Transcript Lec10drs

PHY 184
Spring 2007
Lecture 10
Title: The Electric Potential, V(x)
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Review - Electric Potential V(x)
…a scalar function of position
 The change in electric potential energy U of a charge q
that moves in an electric field is related to the change in
electric potential V
U
V 
q
or
ΔU  qΔΔ
 The unit of electric potential is the volt, V.
 The unit of electric field is V/m.
 For reference state at infinity,
V (x)  
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
x
 
E  ds
2
Electric Potential for a Point Charge
 We’ll derive the electric potential for a point
source q, as a function of distance R from the
source. That is, V(R).
 Remember that the electric field from a point
charge q at a distance r is given by
 The direction of the electric field from a point
charge is always radial.
 We integrate from distance R (distance from the
point charge) along a radial to infinity:
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Electric Potential of a Point Charge (2)

The electric potential V from a point charge q at a distance r is then
kq
V (r ) 
r
Negative point charge
Positive point charge
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Electric Potential from a System of Charges
 We calculate the electric potential from a system
of n point charges by adding the potential functions
from each charge
n
n
kqi
V   Vi  
i 1
i 1 ri
 This summation produces an electric potential at all
points in space – a scalar function
 Calculating the electric potential from a group of
point charges is usually much simpler than
calculating the electric field. (because it’s a scalar)
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Example - Superposition of Electric Potential
 Assume we have a system of
three point charges:
q1 = +1.50 C
q2 = +2.50 C
q3 = -3.50 C.
 q1 is located at (0,a)
q2 is located at (0,0)
q3 is located at (b,0)
a = 8.00 m and b = 6.00 m.
 What is the electric potential at
point P located at (b,a)?
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Example - Superposition of Electric Potential (2)
 The electric potential at point P is
given by the sum of the electric
potential from the three charges
r1
r2
r3
 q1
 q1 q2 q3 
kqi
q2
q3 
V 
 k     k 
 
2
2
a
 r1 r2 r3 
b
a b
i 1 ri
3
 1.50 10 6 C
V  8.99 10 N/C 

 6.00 m

9

2.50 10 6 C
8.00 m   6.00 m 
2
2
3.50 10 6 C 


8.00 m 
V  562 V
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Clicker Question - Electric Potential
 Rank (a), (b) and (c) according to the net electric
potential V produced at point P by two protons.
Greatest first!
A: (b), (c), (a)
B: all equal
C: (c), (b), (a)
D: (a) and (c) tie, then (b)
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Clicker Question - Electric Potential
 Rank (a), (b) and (c) according to the net electric potential V produced
at point P by two protons. Greatest first!
B: all equal
2qd
V
a
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Calculating the Field from the Potential
 We can calculate the electric field from the electric
potential starting with
We,
V 
q
 Which allows us to write
 If we look at the component of the electric field along the
direction of ds, we can write the magnitude of the electric
field as the partial derivative along the direction s
V
ES  
s
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Math Reminder - Partial Derivatives
 Given a function V(x,y,z), the partial derivatives are
act on x, y and z independently
 Example: V(x,y,z)=2xy2+z3
Meaning: partial derivatives
give the slope along the
respective direction
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Calculating the Field from the Potential (2)
 We can calculate any component of the electric field by
taking the partial derivative of the potential along the
direction of that component.
 We can write the components of the electric field in terms
of partial derivatives of the potential as
V
V
V
Ex  
; Ey  
; Ez  
x
y
z
 In terms of graphical representations of the electric
potential, we can get an approximate value for the electric
field by measuring the gradient of the potential
perpendicular to an equipotential line

E  V
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Example - Graphical Extraction of the Field from the Potential
 Assume a system of three point charges
q1  6.00 C
q2  3.00 C
q3  9.00 C
x1, y1   1.5 cm, 9.0 cm  x2 , y2   6.0 cm, 8.0 cm  x3 , y3   5.3 cm, 2.0 cm 
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Example - Graphical Extraction of the Field
from the Potential (2)



We calculate the magnitude of the
electric field at point P.
To perform this task, we draw a
line through point P perpendicular
to the equipotential line reaching
from the equipotential line of +1000
V to the line of –1000V.
The length of this line is 1.5 cm. So
the magnitude of the electric field
can be approximated as
ES  

