Download PH213 – Chapter 29 Solutions Ionic Potentials across Cell

PH213 – Chapter 29 Solutions Ionic Potentials across Cell Membranes Conceptual Question
Description: Short conceptual problem dealing with ionic potentials across cell membranes.
In its resting state, the membrane surrounding a neuron is permeable to potassium ions but only slightly
permeable to sodium ions. Thus, positive K ions can flow through the membrane in an attempt to
equalize K concentration, but Na ions cannot as quickly. This leads to an excess of Na ions outside of
the cell. If the space outside the cell is defined as zero electric potential, then the electric potential of the
interior of the cell is negative. This resting potential is typically about
situation is shown in the figure.
80
. A schematic of this
In response to an electrical stimulus, certain channels in the membrane can become permeable to Na
ions. Due to the concentration gradient, Na ions rush into the cell and the interior of the cell reaches an
electric potential of about 40
. This process is termed depolarization. In response to
depolarization, the membrane again becomes less permeable to Na ions, and the K ions flow out of the
interior of the cell through channels established by the positive electric potential inside of the cell. This
then reestablishing the resting potential. This is termed repolarization. Only a small percentage of the
available Na and K ions participate in each depolarization/repolarization cycle, so the cell can respond
to many stimuli in succession without depleting its "stock" of available Na and K ions. A graph of an
electric potential inside a cell vs. time is shown in the next figure
for a single
depolarization/repolarization cycle.
Part A
During the resting phase, what is the electric potential energy of a typical Na ion outside of the cell?
Hint A.1
The electron volt
Electric potential energy is defined as
.
The electric charge on individual particles is always a multiple of the fundamental charge
charge on a single proton). Rather than substituting a numerical value for
convenient to use the constant
energy
(the
, it is often more
as a unit. Thus, a proton located at a potential of 100
has
,
which can be written as
or
.
Thus, the proton has 100 electron volts of energy. (Electron volts can be converted to the more
traditional unit of energy, the joule, by multiplying by the conversion factor
and recalling that
ANSWER:
40
+40
80
+80
. Thus,
.)
ANSWER:
40
+40
80
+80
0
Part B
During the resting phase, what is the electrical potential energy of a typical K ion inside of the cell?
Hint B.1
The electron volt
Electric potential energy is defined as
.
The electric charge on individual particles is always a multiple of the fundamental charge
charge on a single proton). Rather than substituting a numerical value for
convenient to use the constant
energy
(the
, it is often more
as a unit. Thus, a proton located at a potential of 100
has
,
which can be written as
or
.
Thus, the proton has 100 electron volts of energy. (Electron volts can be converted to the more
traditional unit of energy, the joule, by multiplying by the conversion factor
and recalling that
.)
ANSWER:
40
+40
80
+80
. Thus,
0
Part C
During depolarization, what is the work done (by the electric field) on the first few Na ions that enter
the cell?
Hint
C.1
The electron volt
Electric potential energy is defined as
.
The electric charge on individual particles is always a multiple of the fundamental charge
charge on a single proton). Rather than substituting a numerical value for
convenient to use the constant
energy
(the
, it is often more
as a unit. Thus, a proton located at a potential of 100
has
,
which can be written as
or
.
Thus, the proton has 100 electron volts of energy. (Electron volts can be converted to the more
traditional unit of energy, the joule, by multiplying by the conversion factor
and recalling that
. Thus,
.)
Hint
C.2
Algebraic sign of the work
In general, work is defined as the product of the force applied parallel (or antiparallel) to the
displacement of an object. Thus,
.
The work done by a force is positive if the force and the displacement are parallel; it is negative if
the force and displacement are opposite in direction.
Hint C.3
Magnitude of the work
Work transfers energy into or out of a system. Therefore, in the absence of other energy transfers,
the magnitude of the work done on an object is equal to the magnitude of the object’s change in
energy. Since the primary form of energy present in this example is electric potential energy, the
magnitude of the work done is equal to the change in the ion’s electric potential energy.
