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 27 7:30-9:30 pm Room
Assignments (See Stellar and 8.02t webpage
announcements)
Review Sessions:
(1)  Tuesday Feb 25 from 7:30-9:00 pm in 26-152
(2)  Tuesday Feb 25 from 9:00-10:30 pm in 26-152
PS 3 due Tuesday Tues Feb 25 at 9 pm in boxes
outside 32-082 or 26-152
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Outline
Review E and V
Deriving E from V
Using Gauss s Law to find V from E
Equipotential Surfaces
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Electric Potential and Electric Field
Set of discrete charges:
Continuous charges:
qi
!
V (r) ! V (") = ke # ! !
i r - ri
!
V (r) ! V (") = ke
$
source
d q#
! !
r - r#
If you already know electric field (Gauss Law)
compute electric potential difference using
! !
VB ! V A = ! " E # d s
B
A
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E Field and Potential: Effects
If you put a charged particle, (charge q), in a field:
!
!
F = qE
To move a charged particle, (charge q), in a field
and the particle does not change its kinetic energy
then:
Wext = !U = q!V
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Concept Question: Two Point
Charges
The work done in moving a
positively charged object that
starts from rest at infinity and
ends at rest at the point P
midway between two charges
of magnitude +Q and –Q
1.
2.
3.
4.
is positive.
is negative.
is zero.
can not be determined – not enough info is given.
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E Field and Potential: Creating
A point charge q creates a field and potential around it:
!
q
E = ke 2 r̂
r
q
V (r) ! V (") = ke
r
Use superposition for systems of charges
!
V (r) ! V (") = ke
$
source
d q#
! !
r - r#
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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
 
VB − VA = − ∫ E ⋅ d s
B
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
#
Q
r̂,
a<r<b
! %
2
E = $ 4!" 0 r
%
0, elsewhere
%&
r
! !
Use VB ! V A = ! " E # d s
B
A
Region 1: r > b
r
V (r) ! V
(") = ! # 0 dr = 0
!
=0
"
No field 
No change in V!
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Electric Potential for Nested Shells
Region 2: a < r < b
Q
V (r) ! V (r = b) = ! $ dr
!
#"#
$
b
4"# 0 r 2
r
=0
Q
=
4!" 0 r
r
r
r =b
$ 1 1'
1
=
Q& # )
4!" 0 % r b (
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) = ! ( dr 0 = 0
a
" 1 1%
= kQ $ ! '
# a b&
Q $ 1 1'
V (r) = V (a) =
# )
&
4!" 0 % a b (
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 ! . Find
V (x A ) ! V (0) ; x A > d
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Deriving E from V
! !
!V = " # E $ d s
B
A
A = (x,y,z), B=(x+Δx,y,z)
!
! s = !x î
( x+!x,y,z )
!V = "
#
! !
! !
!
E $ d s % "E $ ! s = "E $(!x î) = " Ex !x
( x,y,z )
#V
%V Ex = Rate of change in V with
Ex ! "
$"
y and z held constant
#x
%x
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Deriving E from V
If we do all coordinates:
!
# "V
"V
"V &
E = !%
î +
ĵ +
k̂ (
"y
"z '
$ "x
# "
"
" &
= !%
î +
ĵ + k̂ ( V
"y
"z '
$ "x
!
!
E = ! "V
Gradient (del) operator:
!
#
#
#
!"
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.  Yes, its kQ/a2 (up)
2.  Yes, its kQ/a2 (down)
3.  Yes in theory, but I don t know how to take a
gradient
4.  No, you can t get E(P) from V(P)
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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 (a,0).
(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: 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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Potential (V)
Group Problem: E from V
10
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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Group Problem: Equipotential Game
Open applet, play with, answer
questions on worksheet and hand in
answers
http://web.mit.edu/viz/EM/visualizations/electrostatics/EandV/VfromE/VfromE.htm
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