Chapter 23
Electric Potential
Lecture 5
Dr. Armen Kocharian
Electrical Potential Energy
„
When a test charge is placed in an
electric field, it experiences a force
„
„
„
F = qoE
The force is conservative
ds is an infinitesimal displacement
vector that is oriented tangent to a path
through space
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
ΔU = UB − U A = −qo ∫ E ⋅ ds
A
Electric Potential Energy, final
„
„
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
Electric Potential
„
The potential energy per unit charge, U/qo,
is the electric potential
„
„
„
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
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
ΔU
ΔV =
= − ∫ E ⋅ ds
A
qo
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
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 = ΔV = q ΔV
Units
„
1 V = 1 J/C
„
„
„
V is a volt
It takes one joule of work to move a 1coulomb 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
Electron-Volts
„
„
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
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 = ΔV = − ∫ E ⋅ ds = −E ∫ ds = −Ed
„
The negative sign indicates that the
electric potential at point B is lower than
at point A
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
More About Directions
„
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
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
Equipotentials
„
„
„
Point B is at a lower
potential than point A
Points A and C are at the
same potential
The name equipotential
surface is given to any
surface consisting of a
continuous distribution of
points having the same
electric potential
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
Potential and Point Charges
„
„
A positive point
charge produces a
field directed radially
outward
The potential
difference between
points A and B will
be
⎡1 1⎤
VB − VA = keq ⎢ − ⎥
⎣ 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
Problem 30.
„
q
V = ke
r
Plot a graph for potential along
x direction
ke ( +Q )
keQ 1 keQ 2
+
=
+
V ( x) =
2
2
r1
r2
x +a
V ( x) =
V ( x)
( keQ a)
⎛
2keQ
kQ
= e ⎜
a ⎜
x2 + a2
⎝
=
2
( x a)2 + 1
⎞
2
⎟
( x a)2 + 1⎟⎠
ke ( +Q )
x2 + ( − a)
2
Problem 30,
„
q
cont. V = ke r
Now let charge at –a to be negative
(-Q) and plot along y direction
keQ 1 keQ 2 ke ( +Q ) ke ( −Q )
+
=
+
V ( y) =
r1
r2
y− a
y+ a
keQ ⎛
1
1 ⎞
−
V ( y) =
⎜
⎟
a ⎝ y a− 1 y a+ 1 ⎠
V ( y)
⎛
1
1 ⎞
= ⎜
−
( keQ a) ⎝ y a− 1 y a+ 1 ⎟⎠
Electric Potential of a Point
Charge
„
„
The electric
potential in the
plane around a
single point charge
is shown
The red line shows
the 1/r nature of the
potential
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 = ∞
Electric Potential of a Dipole
„
„
The graph shows
the potential (y-axis)
of an electric dipole
The steep slope
between the
charges represents
the strong electric
field in this region
Today: Electric Potential
Energy
„
„
„
You should be familiar with the concept of
gravitational potential energy from Physics 1
Let's review
If a force
acts on a particle as the particle
moves from aÆb, then
is the work done by the force
(
is the infinitesimal displacement along the path)
b
a
Careful: the force does not necessarily
line up with the displacement
For example, a block sliding down an inclined
plane under the influence of gravity:
„
Conservative
force
A force is conservative if the work done by the
force is independent of path
„
Only depends on the initial and final points
b
a
„
Then the work done can be written as function
of the difference between some properties of
the begin and final point
„
„
„
U is the potential energy
W = -ΔU
Work energy theorem:
work = change in kinetic energy
Wa→b = K(b) – K(a)
K(a) + U(a) = K(b) + U(b)
