Electric Potential
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Physics for Scientists & Engineers 2, Chapter 23
1
Electric Potential Difference ΔV
! The electric potential difference between an initial point i
and final point f is
U f U i ΔU
ΔV = Vf −Vi =
− =
q
q
q
! Work and potential are related through
We
ΔV = −
q
! The units of electric potential are
1J
1V=
1C
! We express the electric field as
[F] N J/m V
[E]=
= =
=
[q] C
C m
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Physics for Scientists & Engineers 2, Chapter 23
2
General Considerations
! If a charged particle moves perpendicular to electric
field lines, no work is done
 
 
W = qE ⋅ d = 0 if d ⊥ E
! If the work done by the electric field is zero, then the electric
potential must be constant
! Equipotential surfaces and lines must always be
perpendicular to the electric field lines
! General observations:
• The surface of any conductor forms an equipotential surface
• Equipotential surface are always perpendicular to the electric field
lines at any point in space
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Physics for Scientists & Engineers 2, Chapter 23
3
Single Point Charge
! The electric field lines for point charges are radial
! The equipotential surfaces are concentric spheres in 3D
! The equipotential lines take the form of concentric circles in 2D
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Positive charge
Negative charge
Physics for Scientists & Engineers 2, Chapter 23
4
Electric Potential of Charge Distributions
! The electric potential is defined as the work required to
place a unit charge at a point
! Work is a force acting over a distance
! The electric field is defined as the force acting on a unit
charge at a point
! So the electric potential is related to the electric field
• We can determine one given the other
! To determine the electric potential from the electric field we
start with the work done on a charged particle by a force
over a displacement
 
 
dW = F ⋅ds = qE ⋅ds
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Physics for Scientists & Engineers 2, Chapter 23
5
Electric Potential of Charge Distributions
! Now we integrate the work as the charge moves in the field
W=∫
f
i
 
f 

qE ⋅ds = q ∫ E ⋅ds
i
! Now we can relate the work done to the change in electric
potential
f 
We

ΔV = Vf −Vi = −
= − ∫ E ⋅ds
i
q
! Taking the standard convention, we can write the electric
potential in terms of the electric field

r 



V ( r ) −V ( ∞ ) = V ( r ) = − ∫ E ⋅ds
∞
! In other words, the electric potential is the integral of the
electric field
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Physics for Scientists & Engineers 2, Chapter 23
6
Electric Potential for a Point Charge
! Let’s calculate the electric potential due to a point charge
! The magnitude of the electric field is given by
kq
E (r ) = 2
r
! The direction of the electric field is radial from the charge
! Integrate along a radial line from ∞ to R
R
 