ANSWER:
40
+40
80
ANSWER:
40
+40
80
+80
120
+120
0
Part D
During repolarization, what is the work done (by the electric field) on the first few K ions that exit the
cell?
Hint
D.1
The electron volt
Electric potential energy is defined as
.
The electric charge on individual particles is always a multiple of the fundamental charge
charge on a single proton). Rather than substituting a numerical value for
convenient to use the constant
energy
(the
, it is often more
as a unit. Thus, a proton located at a potential of 100
has
,
which can be written as
or
.
Thus, the proton has 100 electron volts of energy. (Electron volts can be converted to the more
traditional unit of energy, the joule, by multiplying by the conversion factor
and recalling that
. Thus,
.)
Hint
D.2
Algebraic sign of the work
In general, work is defined as the product of the force applied parallel (or antiparallel) to the
displacement of an object. Thus,
.
The work done by a force is positive if the force and the displacement are parallel; the work done is
negative if the force and displacement are opposite in direction.
Hint D.3
Magnitude of the work
In general, work is defined as the product of the force applied parallel (or antiparallel) to the
displacement of an object. Thus,
.
The work done by a force is positive if the force and the displacement are parallel; the work done is
negative if the force and displacement are opposite in direction.
Hint D.3
Magnitude of the work
Work transfers energy into or out of a system. Therefore, in the absence of other energy transfers,
the magnitude of the work done on an object is equal to the magnitude of the object’s change in
energy. Since the primary form of energy present in this example is electric potential energy, the
magnitude of the work done is equal to the change in the ion’s electric potential energy.
ANSWER:
40
+40
80
+80
120
+120
0
Electric Potential Ranking Task
Description: Short conceptual problem involving electrical potentials of point charges. (ranking task)
In the figure
there are
two point charges,
and
. There are also six positions, labeled A through F, at various
distances from the two point charges. You will be asked about the electric potential at the different
points (A through F).
Part A
Rank the locations A to F on the basis of the electric potential at each point. Rank positive electric
potentials as higher than negative electric potentials.
Hint
A.1
Definition of electric potential
The electric potential surrounding a point charge is defined by
,
where is the source charge creating the electric potential and is the distance between the
source charge and the point of interest. If more than one source is present, determine the electric
potential from each source and sum the results.
Hint
A.2
Conceptualizing electric potential
Because positive charges create positive electric potentials in their vicinity and negative charges
create negative potentials in their vicinity, electric potential is sometimes visualized as a sort of
"elevation." Positive charges represent mountain peaks and negative charges deep valleys. In this
picture, when you are close to a positive charge, you are "high up" and have a higher positive
potential. Conversely, near a negative charge, you are deep in a "valley" and have a negative
potential. The utility of this picture becomes clearer when we begin to think of charges moving
through a region of space containing an electric potential. Just as particles naturally roll downhill,
converting gravitational potential energy into kinetic energy, positively charged particles naturally
"roll downhill" as well, toward regions of lower electric potential, converting electrical potential
energy into kinetic energy.
Rank the locations from highest to lowest potential. To rank items as equivalent, overlap them.
ANSWER:
"roll downhill" as well, toward regions of lower electric potential, converting electrical potential
energy into kinetic energy.
Rank the locations from highest to lowest potential. To rank items as equivalent, overlap them.
ANSWER:
View
Change in Electric Potential Ranking Task
Description: Short conceptual problem related to the electric potential difference between pairs of
points. (ranking task)
In the diagram below, there are two charges of
distances from the two charges.
and
and six points (a through f) at various
You will be asked to
rank changes in the electric potential along paths between pairs of points.
Part A
Using the diagram to the left, rank each of the given paths on the basis of the change in electric
potential. Rank the largest-magnitude positive change (increase in electric potential) as largest and the
largest-magnitude negative change (decrease in electric potential) as smallest.