„
Potential energy defined up to additive
constant
Today: Electric Potential
Energy
„
„
„
You should be familiar with the concept of
gravitational potential energy from Physics 1
Let's review
If a force
acts on a particle as the particle
moves from aÆb, then
is the work done by the force
(
is the infinitesimal displacement along the path)
„
„
„
U is the potential energy
W = -ΔU
Work energy theorem:
work = change in kinetic energy
Wa→b = K(b) – K(a)
K(a) + U(a) = K(b) + U(b)
„
Potential energy defined up to additive
constant
Remember gravitational
field, force, potential energy
„
„
„
„
Near the surface of the earth, constant
force (F=const)
Think of it as mass times constant
gravitational field
Then gravitational potential energy
U=mgh (uniform field)
Then gravitational potential energy for
two masses is
m1m2
UG = G
r12
Potential Energy of Multiple
Charges
„
„
Consider two
charged particles
The potential energy
of the system is
q1q2
U = ke
r12
More About U of Multiple
Charges
„
„
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
U with Multiple Charges, final
„
„
If there are more than
two charges, then find
U for each pair of
charges and add them
For three charges:
⎛ q1q2 q1q3 q2q3 ⎞
+
+
U = ke ⎜
⎟
r
r
r
13
23 ⎠
⎝ 12
„
The result is
independent of the
order of the charges
Finding E From V
„
„
„
Assume, to start, that E 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
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
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
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
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
Equipotentials and Conductors
„
„
„
We argued previously that the electric field
must be perpendicular to the surface of a
conductor
This is because otherwise the charges on the
surface of the conductor would be moving
It then follows that the surface of a conductor
is an equipotential
Potential Gradient
„
„
„
„
We will now derive a fundamental relationship
between potential and electric field
But
This must hold for any (a,b) pair and any path
between the two
For this to be true then
„
Write out the components
„
Then, in terms of components:
„
„
Suppose the displacement is in the x-direction
Then dy=dz=0, and –dV=Exdx, or
„
„
„
Can do same thing for the other two
components
Or in short-hand notation
is called the "gradient" or the "grad"
„
„
„
If we shift V Æ V + Const. the E-field does not
change
Makes sense, since V is only defined up-to
arbitrary constant
The expression above for the gradient is in
"Cartesian Coordinates"
„
„
Cartesian coordinates: x,y,z
One important result:
„
If V is a function of r (and not of angle) then
Simple applications of gradient
law
„
Point charge:
„
Infinite line of charge
Slide 16, this lecture
constant
Electric Potential for a
Continuous Charge Distribution
„
Consider a small
charge element dq
„
„
Treat it as a point
charge
The potential at
some point due to
this charge element
is
dq
dV = ke
r
V for a Continuous Charge
Distribution, cont.
„
To find the total potential, you need to
integrate to include the contributions
from all the elements
dq
V = ke ∫
r
„
This value for V uses the reference of V =
0 when P is infinitely far away from the
charge distributions
V for a Uniformly Charged
Ring
„
P is located on the
perpendicular central
axis of the uniformly
charged ring
„
The ring has a radius
a and a total charge Q
dq
dq
= ke ∫
=
V = ke ∫
r
x 2 + a2
ke
x 2 + a2
∫ dq
V for a Uniformly Charged
Disk
„
V=
The ring has a
radiuses b and a
and surface charge
density of σ
(x
ke dq
2
+r
a
V = πke σ ∫
)
1
2
2
=
ke σ 2πrdr
(x
2rdr
(
)
1
2
2
x2 + r
⎡ 2
V = 2πke σ ⎢ x + a 2
⎣
b
2
(
)
+r
2
)
1
2
a
(
