Electrical Potential Energy
Chapter 25
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Chapter 25
When a test charge, q0 is placed in an
electric field E, it experiences a force:
F = q0 E
Electric Potential
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Is this force conservative?
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2
Electric Potential Energy
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Electric Potential Energy, final
What is the work done by the electric
field?
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!
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Because qoE is conservative, the line
integral does not depend on the path
taken by the charge
This is the change in potential energy of
the system.
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Electric Potential
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The potential energy per unit charge, U/qo,
is the electric potential
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Electric Potential, cont.
The potential is independent of the value of qo
The potential has a value at every point in an
electric field
The electric potential is V =
!
The potential is a scalar quantity
!
!
U
qo
Since energy is a scalar
As a charged particle moves in an
electric field, it will experience a change
in potential
∆V =
B
∆U
= − ∫ E ⋅ ds
A
qo
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Work and Electric Potential
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!
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Units
Assume a charge moves in an electric
field without any change in its kinetic
energy
The work performed on the charge is
!
1 V = 1 J/C
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!
W = ∆U = q ∆V
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In addition, 1 N/C = 1 V/m
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V is a volt
It takes one joule of work to move a 1coulomb charge through a potential
difference of 1 volt
This indicates we can interpret the electric
field as a measure of the rate of change
with position of the electric potential
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2
Electron-Volts
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Illustration by Debra Solomon
Another unit of energy that is commonly used
in atomic and nuclear physics is the electronvolt
One electron-volt is defined as the energy a
charge-field system gains or loses when a
charge of magnitude e (an electron or a
proton) is moved through a potential
difference of 1 volt
!
1 eV = 1.60 x 10-19 J
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Potential Difference in a
Uniform Field
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Problem 1
The equations for electric potential can
be simplified if the electric field is
uniform:
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2. A uniform electric field of magnitude 325 V/m is
directed in the negative y direction as shown in the
figure. The coordinates of point A are (-0.200, -0.300)
m, and those of point B are (0.400, 0.500) m. Calculate
the potential difference VB – VA , using the blue path.
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3
Energy and the Direction of
Electric Field
+
∆PE is ____
Problem 2
_
1. Calculate the speed in km/s of a proton that is
accelerated from rest through a potential difference of
133 V.
∆PE is ____
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More About Directions
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Directions, cont.
A system consisting of a positive charge and
an electric field loses electric potential
energy when the charge moves in the
direction of the field
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An electric field does work on a positive charge
when the charge moves in the direction of the
electric field
The charged particle gains kinetic energy
equal to the potential energy lost by the
charge-field system
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!
!
If qo is negative, then ∆U is positive
A system consisting of a negative
charge and an electric field gains
potential energy when the charge
moves in the direction of the field
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In order for a negative charge to move in
the direction of the field, an external agent
must do positive work on the charge
Another example of Conservation of Energy
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Equipotentials
Potential and Point Charges
Point B is at _____
potential than point A
Points B and C ?
The name equipotential
surface is given to any
surface consisting of a
continuous distribution of
points having the same
electric potential
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!
A positive point
charge produces a
field directed radially
outward
The potential
difference between
points A and B is
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Potential and Point Charges,
cont.
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Potential Energy of Multiple
Charges
The electric potential is independent of
the path between points A and B
It is customary to choose a reference
potential of V = 0 at rA = ∞
Then the potential at some point r is
q
V = ke
r
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!
!
Consider two
charged particles
The potential energy
of the system is
U = ke
q1q2
r12
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More About U of Multiple
Charges
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U with Multiple Charges, final
If the two charges are the same sign, U
is positive and work must be done to
bring the charges together
If the two charges have opposite signs,
U is negative and work is done to keep
the charges apart
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!
If there are more than
two charges, then find
U for each pair of
charges and add them
For three charges:
qq
qq
qq 
U = ke  1 2 + 1 3 + 2 3 
r
r
r23 
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 12
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The result is
independent of the
order of the charges
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E and V for an Infinite Sheet
of Charge
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E and V for a Point Charge
The equipotential lines
are the dashed blue
lines
The electric field lines
are the brown lines
The equipotential lines
are everywhere
perpendicular to the
field lines
!
!
!
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The equipotential lines
are the dashed blue
lines
The electric field lines
are the brown lines
The equipotential lines
are everywhere
perpendicular to the
field lines
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Electric Field from Potential,
General
E and V for a Dipole
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!
!
The equipotential lines
are the dashed blue
lines
The electric field lines
are the brown lines
The equipotential lines
are everywhere
perpendicular to the
field lines
!
!
In general, the electric potential is a
function of all three dimensions
Given V (x, y, z) you can find Ex, Ey and
Ez as partial derivatives
Ex = −
∂V
∂x
Ey = −
∂V
∂y
Ez = −
∂V
∂z
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V for a Continuous Charge
Distribution (Extra material)
Problem 3
3. Over a certain region of space, the electric potential is V
= 2x - x2y + 2yz2. Find the expression for the x, y, and z
component of the electric field over this region. (Use x, y,
and z as necessary.)
!
To find the total potential, you need
to integrate to include the
contributions from all the elements
V = ke ∫
!
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dq
r
This value for V uses the
reference of V = 0 when P is
infinitely far away from the charge
distributions
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V for a Uniformly Charged
Ring (Extra material)
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V for a Uniformly Charged
Disk (Extra material)
P is located on the
perpendicular central
axis of the uniformly
charged ring
!
!
V = 2πkeσ  x 2 + a 2

(
The ring has a radius
a and a total charge Q
V = ke ∫
dq
=
r
The ring has a
radius a and surface
charge density of σ
)
1
2
− x 

k eQ
x 2 + a2
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V for a Finite Line of Charge
(Extra material)
!
V for a Uniformly Charged
Sphere (Extra material)
A rod of line ℓ has a
total charge of Q
and a linear charge
density of λ
k Q  ! + !2 + a 2
V = e ln 

!
a

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!
!




!
A solid sphere of
radius R and total
charge Q
Q
For r > R, V = k e
r
For r < R,
VD − VC =
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k eQ 2
R −r2
2R 3
(
)
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The Xerographic Process
Electrostatic Precipitator
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An application of electrical discharge in
gases is the electrostatic precipitator
It removes particulate matter from
combustible gases
The air to be cleaned enters the duct and
moves near the wire
As the electrons and negative ions created
by the discharge are accelerated toward the
outer wall by the electric field, the dirt
particles become charged
Most of the dirt particles are negatively
charged and are drawn to the walls by the
electric field
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Potentials Due to Various
Charge Distributions
Application – Laser Printer
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The steps for producing a document on a laser
printer is similar to the steps in the xerographic
process
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Steps a, c, and d are the same
The major difference is the way the image forms on
the selenium-coated drum
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A rotating mirror inside the printer causes the beam of the
laser to sweep across the selenium-coated drum
The electrical signals form the desired letter in positive
charges on the selenium-coated drum
Toner is applied and the process continues as in the
xerographic process
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