Hint
A.1
Change in electric potential
Determining the change in electric potential along some path involves determining the electric
potential at the two end points of the path, and subtracting:
Hint
A.2
Determine the algebraic sign of the change in potential
The path from point d to point a results in a positive change in electric potential. Which of the other
potential at the two end points of the path, and subtracting:
Hint
A.2
Determine the algebraic sign of the change in potential
The path from point d to point a results in a positive change in electric potential. Which of the other
paths also involves a positive change in electric potential (i.e., electric potential that increases along
the path)?
ANSWER:
from b to a
from f to e
from c to d
from c to e
from c to b
Hint
A.3
Conceptualizing changes in electric potential
Since positive charges create large positive electric potentials in their vicinity and negative charges
create negative potentials in their vicinity, electric potential is sometimes visualized as a sort of
“elevation.” Positive charges represent mountain peaks and negative charges deep valleys. In this
picture, when you are close to a positive charge, you are high “up” and have a large positive
potential. Conversely, near a negative charge you are deep in a “valley” and have a negative
potential. Thus, changes in electric potential can be thought of as changes in elevation. The change
is positive if you are moving “uphill” and the change is negative if you move “downhill.” The
farther you travel either uphill or downhill, the larger the magnitude of the change in electric
potential.
Rank from largest to smallest. To rank items as equivalent, overlap them.
ANSWER:
View
Not So Fast!
Description: A charge is moving in the electric field of other point charges; use energy considerations
to find its initial velocity.
Four point charges, fixed in place, form a square with side length
.
Part A
The particle with charge is now released and given a quick push; as a result, it acquires speed .
Eventually, this particle ends up at the center of the original square and is momentarily at rest. If the
mass of this particle is
Hint
A.1
, what was its initial speed
?
How to approach the problem
Use the law of conservation of energy for the particle with charge
Hint
A.2
.
Finding the potential energy
Find the potential energy of the particle with charge due to each of the other three charged
particles; then use the principle of superposition. You have to complete the entire procedure twice,
of course: for the initial and the final moments.
Hint
A.3
Find the initial potential energy
Find the initial potential energy
Hint
A.3.1
of the particle with charge
Formula for the electric potential energy
The formula for the electric potential energy
distance
.
between charges
is
,
and
separated by
The formula for the electric potential energy
distance
between charges
and
separated by
is
,
where
.
Express your answer in terms of , , and appropriate constants. Use
numeric coefficient should be a decimal with three significant figures.
instead of
. The
ANSWER:
=
Hint
A.4
Find the final potential energy
Find the potential energy
Hint
A.4.1
of the particle with charge
at the center of the original square.
Formula for the electric potential energy
The formula for the electric potential energy
distance
between charges
and
separated by
is
,
where
.
Express your answer in terms of , , and appropriate constants. Use
numeric coefficient should be a decimal with three significant figures.
ANSWER:
=
instead of
. The
ANSWER:
=
Hint
A.5
Find the kinetic energy
Using conservation of energy, find the initial kinetic energy
.
Hint
A.5.1
of the particle with charge
Conservation of energy
Conservation of energy implies that the sum of the initial kinetic and potential energies is equal
to the sum of the final kinetic and potential energies. Since the final kinetic energy, when the
particle is momentarily at rest, is zero, you can express the initial kinetic energy in terms of the
initial and final potential energies.
Express your answer in terms of , , and appropriate constants. Use
numeric coefficient should be a decimal with three significant figures.
instead of
. The
ANSWER:
=
Hint
A.6
Formula for kinetic energy
To obtain the inital velocity of the charged particle, you will need to recall the formula for the
kinetic energy
of a particle with mass
and velocity
:
.
Express your answer in terms of , ,
, and appropriate constants. Use
The numeric coefficient should be a decimal with three significant figures.
ANSWER:
=
instead of
.
Part B
When the particle with charge reaches the center of the original square, it is, as stated in the
problem, momentarily at rest. Is the particle at equilibrium at that moment?