= πke σ ∫ x 2 + r 2
)
−
1
2
2rdr
b
1
2
⎤
⎡ 2
2
−
+
2
πk
σ
x
b
e ⎢
⎥⎦
⎣
(
)
1
2
⎤
⎥⎦
V for a Uniformly Charged
Disk
„
V=
The ring has a
radiuses a and b
and surface charge
density of σ
(x
ke dq
2
+r
b
V = πke σ ∫
)
1
2
2
=
ke σ 2πrdr
(x
2rdr
(
)
1
2
2
x2 + r
⎡ 2
V = 2πke σ ⎢ x + b 2
⎣
a
2
(
)
+r
2
)
1
2
b
(
= πke σ ∫ x 2 + r 2
)
−
1
2
2rdr
a
1
2
⎤
⎡ 2
2
−
+
2
πk
σ
x
a
e ⎢
⎥⎦
⎣
(
)
1
2
⎤
⎥⎦
V for a Uniformly Charged
Disk, cont
„
The ring has a
radius a and radius
b=0. Surface charge
density of σ
a
(
V = πke σ ∫ x 2 + r 2
)
−
1
2
2rdr
0
⎡ 2
V = 2πke σ ⎢ x + a 2
⎣
(
)
1
2
⎤
− x⎥
⎦
V for a Finite Line of Charge
„
A rod of line ℓ has a total charge
of Q and a linear charge density
of λ
l
V = ke λ ∫
0
l
dx
(
x 2 + a2
)
1
2
Q ⎛
ke ln ⎜ x + x 2 + a 2
l ⎝
(
Q
= ke ∫ x 2 + r 2
l 0
)
1
2
(
)
−
1
2
dx =
⎞
⎟
⎠
keQ ⎛ A + A 2 + a 2
ln ⎜
V=
⎜
A
a
⎝
⎞
⎟
⎟
⎠
V for a Uniformly Charged
Sphere
„
„
„
A solid sphere of
radius R and total
charge Q
Q
For r > R, V = ke
r
For r < R,
keQ
Er = 3 r
R
V for a Uniformly Charged
Sphere
„
For r < R,
keQ
Er = 3 r
R
r
keQ r
VD − VC = − ∫ Er dr = 3 ∫ rdr =
R R
R
keQ 2
2
−
R
r
2R 3
(
)
keQ 2
2
VD − VC =
R
−
r
2R 3
(
keQ
VC =
R
)
keQ keQ 2
keQ
2
2
2
+
−
=
3
−
VD =
R
r
R
r
2R 3
2R 3
R
(
)
(
)
V for a Uniformly Charged
Sphere, Graph
„
The curve for VD is
for the potential
inside the curve
„
„
„
It is parabolic
It joins smoothly with
the curve for VB
The curve for VB is
for the potential
outside the sphere
„
It is a hyperbola
V Due to a Charged
Conductor
„
„
„
„
Consider two points on
the surface of the
charged conductor as
shown
E is always
perpendicular to the
displacement ds
Therefore, E · ds = 0
Therefore, the potential
difference between A
and B is also zero
V Due to a Charged
Conductor, cont.
„
V is constant everywhere on the surface of a
charged conductor in equilibrium
„
„
„
ΔV = 0 between any two points on the surface
The surface of any charged conductor in
electrostatic equilibrium is an equipotential
surface
Because the electric field is zero inside the
conductor, we conclude that the electric
potential is constant everywhere inside the
conductor and equal to the value at the
surface
E Compared to V
„
„
„
The electric potential is
a function of r
The electric field is a
function of r2
The effect of a charge
on the space
surrounding it:
„
„
The charge sets up a
vector electric field which
is related to the force
The charge sets up a
scalar potential which is
related to the energy
Irregularly Shaped Objects
„
The charge density is high
where the radius of
curvature is small
„
„
And low where the radius of
curvature is large
The electric field is large
near the convex points
having small radii of
curvature and reaches
very high values at sharp
points
Irregularly Shaped Objects,
cont.
„
„
The field lines are
still perpendicular to
the conducting
surface at all points
The equipotential
surfaces are
perpendicular to the
field lines
everywhere
Cavity in a Conductor
„
„
„
Assume an
irregularly shaped
cavity is inside a
conductor
Assume no charges
are inside the cavity
The electric field
inside the conductor
must be zero
Cavity in a Conductor, cont
„
„
The electric field inside does not depend on
the charge distribution on the outside surface
of the conductor
For all paths between A and B,
B
VB − VA = − ∫ E ⋅ ds = 0
A
„
A cavity surrounded by conducting walls is a
field-free region as long as no charges are
inside the cavity
Corona Discharge
„
„
If the electric field near a conductor is
sufficiently strong, electrons resulting
from random ionizations of air
molecules near the conductor
accelerate away from their parent
molecules
These electrons can ionize additional
molecules near the conductor
Corona Discharge, cont.