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V
2000 V  0 V  1.3105 V/m

s
1.5 cm
The direction of the electric field
points from the positive
equipotential line to the negative
potential line.
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Clicker Question - E Field from Potential
Pairs of parallel plates with the same separation and a given V of
each plate. The E field is uniform between plates and
perpendicular to them. Rank the magnitude of the electric field
E between them. Greatest first!
A: (1), (2), (3)
B: (3) and (2) tie, then (1)
C: all equal
D: (2), then (1) and (3) tie
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Clicker Question - E Field from Potential
Pairs of parallel plates with the same separation d and a given V
of each plate. The E field is uniform between plates and
perpendicular to them. Rank the magnitude of the electric field
E between them. Greatest first!
D: (2), then (1) and (3) tie
Use
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and take the magnitude only
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Electric Potential Energy for a
System of Particles
 So far, we have discussed the electric potential energy of a point
charge in a fixed electric field.
 Now we introduce the concept of the electric potential energy of a
system of point charges.
 In the case of a fixed electric field, the point charge itself did not
affect the electric field that did work on the charge.
 Now we consider a system of point charges that produce the electric
potential themselves.
 We begin with a system of charges that are infinitely far apart.
Reference state, U = 0.
 To bring these charges into proximity with each other, we must do work
on the charges, which changes the electric potential energy of the
system.
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Electric Potential Energy for a
System of Particles (2)
 To illustrate the concept of the electric potential energy of a system of
particles we calculate the electric potential energy of a system of two
point charges, q1 and q2 .
 We start our calculation with the two charges at infinity.
 We then bring in point charge q1; ; that requires no work.
 Because there is no electric field and no corresponding electric force,
this action requires no work to be done on the charge.
 Keeping this charge (q1) stationary, we bring the second point charge
(q2) in from infinity to a distance r from q1; that requires work q2 V1(r).
q1
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r
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q2
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Electric Potential Energy for a
System of Particles (3)
 So, the electric potential energy of this two charge system is
U  q2V1 (r )
kq1
where V1 ( r ) 
r
 Hence the electric potential of the two charge system is
q1
r
kq1q 2
U
r
q2
 If the two point charges have the same sign, then we must do positive
work on the particles to bring them together.
 If the two charges have opposite signs, we must do negative work on the
system to bring them together from infinity.
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Example - Electric Potential Energy (1)
 Consider three point charges at
fixed positions. What is the
electric potential energy U of the
assembly of charges?
Key Idea: The potential energy is
equal to the work we must do to
assemble the system, bringing in
each charge from an infinite
distance.
Strategy: Let’s build the system by
starting with one charge in place
and bringing in the others from
infinity.
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q1=+q, q2=-4q, q3=+2q
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Example - Electric Potential Energy (2)
Strategy: Let’s say q1 is in place and
we bring in q2.
We then bring in charge 3. The
work we must do to bring 3 to its
place relative to q1 and q2 is then:
q1=+q, q2=-4q, q3=+2q
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Example - Electric Potential Energy (3)
Total potential energy: Sum over U’s
for all pairs of charges
Picture D
Answer :
q1=+q, q2=-4q, q3=+2q
The negative potential energy means that negative work would
have to be done to assemble the system starting with three
charges at infinity.
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Example - Four Charges
 Consider a system of four
point charges as shown. The
four point charges have the
values q1 =+1.0 C, q2 = +2.0 C,
q3 = -3.0 C, and q4 = +4.0 C.
The charges are placed such
that a = 6.0 m and b = 4.0 m.
 What is the electric potential
energy of this system of four
point charges?
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Example - Four Charges (2)
Energy of the complete assembly
 q1q2 q1q3 q1q4
U  k


b
D
 a
q2q3 q2q4 q3q4 




D
b
a 
= sum of pairs
Answer: 1.2 x 10-3 J
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Example - 12 Electrons on a Circle
 Consider a system of 12 electrons arranged
on a circle with radius R as indicated in the
figure. Relative to V=0 at infinity, what are
the electric potential V and the electric
field E at point C?
 Superposition principle:
12
V (C )  
i 1


 e
 12e 
k
k
R
R
 Symmetry: E=0
The electric field of any given electron is
canceled by the field due to the electron
located at the diametrically opposite
position.

E (C )  0
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Clicker Question
 Consider a rearrangement of the 12 electrons as
indicated in the figure. Relative to previous
arrangement of the electrons, what are the
electric potential V and the electric field E at
point C?
A:
B:
C:
D:
V
V
V
V
is
is
is
is
unchanged, E not 0 anymore
bigger, still E=0
smaller, E not 0 anymore
the same, still E=0
The symmetry is lost, there is no total
cancellation of the electric fields anymore
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