Hint B.1
How to approach the problem
Equilibrium means zero acceleration (not zero velocity). Zero acceleration, in turn, means that the
net force applied to the particle must be zero. Sketch the particle at its final position and the vectors
representing the forces applied to it by the other three particles to estimate the net force on the
particle.
ANSWER:
yes
no
The exact value of the net force can be found by a calculation.
Potential of a Charged Annulus
Description: Find the potential along the axis of a charged annulus.
An annular ring with a uniform surface charge density
of the coordinate axes. The annulus has an inner radius
sits in the xy plane, with its center at the origin
and outer radius
.
Part A
If you can find symmetries in a physical situation, you can often greatly simplify your calculations. In
this part you will find a symmetry in the annular ring before calculating the potential along the axis
through the ring's center in Part B.
Consider three sets of points: points lying on the vertical line A; those on circle B; and those on the
horizontal line C, as shown in the figure. Which set of points makes the same contribution toward the
potential calculated at any point along the axis of the annulus?
through the ring's center in Part B.
Consider three sets of points: points lying on the vertical line A; those on circle B; and those on the
horizontal line C, as shown in the figure. Which set of points makes the same contribution toward the
potential calculated at any point along the axis of the annulus?
Hint A.1
Definition of the potential due to a point charge
The potential
due to a point charge
at a distance
from it is given by
,
where
.
ANSWER:
points on line A
points on circle B
points on line C
Part B
By exploiting the above symmetry, or otherwise, calculate the electric potential
axis of the annulus a distance
Hint
at a point on the
from its center.
How to exploit the angular symmetry of the problem
The total potential
at a point on the axis of the annulus can be written as
,
B.1
The total potential
at a point on the axis of the annulus can be written as
,
where
is the distance from a point on the annulus to the point at which the potential is to be
determined. However, on account of the angular symmetry of this problem, it is more convenient to
write this integral in terms of polar coordinates:
.
The integral over is easy and should be done first, since the integrand has no dependence on
This will put the integral in the form
.
,
where
Hint
B.2
is the area of a thin annular slice of thickness
and radius
.
and radius
?
Find the area of an annular slice
What is
, the area of a thin annular slice of thickness
Express your answer in terms of
and
.
ANSWER:
=
Now do the integral over
variable substitution.
. Don't forget that
is a function of
. You will need to use a
Hint
B.3
Doing the integral
Set
. What is
Express your answer in terms of
?
and
?
ANSWER:
=
Substitute for
carefully.
and
in terms of
and
to do the integral. Find the new limits of integration
Substitute for
carefully.
Hint
B.4
and
in terms of
and
to do the integral. Find the new limits of integration
A formula for the integral
The integral
has antiderivative
.
Express your answer in terms of some or all of the variables
,
,
, and
. Use
.
ANSWER:
=
It is interestering to note that the potential at any point on the axis of a disk of radius
obtained from the expression above by setting
and
can be
. Doing so, one obtains
.
Conversely, the annulus can be thought of as the superposition of two disks, one with charge density
and radius
, and the other with charge density
and radius
. In the region from the
center to
, the opposite charge densities cancel out, so the net charge distribution would be just
like that of the annulus. Moreover, by adding the potentials due to these two disks, using the formula
above, you would recover the potential of the annulus.
It is also instructive to look at the general behavior of these potentials as a function of the parameters.
Clearly, the potential increases with increasing charge densities, as well as with increasing areas (if
the charge density is held constant), which intuitively seems reasonable. However, if the distance
increases, it is not clear whether the potential should grow, since appears in both terms, of which
one is subtracted from the other. If you are far from the disk, the disk looks like a point, and the
potential should drop off, just like the potential due to a point charge. Indeed, on account of the
negative second term in the expressions, this is the case. Try some values or check that the derivative
of
is indeed negative. You can also check that the above expression actually reduces to the
potential due to a point charge for
.