„
„
„
This creates more free electrons
The corona discharge is the glow that
results from the recombination of these
free electrons with the ionized air
molecules
The ionization and corona discharge are
most likely to occur near very sharp
points
Millikan Oil-Drop Experiment –
Experimental Set-Up
Millikan Oil-Drop Experiment
„
„
„
Robert Millikan measured e, the
magnitude of the elementary charge on
the electron
He also demonstrated the quantized
nature of this charge
Oil droplets pass through a small hole
and are illuminated by a light
Oil-Drop Experiment, 2
„
„
With no electric field
between the plates,
the gravitational
force and the drag
force (viscous) act
on the electron
The drop reaches
terminal velocity
with FD = mg
Oil-Drop Experiment, 3
„
When an electric field
is set up between the
plates
„
„
The upper plate has a
higher potential
The drop reaches a
new terminal velocity
when the electrical
force equals the sum
of the drag force and
gravity
Oil-Drop Experiment, final
„
„
The drop can be raised and allowed to
fall numerous times by turning the
electric field on and off
After many experiments, Millikan
determined:
„
„
q = ne where n = 1, 2, 3, …
e = 1.60 x 10-19 C
Van de Graaff
Generator
„
„
„
„
Charge is delivered continuously to
a high-potential electrode by means
of a moving belt of insulating
material
The high-voltage electrode is a
hollow metal dome mounted on an
insulated column
Large potentials can be developed
by repeated trips of the belt
Protons accelerated through such
large potentials receive enough
energy to initiate nuclear reactions
Electrostatic Precipitator
„
„
„
„
„
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
Application – Xerographic
Copiers
„
„
The process of xerography is used for
making photocopies
Uses photoconductive materials
„
A photoconductive material is a poor
conductor of electricity in the dark but
becomes a good electric conductor when
exposed to light
The Xerographic Process
Application – Laser Printer
„
The steps for producing a document on a laser
printer is similar to the steps in the xerographic
process
„
„
Steps a, c, and d are the same
The major difference is the way the image forms on
the selenium-coated drum
„
„
„
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
Potentials Due to Various
Charge Distributions
Application – Laser Printer
„
The steps for producing a document on a laser
printer is similar to the steps in the xerographic
process
„
„
Steps a, c, and d are the same
The major difference is the way the image forms on
the selenium-coated drum
„
„
„
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
Problem, Method 1.
„
Alternative way to get term is to recognize
that there are 4 side pairs and 2 face
diagonal pairs.
„
4 sides of length s and 2 diagonals with length s√2
U 12 = U 23 = U 34 = U 14
U 13 = U 24
U = 4U 12 + 2U 13
4keQ 2 2keQ 2 1
+
U =
s
s
2
keQ 2
U =
s
keQ 2
⎛
2 ⎞
⎜4+
⎟ = 5.41
s
2⎠
⎝
Problem, Method 2.
„
Alternative way to get term is to recognize
that there are 4 side pairs and 2 face
diagonal pairs.
„
4 sides of length s and 2 diagonals with length s√2
U = U1+U 2 +U 3 +U 4
U = 0 + U 12 + (U 13 + U 23 ) + (U 14 + U 24 + U 34 )
keQ 2 keQ 2 ⎛ 1
1
⎞ keQ 2 ⎛
⎞
+
+
+
+
+
1
1
1
U = 0+
⎟
s
s ⎜⎝ 2 ⎟⎠
s ⎜⎝
2 ⎠
keQ 2 ⎛
keQ 2
2⎞
= 5.41
4+
U =
⎜
⎟
s ⎝
s
2⎠
Problem.
„
Find potential energy.
No energy is required for the 20-nC charge
placed at its location
(
)(
)
8.99 × 109 N ⋅ m 2 C 2 20 × 10−9 C
keq1
V1 =
=
= 4.50 kV
r
0.04 m
To place the 10-nC charge there we must put in energy
(
)(
)
U 12 = q2V1 = 10 × 10−9 C 4.5 × 103 V = 4.50 × 10−5 J
Next, to bring up the –20-nC charge requires energy
U 23 + U 13 = q3V 2 + q3V1 = q3 (V 2 + V1 )
.
−9
= −20 × 10
(
C 8.99 × 10 N ⋅ m
9
= −4.50 × 10−5 J− 4.50 × 10−5 J
2
C
2
)
⎛ 10 × 10−9 C 20 × 10−9 C ⎞
⎜ 0.04 m + 0.08 m ⎟
⎝
⎠
U 12 + U 23 + U 13 = −4.50 × 10−5 J
Problem, cont.
„
Find 40 nC charge is released.
The three fixed charges create potential at the location
where the fourth is released:
⎛ 20×10−9
10×10−9 20×10−9⎞
+
−
V = V1 +V2 +V3 = 8.99×10 N ⋅ m C ⎜
⎟ Cm
2
2
0
.
0
3
0
.
0
5
⎝ 0.04 +0.03
⎠
(
9
2
)
2
V = 3.00×103 V
Energy of the system of four charged objects is conserved
as the fourth charge flies away:
⎛1 2
⎞ ⎛1 2
⎞
⎜⎝ m v + qV ⎟⎠ = ⎜⎝ m v + qV ⎟⎠
2
2
i
f
(
)(
)
0 + 40 × 10−9 C 3.00 × 103 V =
v=
(
)=
2 1.20 × 10−4 J
−13
2 × 10
kg
(
)
1
2.00 × 10−13 kg v2 + 0
2
3.46 × 104 m s
Quick Quiz 1
In the figure below, two points A and B are located within a
region in which there is an electric field. The potential
difference ΔV = VB – VA is
(a) positive
(b) negative
(c) zero
Quick Quiz 1
Answer: (b). When moving straight from A to B, E and ds
both point toward the right. Thus, the dot product E · ds is
positive and ΔV is negative.