R kq
kq
kq
⎡ ⎤
V ( R ) = − ∫ E ⋅ds = − ∫ 2 dr = ⎢ ⎥ =
∞
∞ r
⎣ r ⎦∞ R
R
! So the electric potential due to a point charge is
kq
V=
R
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Physics for Scientists & Engineers 2, Chapter 23
7
Electric Potential for a Point Charge
! A positive point charge produces a positive potential
! A negative point charge produces a negative potential
! We can calculate the electric potential for all points in an
xy-plane
V ( x, y ) =
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kq
x2 + y2
Physics for Scientists & Engineers 2, Chapter 23
8
Example - Charge moves in E field
! Given is a uniform electric field E. Find
the potential difference Vf-Vi by moving a
test charge q0 along the path icf.
i
c
! Idea: Integrate
along the path
connecting ic and cf. Imagine, we move a
test charge q0 from i to c and from c to f.
f
January 29, 2014
Physics for Scientists&Engineers 2
9
Example - Charge moves in E field
! Given is a uniform electric field E. Find
the potential difference Vf-Vi by moving a
test charge q0 along the path icf.
i
c
! Along ic, ds and E are perpendicular
f
With the distance from c to f: d/sin(45o)
January 29, 2014
Physics for Scientists&Engineers 2
10
Fixed and Moving Positive Charges
PROBLEM:
! A positive charge of 4.50 μC is fixed in place.
! A particle of mass 6.00 g and charge +3.00 μC is fired with
an initial speed of 66.9 m/s directly toward the fixed charge
from a distance of 4.20 cm.
! How close does the moving charge get to the fixed charge?
SOLUTION:
Think
! The moving charge will gain potential energy as it near the
fixed charge.
! The negative of the change in potential energy of the moving
charge is equal to the change in kinetic energy of the
moving charge because ΔK + ΔE = 0.
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Chapter 23
13
Fixed and Moving Positive Charges
Sketch
Research
! The moving charge gains electric potential energy and loses
kinetic energy until it stops.
! At that point, all the original kinetic energy has been
converted into electric potential energy.
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Chapter 23
14
Fixed and Moving Positive Charges
! We can express this energy conservation as:
ΔK + ΔU = 0 ⇒ ΔK = −ΔU
1 2
1 2
0− mv0 = −qmovingΔV ⇒
mv0 = qmovingΔV
2
2
! The electric potential’s effect on the moving charge is due to
the fixed charge:
⎛ 1 1 ⎞⎟
qfixed
qfixed
ΔV = Vf −Vi = k
−k
= kqfixed ⎜⎜ − ⎟⎟
⎜⎝ df di ⎟⎠
df
di
Simplify
! Substitute in our expression for the potential difference:
⎛ 1 1 ⎞⎟
1 2
mv0 = qmovingΔV = kqmoving qfixed ⎜⎜ − ⎟⎟ ⇒
⎜⎝ df di ⎟⎠
2
1
1
mv02
= +
df di 2kqmoving qfixed
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Chapter 23
15
Fixed and Moving Positive Charges
Calculate
! Putting in our numerical values we get:
(0.00600 kg)(66.0 m/s)
1
1
=
+
df 0.0420 m 2(8.99⋅109 N m 2 /C2 )(3.00⋅10−6 C)(4.50⋅10−6 C)
2
1
= 131.485 m−1 ⇒ df = 0.00760545 m
df
Round
! We round our result to three significant figures:
df = 0.00761 m = 0.761 cm
Double-check
! The final distance of 0.761 cm is less than 4.20 cm.
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Chapter 23
16
Fixed and Moving Positive Charges
! At the final distance, the electric potential energy of the
moving charge is:
⎛ q ⎞⎟
qmoving qfixed
⎜
fixed ⎟
U = qmovingV = qmoving ⎜⎜ k
=k
⎟
⎟
⎜⎝ d ⎠
d
f
f
2 ⎞ 3.00⋅10−6 C 4.50⋅10−6 C
⎛
(
)
(
)
9 N m ⎟
⎜
⎟
U = ⎜⎜8.99⋅10
= 16.0 J
2 ⎟
⎜⎝
0.00761 m
C ⎟⎠
! At the initial distance, the electric potential energy of the
moving charge is:
⎛ q ⎞⎟
qmoving qfixed
⎜
fixed ⎟
U = qmovingV = qmoving ⎜⎜ k
=k
⎟
⎜⎝ d ⎟⎠
d
i
i
2 ⎞ 3.00⋅10−6 C 4.50⋅10−6 C
⎛
(
)
(
)
9 N m ⎟
⎜
⎟
U = ⎜⎜8.99⋅10
= 2.90 J
2 ⎟
⎜⎝
0.0420 m
C ⎟⎠
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Chapter 23
17
Fixed and Moving Positive Charges
! The initial kinetic energy is:
1 2 (0.00600 kg)(66.0 m/s)
K = mv =
= 13.1 J
2
2
2
! We can see the energy conservation is satisfied:
K = ΔU
13.1 J = 16.0 J − 2.90 J = 13.1 J
! We are confident in our result.
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Chapter 23
18
Finding the Field from the Potential
! We can calculate the electric field from the electric potential
starting with:
 
 
−qdV = qE i ds
⇒ E i ds = −dV
! 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 = −
in the direction of ds
∂s
! We can write the components of the electric field in terms of
partial derivatives of the potential:
∂V
∂V
∂V
Ex = −
; Ey = −
; Ez = −
∂x
∂y
∂z