Charged Mercury Droplets
Description: A large mercury drop, at known electric potential, breaks into n smaller drops. Find the
ratio of the electric potentials of the original and final drops.
A uniformly charged spherical droplet of mercury has electric potential
The droplet then breaks into
throughout the droplet.
identical spherical droplets, each of which has electric potential
throughout its volume. The
not interact significantly.
small droplets are far enough apart form one another that they do
Part A
Find
, the ratio of
, the electric potential throughout the initial drop, to
electric potential throughout one of the smaller drops.
Hint
A.1
, the
How to approach the problem
Mercury is a metal and therefore a conductor. It also has very high surface tension, so that droplets
of mercury are roughly spherical. The electric potential
conducting droplet with charge
both electric potentials
and radius
and
charge and radius of the big drop and
Hint
A.2
is given by
. Write expressions for
. Express all the quantities involved in terms of the
, and then compute the ratio.
Find the charge on the small droplets
Keeping conservation of charge in mind, find the charge
Express your answer in terms of
=
Find the radius of a small droplet
What is the radius
on each small droplet.
, the charge on the big droplet, and
ANSWER:
Hint
A.3
throughout the volume of a spherical
of each small droplet?
.
ANSWER:
=
Hint
A.3
Find the radius of a small droplet
What is the radius
Hint
A.3.1
of each small droplet?
Consider volume
The volume of each small spherical droplet is
of a sphere of radius
is equal to
Express your answer in terms of
that of the large droplet. In general, the volume
.
, the radius of the big droplet, and
.
ANSWER:
=
The ratio should be dimensionless and should depend only on
ANSWER:
=
± The Geiger Counter
Description: ± Includes Math Remediation. Calculate the potential difference between the inner wire
and the outer cylinder in a Geiger counter, given the electric field at a particular point.
A Geiger counter detects radiation such as alpha particles by using the fact that the radiation ionizes the
air along its path. A thin wire lies on the axis of a hollow metal cylinder and is insulated from it.
A large potential
difference is established between the wire and the outer cylinder, with the wire at a higher potential; this
sets up a strong electric field directed radially outward. When ionizing radiation enters the device, it
ionizes a few air molecules. The free electrons produced are accelerated by the electric field toward the
wire and, on the way there, ionize many more air molecules. Thus a current pulse is produced that can
be detected by appropriate electronic circuitry and converted into an audible click. Suppose the radius of
the central wire is 145 micrometers and the radius of the hollow cylinder is 1.80 centimeters.
Part A
What potential difference
between the wire and the cylinder produces an electric field of
volts per meter at a distance of 1.20 centimeters from the axis of the wire? (Assume
that the wire and cylinder are both very long in comparison to their radii.)
Hint
A.1
How to approach the problem
According to the introduction, and looking at the illustration, there is a positive charge on the inner
wire and a negative charge of equal magnitude on the outer cylinder, which creates an electric field
inside the cylinder that points radially outward, or away from the wire. Therefore, the first thing to
do is to write a generic equation for the electric field between the cylinder and the wire. Using this,
find a corresponding equation for the potential difference between the wire and cylinder. Finally,
combine the two equations and calculate the potential difference at the given point.
Hint
A.2
Find an expression for the electric field
Assume that the inner wire of a Geiger counter has a charge per unit length of
cylinder has an equal but opposite charge per unit length,
electric field
wire.
Hint
A.2.1
. Find a generic expression for the
inside the Geiger counter as a function of the radial distance
Using Gauss's law
, and the outer
from the inner
cylinder has an equal but opposite charge per unit length,
electric field
wire.