Quick Quiz 2
In this figure, a negative charge is placed at A and then
moved to B. The change in potential energy of the charge–
field system for this process is
(a) positive
(b) negative
(c) zero
Quick Quiz 2
Answer: (a). From ΔU = q0 ΔV, if a negative test charge is
moved through a negative potential difference, the potential
energy is positive. Work must be done to move the charge in
the direction opposite to the electric force on it.
Quick Quiz 3
The labeled points of the figure below 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 along the following
transitions.
(a) A -> B, B -> C, C -> D, D -> E
(b) A -> B, D -> E, B -> C, C -> D
(c) B -> C, C -> D, A -> B, D -> E
(d) D -> E, C -> D, B -> C, A -> B
Quick Quiz 3
Answer: (c). 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, and thus the field does –1 J of work per unit of
positive charge that moves.
Quick Quiz 4
For the equipotential surfaces in this figure, what is the
approximate direction of the electric field?
(a) Out of the page
(b) Into the page
(c) Toward the right edge
of the page
(d) Toward the left edge of
the page
(e) Toward the top of the
page
(f) Toward the bottom of
the page
Quick Quiz 4
Answer: (f). The electric field points in the direction of
decreasing electric potential.
Quick Quiz 5a
A spherical balloon contains a positively charged object at
its center. As the balloon is inflated to a greater volume
while the charged object remains at the center, the electric
potential at the surface of the balloon will
(a) increase
(b) decrease
(c) remain the same.
Quick Quiz 5a
Answer: (b). The electric potential is inversely proportional
to the radius (see Eq. 25.11).
Quick Quiz 5b
Recall that the spherical balloon from part a) contains a
positively charged object at its center. As the balloon is
inflated to a greater volume while the charged object
remains at the center, the electric flux through the surface of
the balloon will
(a) increase
(b) decrease
(c) remain the same.
Quick Quiz 5b
Answer: (c). Because the same number of field lines passes
through a closed surface of any shape or size, the electric
flux through the surface remains constant.
Quick Quiz 6
In Figure 25.10a, take q1 to be a negative source charge and
q2 to be the test charge. If q2 is initially positive and is
changed to a charge of the same magnitude but negative, the
potential at the position of q2 due to q1
(a) increases
(b) decreases
(c) remains the same
Quick Quiz 6
Answer: (c). The potential is established only by the source
charge and is independent of the test charge.
Quick Quiz 7
Consider the situation in question 6 again. When q2 is
changed from positive to negative, the potential energy of
the two-charge system
(a) increases
(b) decreases
(c) remains the same
Quick Quiz 7
Answer: (a). The potential energy of the two-charge system
is initially negative, due to the products of charges of
opposite sign in Equation 25.13. When the sign of q2 is
changed, both charges are negative, and the potential energy
of the system is positive.
Quick Quiz 8
In a certain region of space, the electric potential is zero
everywhere along the x axis. From this we can conclude that
the x component of the electric field in this region is
(a) zero
(b) in the x direction
(c) in the –x direction.
Quick Quiz 8
Answer: (a). If the potential is constant (zero in this case), its
derivative along this direction is zero.
Quick Quiz 9
In a certain region of space, the electric field is zero. From
this we can conclude that the electric potential in this region
is
(a) zero
(b) constant
(c) positive
(d) negative
Quick Quiz 9
Answer: (b). If the electric field is zero, there is no change in
the electric potential and it must be constant. This constant
value could be zero but does not have to be zero.
Quick Quiz 10
Consider starting at the center of the left-hand sphere (sphere 1, of radius
a) in the figure below and moving to the far right of the diagram, passing
through the
center of the
right-hand
sphere (sphere
2, of radius c)
along the way.
The centers of
the spheres are
a distance b
apart. Draw a
graph of the
electric
potential
as a function
of position
relative to the center of the left-hand sphere.
Quick Quiz 10
Answer: The graph would look like the sketch below. Notice
the flat plateaus at each conductor, representing the constant
electric potential inside a solid conductor.
