# chapter25 ```Chapter 25
Electric Potential
Specific goals
1.
Derive equations for the difference in potential energy, and the potential
difference, between two points.
2.
Define the potential difference between two points, the potential at a point,
the volt, and the electron-volt.
3.
Derive an expression for the potential difference between two points in a
uniform electric field.
4.
Find the potential at any point due to a group of point charges.
5.
Define the electric potential energy due to a group of point charges.
6.
Solve problems on the above.
Electrical Potential Energy
⚫
When a test charge is placed in an electric field, it
experiences a force
⚫ F = q E
o
⚫
⚫
⚫
The force is conservative
If the test charge is moved in the field by some
external agent, the work done by the field is the
negative of the work done by the external agent
ds is an infinitesimal displacement vector that is
oriented tangent to a path through space
Section 25.1
Electric Potential Energy, cont
⚫
The work done by the electric field is
F  ds = qoE  ds
⚫
As this work is done by the field, the potential energy of
the charge-field system is changed by
ΔU = −qoE  ds
⚫
For a finite displacement of the charge from A to B,
B
U = UB − U A = −qo  E  ds
A
Section 25.1
Electric Potential Energy, final
⚫
⚫
Because the force 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
Section 25.1
Electric Potential
⚫
The potential energy per unit charge, U/qo, is
the electric potential
⚫
The potential is characteristic of the field only
⚫
⚫
⚫
⚫
The potential energy is characteristic of the charge-field
system
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
U
V=
qo
Section 25.1
Electric Potential, cont.
⚫
The potential is a scalar quantity
⚫
⚫
Since energy is a scalar
As a charged particle moves in an electric
field, it will experience a change in potential
B
U
V =
= −  E  ds
A
qo
Section 25.1
Electric Potential, final
⚫
⚫
⚫
The difference in potential is the
meaningful quantity
We often take the value of the potential to
be zero at some convenient point in the
field
Electric potential is a scalar characteristic
of an electric field, independent of any
charges that may be placed in the field
Section 25.1
Work and Electric Potential
⚫
⚫
Assume a charge moves in an electric field
without any change in its kinetic energy
The work performed on the charge is
W = ΔU = q ΔV
Section 25.1
Units
⚫
1 V = 1 J/C
⚫
⚫
⚫
V is a volt
It takes one joule of work to move a 1-coulomb
charge through a potential difference of 1 volt
In addition, 1 N/C = 1 V/m
⚫
This indicates we can interpret the electric field as
a measure of the rate of change with position of
the electric potential
Section 25.1
Electron-Volts
⚫
⚫
Another unit of energy that is commonly used in
atomic and nuclear physics is the electron-volt
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
Section 25.1
Quick Quiz 25.1 Part I
In the figure, two points A and B are located within a
region in which there is an electric field. How would
you describe the potential difference V = VB − VA?
(a) It is positive.
(b) It is negative.
(c) It is zero.
Section 25.1
Quick Quiz 25.1 Part II
A negative charge is placed at A and then moved to B.
How would you describe the change in potential energy
of the charge–field system for this process?
(a) It is positive.
(b) It is negative.
(c) It is zero.
ΔU = q0&middot;ΔV, so if a negative test charge is moved through a negative potential difference, the
change in potential energy is positive. Work must be done to move the charge in the direction
opposite to the electric force on it.
Section 25.1
Potential Difference in a
Uniform Field
⚫
The equations for electric potential can be
simplified if the electric field is uniform:
B
B
A
A
VB − VA = V = −  E  ds = −E  ds = −Ed
⚫
potential at point B is lower than at point A
⚫
Electric field lines always point in the direction of
decreasing electric potential
Section 25.2
Energy and the Direction of
Electric Field
⚫
When the electric field is
directed downward, point
B is at a lower potential
than point A
⚫
When a positive test
charge moves from A to B,
the charge-field system
loses potential energy
Section 25.2
⚫
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
⚫
⚫
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
⚫
Another example of Conservation of Energy
Section 25.2
Directions, cont.
⚫
⚫
If qo is negative, then ΔU 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
⚫
In order for a negative charge to move in the
direction of the field, an external agent must do
positive work on the charge
Section 25.2
Equipotentials
⚫
⚫
Point B is at a lower
potential than point A
Points B and C are at the
same potential
⚫
⚫
All points in a plane
perpendicular to a uniform
electric field are at the
same electric potential
The name equipotential
surface is given to any
surface consisting of a
continuous distribution of
points having the same
electric potential
Section 25.2
Charged Particle in a Uniform
Field, Example
⚫
⚫
⚫
⚫
⚫
A positive charge is
released from rest and
moves in the direction of the
electric field
The change in potential is
negative
The change in potential
energy is negative
The force and acceleration
are in the direction of the
field
Conservation of Energy can
be used to find its speed
Section 25.2
Quick Quiz 25.2
The labeled points in the figure are on a series of equipotential surfaces
associated with an electric field. Rank (from greatest to least) the work done
by the electric field on a positively charged particle that moves from A to B,
from B to C, from C to D, and from D to E.
B to C,
C to D,
A to B,
D to E
Moving from B to C decreases the electric potential by 2 V, so the electric field performs 2 J of
work on each coulomb of positive charge that moves. Moving from C to D decreases the
electric potential by 1 V, so 1 J of work is done by the field. It takes no work to move the charge
from A to B because the electric potential does not change. Moving from D to E increases the
electric potential by 1 V; therefore, the field does −1 J of work per unit of positive charge that
moves.
Section 25.2
Example 25.1: The Electric Field Between
Two Parallel Plates of Opposite Charge
A battery has a specified potential difference V between its
terminals and establishes that potential difference between conductors
attached to the terminals. A 12-V battery is connected between two
parallel plates as shown in the figure. The separation between the
plates is d = 0.30 cm, and we assume the electric field between the
plates to be uniform. (This assumption is reasonable
if the plate separation is small
relative to the plate dimensions
and we do not consider locations
near the plate edges.) Find the
magnitude of the electric field
between the plates.
Section 25.2
Example 25.1: The Electric Field Between
Two Parallel Plates of Opposite Charge
E=
VB − VA
d
12 V
3
=
=
4.0