Hint
A.2.1
. Find a generic expression for the
inside the Geiger counter as a function of the radial distance
from the inner
Using Gauss's law
The easiest way to solve this is to use Gauss's law. Because of symmetrical properties of the
wire, the Gaussian surface needed will be a cylinder with its ends perpendicular to the wire, and
with arbitrary length
and arbitrary radius
radius of the inner wire and
, such that
, where
is the
that of the metal cylinder. Again, because of symmetry, the
electric field will have only a radial component
, and only the flux through the side walls will
contribute to the total flux through the cylinder. Also note that since only the enclosed charge
will give rise to the electric field, only the charge per unit length
used in the calculations. From Gauss's law, then,
on the central wire will be
,
where the surface integral is calculated over the side walls of the Gaussian cylinder.
Use
as the permittivity of free space and express your answer in terms of some or all the
variables
,
, and any appropriate constants.
ANSWER:
=
Find an expression for the potential difference
Hint
A.3
Consider a Geiger counter whose central wire has radius
and outer cylinder has radius
let be the charge per unit length on the central wire. Using the equation for the electric field
inside a Geiger counter found in Hint 2, write a general expression for the potential difference
between the inner wire and the outer cylinder. Recall that
=
Use
,
as the permittivity of free space and express your answer in terms of some or all the
variables
,
,
, and any appropriate constants.
ANSWER:
=
, and
ANSWER:
=
Hint
A.4
Find a new expression for the electric field
Using the equation for the potential difference
that was found in Hint 3, rewrite the
expression for the
and the radial distance
in terms of electric field
Express your answer in terms of some or all the variables
constants.
,
,
,
.
, and any appropriate
ANSWER:
=
By combining the equations for the electric field and the potential difference, the charge per unit
length on the wire has been eliminated from the equations. This is necessary, since no
information about it is given in this problem.
Hint
A.5
Putting it all together
Information about the electric field at a particular point was given in the introduction, so the
potential difference between the wire and the cylinder can now be calculated at that point.
Express your answer numerically in volts.
ANSWER:
=
It is also possible to find the potential difference between the inner wire and the outer cylinder by
integrating the electric field from
to
, and then from
to
, using the given value of the field at
as an
intermediate integration limit. However, this is much more difficult to do, since the charge per unit
length on the wire is not known, and is not necessary if you instead find an expression for the electric
field in terms of the potential difference.
Speed of an Electron in an Electric Field
Description: Calculate the final speed of an electron released from rest between two stationary positive
point charges of given magnitudes. The distance between the stationary charges and the final distance
between the electron and one of the positive charges are also given.
Two stationary positive point charges, charge 1 of magnitude 3.00
and charge 2 of magnitude 1.75
, are separated by a distance of 54.0
. An electron is released from rest at the point midway
between the two charges, and it moves along the line connecting the two charges.
Part A
What is the speed
Hint
A.1
of the electron when it is 10.0
from charge 1?
How to approach the problem
Only the final speed of the electron is needed, so an easy way to solve this problem is by using energy
techniques. Since the only force that acts on the electron is the conservative electric force, mechanical
energy is conserved. Thus, find the potential energy at the initial and final positions of the electron
from the expression of the potential due to a collection of point charges. Also, you can easily
determine the initial kinetic energy of the electron, since it is released from rest. Now you have enough
information to write the equation of conservation of energy and calculate the speed of the electron
from its final kinetic energy.
Hint
A.2
Calculate the potential at the midpoint
What is the potential
Hint
A.2.1
at the midpoint between the two stationary positive charges?
Electric potential
The potential
due to a single point charge is
,
where
= 8.85×10−12
,
is the distance from the point charge to the point at
which the potential is calculated, and is the charge. If instead of a single point charge, there is a
collection of point charges, the total potential is given by the sum of the potentials due to each
charge.
Hint
A.2.2
Find the potential due to charge 1
What is the potential
at the midpoint due to charge 1 alone?
Express your answer numerically in volts.
ANSWER:
=
For a collection of point charges, the total potential at any point is the sum of the potentials due to
each charge.
Express your answer numerically in volts.
ANSWER:
=
For a collection of point charges, the total potential at any point is the sum of the potentials due to
each charge.
Express your answer numerically in volts.