⎛ ∂V ∂V ∂V ⎞⎟ 
⎟ ∇ is the gradient
E = −∇V ≡ −⎜⎜
,
,
⎜⎝ ∂x ∂ y ∂z ⎟⎟⎠
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Chapter 23
19
Graphical Extraction of the Electric Field
! 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 line of +2000 V
to the line of 0 V.
! 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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ΔV
Δs
Chapter 23
20
Graphical Extraction of the Electric Field
! Putting in our numbers give us:
ES =
(+2000 V)−(0 V)
1.5 cm
ES = 1.3⋅105 V/m
! The direction of the electric
field points from the
positive equipotential line
to the negative potential line.
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Chapter 23
21
System of Point Charges
! We calculate the electric potential from a system of n point
charges by adding the potential functions from each charge
kqi
V = ∑Vi = ∑
i=1
i=1 ri
n
n
! 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
• It’s a scalar
1/29/14
Physics for Scientists & Engineers 2, Chapter 23
23
Superposition of Electric Potential
! Assume we have a system of three point charges:
q1 = +1.50 μC
q2 = +2.50 μC
q3 = -3.50 μC
! 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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Chapter 23
24
Superposition of Electric Potential
! The electric potential at point P is given by the sum of the
electric potential from the three charges:
kqi
V =∑
i=1 ri
3
⎛q
⎞⎟
⎛q q
⎞⎟
q
q
q
⎜
⎜
2
V = k ⎜⎜ 1 + 2 + 3 ⎟⎟ = k ⎜⎜ 1 +
+ 3 ⎟⎟
⎜⎝ r1 r2 r3 ⎟⎠
⎜⎝ b
a ⎟⎟⎠
a 2 + b2
⎛
⎛
⎞⎟⎜⎜1.50⋅10−6 C
N
m
9
⎜⎜
⎟
V = ⎜⎜⎜8.99⋅10
+
⎟
2 ⎟⎜
⎜⎝
C ⎠⎜ 6.00 m
⎜⎝
−6
2
2.50⋅10
C
(8.00 m) + (6.00 m)
2
⎞⎟
−3.50⋅10 C ⎟⎟
⎟⎟
+
⎟⎟
8.00 m
⎟⎠
−6
2
V = 562 V
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Chapter 23
25
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, U = 0, by convention.
! 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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Chapter 23
26
Two Charges
! 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. 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.
! This requires work q2V1(r).
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Chapter 23
27
Two Charges
! The electric potential energy of this two charge system is:
kq1
U = q2V1 (r) where V1 (r) =
r
! The electric potential of the two charge system is:
kq1q2
U=
r
! If the two point charges have the same sign, then we must
do positive work on the particles to bring them together
from infinity (i.e., we must put energy into the system).
! If the two charges have opposite signs, we must do negative
work on the system to bring them together from infinity
(i.e., energy is released from the system).
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Chapter 23
28
Three Charges
! Now let’s try three charges.
! Consider three point charges, q1, q2,
and q3, at fixed positions arranged an
equal distance d from each other.
! What is the electric potential energy
U of the assembly of these charges?
! The potential energy is equal to
the work we must do to assemble
the system, bringing in each
charge from an infinite
distance.
! Let’s build the system by bringing the charges in from
infinity, one at a time.
© W. Bauer and G.D. Westfall
1/29/14
Chapter 23
29
Three Charges
! Bringing in q1 requires no work.
! With q1 in place, bring in q2:
q1
U12 = W = q2V = q2 k
d
© W. Bauer and G.D. Westfall
! Now bring in q3:
W13 +W23 = U13 +U 23 = q3 (V1 +V2 )
⎛ q
q2 ⎞⎟
1
⎜
U13 +U 23 = q3 ⎜⎜ k + k ⎟⎟
d ⎟⎠
⎝ d
! The total electric potential of the
system is then:
U123 = U12 +U13 +U 23
⎛ q
q1
q2 ⎞⎟ k
1
⎜
U123 = q2 k + q3 ⎜⎜ k + k ⎟⎟ = (q1q2 + q1q3 + q2 q3 )
d
d ⎟⎠ d
⎝ d
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Chapter 23
30