10
V/m
−2
0.30 10 m
Section 25.2
Example 25.2:
Motion of a Proton in a Uniform Electric Field
A proton is released from rest at point A in a
uniform electric field that has a magnitude of
4
8.0  10 V/m, as shown in the figure. The
proton undergoes a displacement of magnitude
d = 0.50 m to point B in the
direction of E. Find the speed
of the proton after completing
the displacement.
Section 25.2
Example 25.2:
Motion of a Proton in a Uniform Electric Field
1 2

  mv − 0  + eV = 0
2

K + U E = 0
v=
v=
−2eV
=
m
−2e ( − Ed )
m
=
2eEd
m
2 (1.6  10 −19 C )( 8.0 10 4 V ) ( 0.50 m )
1.67  10 −27 kg
= 2.8  106 m/s
Section 25.2
Potential and Point Charges
⚫
A positive point charge produces a field directed radially outward
B
VB − VA = −  E  d s
A
E  d s = ke
q
rˆ  d s
2
r
q
E  d s = ke 2 dr
r
ke q
E= 2 r
r
rˆ  d s = ds cos 
ds cos  = dr
dr
q
VB − VA = −ke q  2 = ke
rA r
r
1 1
VB − VA = ke q  − 
 rB rA 
rB
VA = 0 at rA = 
q
V = ke
r
Section 25.3
rB
rA
Potential and Point Charges,
cont.
⚫
⚫
⚫
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
Section 25.3
Electric Potential with Multiple
Charges
⚫
The electric potential due to several point
charges is the sum of the potentials due to
each individual charge
⚫
⚫
This is another example of the superposition
principle
The sum is the algebraic sum
⚫
qi
V = ke 
i ri
V = 0 at r = ∞
Section 25.3
Potential Energy of Multiple charges
⚫
The potential energy of the system is
.
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.
⚫
Section 25.3
U with Multiple Charges, final
⚫
⚫
If there are more than
two charges, then find
U for each pair of
For three charges:
 q1q2 q1q3 q2q3 
U = ke 
+
+

r
r
r
13
23 
 12
⚫
The result is independent
of the order of the
charges
Section 25.3
Quick Quiz 25.3 Part I
In the figure, take q2 to be a negative source charge and q1 to be a
second charge whose sign can be changed. If q1 is initially
positive and is changed to a charge of the same magnitude but
negative, what happens to the potential at the position of q1 due
to q2?
(a) It increases.
(b) It decreases.
(c) It remains the same.
Section 25.3
Quick Quiz 25.3 Part II
In the figure, take q2 to be a negative source charge and
q1 to be a second charge whose sign can be changed.
When q1 is changed from positive to negative, what
happens to the potential energy of the two-charge
system?
(a) It increases.
(b) It decreases.
(c) It remains the same.
Section 25.3
Example 25.3: The Electric Potential
Due to Two Point Charges
As shown in the figure, a charge q1 = 2.00 C is
located at the origin and a charge q2 = −6.00 C is
located at (0, 3.00) m.
(A) Find the total electric potential due to these charges
at the point P, whose coordinates are (4.00, 0) m.
 q1 q2 
VP = ke  + 
 r1 r2 
VP = ( 8.988  109 N  m 2 /C 2 )
 2.00 106 C −6.00 10 −6 C 

+

4.00
m
5.00
m


= −6.29 103 V
Section 25.3
Example 25.3: The Electric Potential
Due to Two Point Charges
(B) Find the change in potential energy of the system of two
charges plus a third charge q3 = 3.00 C as the latter charge moves
from infinity to point P.
U f = q3VP
U E = U f − U i
= q3VP − 0
= ( 3.00 10−6 C )( −6.29 103 V )
= −1.89  10−2 J
Section 25.3
Finding E From V
⚫
Assume, to start, that the field has only an x
component
dV
Ex = −
dx
⚫
Similar statements would apply to the y and z
components
⚫
Equipotential surfaces must always be
perpendicular to the electric field lines passing
through them
Section 25.4
E and V for an Infinite Sheet of
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
Section 25.4
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
Section 25.4
E and V for a Dipole
⚫
⚫
⚫
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
Section 25.4
Electric Field from Potential,
General
⚫
⚫
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
V
Ex = −
x
V
Ey = −
y
V
Ez = −
z
Section 25.4
Quick Quiz 25.4 Part I
In a certain region of space, the electric potential is zero
everywhere along the x axis. From this information, you
can conclude that the x component of the electric field
in this region is
(a) zero,
(b) in the positive x direction, or
(c) in the negative x direction.
Section 25.4
Quick Quiz 25.4 Part II
Suppose the electric potential is +2 V everywhere along
the x axis. From this information, you can conclude that
the x component of the electric field in this region is
(a) zero,
(b) in the positive x direction, or
(c) in the negative x direction.
Section 25.4
Tutorial Problems
NB: All problems are from the prescribed text book (9th Edition)
P 25.1
P 25.5
P 25.14
P 25.25
P 25.27
```