Conceptual questions
for practice
on chapter 23
The positive charge is the end
view of a positively charged
glass rod. A negatively charged
particle moves in a circular arc
around the glass rod. Is the work
done on the charged particle by
the rod’s electric field positive,
negative or zero?
1. Positive
2. Negative
3. Zero
The positive charge is the end
view of a positively charged
glass rod. A negatively charged
particle moves in a circular arc
around the glass rod. Is the work
done on the charged particle by
the rod’s electric field positive,
negative or zero?
1. Positive
2. Negative
3. Zero
Rank in order, from largest to smallest, the potential
energies Ua to Ud of these four pairs of charges.
Each + symbol represents the same amount of charge.
1.
2.
3.
4.
5.
Ua = Ub > Uc = Ud
Ua = Uc > Ub = Ud
Ub = Ud > Ua = Uc
Ud > Ub = Uc > Ua
Ud > Uc > Ub > Ua
Rank in order, from largest to smallest, the potential
energies Ua to Ud of these four pairs of charges.
Each + symbol represents the same amount of charge.
1.
2.
3.
4.
5.
Ua = Ub > Uc = Ud
Ua = Uc > Ub = Ud
U b = Ud > Ua = Uc
Ud > Ub = Uc > Ua
Ud > Uc > Ub > Ua
A proton is released
from rest at point B,
where the potential
is 0 V. Afterward,
the proton
1. moves toward A with an increasing speed.
2. moves toward A with a steady speed.
3. remains at rest at B.
4. moves toward C with a steady speed.
5. moves toward C with an increasing speed.
A proton is released
from rest at point B,
where the potential
is 0 V. Afterward,
the proton
1. moves toward A with an increasing speed.
2. moves toward A with a steady speed.
3. remains at rest at B.
4. moves toward C with a steady speed.
5. moves toward C with an increasing speed.
Rank in order, from
largest to smallest, the
potentials Va to Ve at the
points a to e.
1.
2.
3.
4.
5.
Va = Vb = Vc = Vd = Ve
Va = Vb > Vc > Vd = Ve
Vd = Ve > Vc > Va = Vb
Vb = Vc = Ve > Va = Vd
Va = Vb = Vd = Ve > Vc
Rank in order, from
largest to smallest, the
potentials Va to Ve at the
points a to e.
1.
2.
3.
4.
5.
Va = Vb = Vc = Vd = Ve
Va = Vb > Vc > Vd = Ve
Vd = Ve > Vc > Va = Vb
Vb = Vc = Ve > Va = Vd
Va = Vb = Vd = Ve > Vc
Rank in order, from largest to
smallest, the potential
differences ∆V12, ∆V13, and
∆V23 between points 1 and 2,
points 1 and 3, and points 2
and 3.
1.
2.
3.
4.
5.
∆V12 > ∆V13 = ∆V23
∆V13 > ∆V12 > ∆V23
∆V13 > ∆V23 > ∆V12
∆V13 = ∆V23 > ∆V12
∆V23 > ∆V12 > ∆V13
Rank in order, from largest to
smallest, the potential
differences ∆V12, ∆V13, and
∆V23 between points 1 and 2,
points 1 and 3, and points 2
and 3.
1.
2.
3.
4.
5.
∆V12 > ∆V13 = ∆V23
∆V13 > ∆V12 > ∆V23
∆V13 > ∆V23 > ∆V12
∆V13 = ∆V23 > ∆V12
∆V23 > ∆V12 > ∆V13
What are the units of potential difference?
1. Amperes
2. Potentiometers
3. Farads
4. Volts
5. Henrys
What are the units of potential difference?
1. Amperes
2. Potentiometers
3. Farads
4. Volts
5. Henrys
New units of the electric field were introduced in
this chapter. They are:
1. V/C.
2. N/C.
3. V/m.
4. J/m2.
5. W/m.
New units of the electric field were introduced in
this chapter. They are:
1. V/C.
2. N/C.
3. V/m.
4. J/m2.
5. W/m.
The electric potential inside a capacitor
1. is constant.
2. increases linearly from the negative to
the positive plate.
3. decreases linearly from the negative to
the positive plate.
4. decreases inversely with distance from
the negative plate.
5. decreases inversely with the square of the
distance from the negative plate.
The electric potential inside a capacitor
1. is constant.
2. increases linearly from the negative to
the positive plate.
3. decreases linearly from the negative to
the positive plate.
4. decreases inversely with distance from
the negative plate.
5. decreases inversely with the square of the
distance from the negative plate.