
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Electric Potential for Nested Shells
Region 2: a < r < b
r
Q
b
4pe 0 r 2
V (r) - V (r = b) = - ò dr
=0
=
Q
4pe 0 r
r
r
r =b
æ 1 1ö
=
Qç - ÷
4pe 0 è r b ø
1
Electric field is just a point charge.
Electric potential is DIFFERENT – surroundings matter
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Electric Potential for Nested Shells
Region 3: r < a
r
V (r) - V (a) = - ò dr 0 = 0
æ 1 1ö
= kQ ç - ÷
è a bø
a
Q æ 1 1ö
V (r) = V (a) =
- ÷
ç
4pe 0 è a b ø
r
Again, potential is CONSTANT since E = 0, but the
potential is NOT ZERO for r < a.
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Group Problem: Charge Slab
Infinite slab of thickness 2d, centered at x = 0 with
uniform charge density r . Find
V (x A ) - V (0) ; x A > d
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B
Deriving E from V
DV = - ò E × d s
A
A = (x,y,z), B=(x+Δx,y,z)
D s = Dx î
( x+Dx,y,z )
DV = -
ò
E × d s @ -E × D s = -E ×(Dx î) = - Ex Dx
( x,y,z )
DV
¶V Ex = Rate of change in V with
Ex @ ®y and z held constant
Dx
¶x
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Deriving E from V
If we do all coordinates:
æ ¶V
¶V
¶V ö
E = -ç
î +
ĵ +
k̂ ÷
¶y
¶z ø
è ¶x
æ ¶
¶
¶ ö
= -ç
î +
ĵ + k̂ ÷ V
¶y
¶z ø
è ¶x
E = -ÑV
Gradient (del) operator:
¶
¶
¶
Ѻ
î +
ĵ + k̂
¶x
¶y
¶z
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Concept Question: E from V
Consider the point-like
charged objects arranged in
the figure below. The electric
potential difference between
the point P and infinity and is
V (P) = - kQ a
From that can you derive E(P)?
1.
2.
3.
4.
Yes, its kQ/a2 (up)
Yes, its kQ/a2 (down)
Yes in theory, but I don’t know how to take a
gradient
No, you can’t get E(P) from V(P)
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Concept Question Answer: E from V
4. No, you can’t get E(P) from V(P)
The electric field is the gradient (spatial
derivative) of the potential. Knowing the
potential at a single point tells you nothing
about its derivative.
People commonly make the mistake of
trying to do this. Don’t!
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Group Problem: E from V
Consider two point like charged
objects with charge –Q located at
the origin and +Q located at the
point (0,a).
(a)Find the electric potential
V(x,y)at the point P located at (x,y).
(b)Find the x-and y-components of
the electric field at the point P using
¶V
¶V
Ex (x, y) = , E y (x, y) = ¶x
¶y
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Concept Question: E from V
The graph above shows a potential V as a function
of x. The magnitude of the electric field for x > 0 is
1. larger than that for x < 0
2. smaller than that for x < 0
3. equal to that for x < 0
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Concept Question Answer: E from V
Answer: 2. The magnitude of the electric field for
x > 0 is smaller than that for x < 0
The slope is smaller for x > 0 than x < 0
Translation: The hill is steeper on the
left than on the right.
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Concept Question: E from V
The above shows potential V(x). Which is true?
1.
2.
3.
4.
Ex > 0 is positive and Ex < 0 is positive
Ex > 0 is positive and Ex < 0 is negative
Ex > 0 is negative and Ex < 0 is negative
Ex > 0 is negative and Ex < 0 is positive
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Concept Question Answer: E from V
Answer: 2. Ex > 0 is positive and Ex < 0 is negative
E is the negative slope of the potential, positive
on the right, negative on the left,
Translation: “Downhill” is to the left on the left
and to the right on the right.
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Potential (V)
Group Problem: E from V
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5
0
-5
0
5
Z Position (mm)
A potential V(x,y,z) is plotted above. It does
not depend on x or y.
What is the electric field everywhere?
Are there charges anywhere? What sign?
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Equipotentials
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Topographic Maps
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Equipotential Curves: Two Dimensions
All points on equipotential curve are at same potential.
Each curve represented by V(x,y) = constant
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Direction of Electric Field E
E is perpendicular to all equipotentials
Constant E field
Point Charge
Electric dipole
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Direction of Electric Field E
E is perpendicular to all equipotentials
Field of 4 charges
Equipotentials of 4 charges
http://web.mit.edu/viz/EM/visualizations/electrostatics/InteractingCharges/zoo/zoo.htm
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Properties of Equipotentials
E field lines point from high to low potential
E field lines perpendicular to equipotentials
E field has no component along equipotential
The electrostatic force does zero work to
move a charged particle along equipotential
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