ANSWER:
=
Hint
A.3
Calculate the initial potential energy
Calculate the potential energy
of the electron at the midpoint between the two stationary
positive charges, using the potential at that point.
Hint
A.3.1
How to find the potential energy from the potential
Recall that the potential energy
the equation
associated with a test charge
is related to the potential
by
. In this case, the test charge is the electron.
Express your answer numerically in joules.
ANSWER:
=
Hint
A.4
Calculate the initial kinetic energy
Calculate the initial kinetic energy
positive charges.
Hint
A.4.1
of the electron at the midpoint between the two stationary
Initial velocity of the electron
Recall that the electron is initially at rest; thus its inital velocity is zero.
Express your answer numerically in joules.
ANSWER:
=
Express your answer numerically in joules.
ANSWER:
=
Hint
A.5
Calculate the final potential energy
Calculate the potential energy
of the electron at its final position, a point along the line
connecting the positive stationary charges at distance 10.0
Hint
A.5.1
from charge 1.
Calculate the potential at the final position of the electron
Calculate the potential
at the final position of the electron, that is, at a point along the line
connecting the positive stationary charges 10.0
Express your answer numerically in volts.
from charge 1.
ANSWER:
=
Express your answer in joules.
ANSWER:
=
Hint
A.6
Putting it all together
Once you know the initial kinetic energy
, the initial potential energy
, and the final
potential energy
of the electron, you can calculate its final kinetic energy
conservation of energy:
by applying
.
Finally, use the equation for the kinetic energy of a particle (in this case, the moving electron) to
calculate its speed, since the mass of the electron is a known constant, 9.11×10−31
Express your answer in meters per second.
.
ANSWER
:
=
Note that the electric field between the two charges is not constant, so the easiest way to do these
calculations is to use conservation of energy. It is possible to integrate along the path of the electron,
using the electric field as a function of the distance from each charge, but this is much more difficult to
do and not necessary for the problem.
PROBLEMS: 29.4. Model: The mechanical energy of the charged particles is conserved. A parallel-plate capacitor has a
uniform electric field.
Visualize:
The figure shows the before-and-after pictorial representation.
Solve: The potential energy is defined as U = U0 + qEx, where x is the distance from the negative plate and U0 is
the potential energy at the negative plate (at x = 0 m). Thus, the change in the potential energy of the proton as it
moves from the positive plate to the negative plate is
(
) (
)
ΔU p = U f − U i = U 0 + 0 J − U 0 + eEd = −eEd
This decrease in potential energy appears as an increase in the proton’s kinetic energy:
ΔK = K f − K i = 12 mp vf2 p − 12 mp vi2p = 12 mp vf2 p
Applying the law of conservation of mechanical energy ΔK + ΔUp = 0 J, we have
1
2
(
)
mvf2p + −eEd = 0 J ⇒ vf2 p =
2eEd
mp
When the proton is replaced with a helium ion and the same experiment is repeated,
vf2ion =
2eEd
mion
Dividing the two equations,
vf ion =
mp
mion
vf p =
1
4
(50,000 m/s) = 25,000 m/s
Assess: Being a heavier particle, the helium ion’s velocity is expected to be smaller compared to the proton’s
velocity.
29.15. Model: Energy is conserved. The potential energy is determined by the electric potential.
Visualize:
The figure shows a before-and-after pictorial representation of a proton moving through a potential difference.
Solve: (a) Because the proton is a positive charge and it slows down as it travels, it must be moving from a
region
of lower potential to a region of higher potential.
(b) Using the conservation of energy equation,
Kf + Uf = Ki + Ui ⇒ Kf + qVf = Ki + qVi
⇒ Vf − Vi =
1
1
K − Kf =
q i
e
(
( )(
)
(
)(
1
2
mvi2 − 0 J
1.67 × 10−27 kg 800,000 m/s
mvi2
⇒ ΔV =
=
2e
2 1.60 × 10−19 C
(
)
)
)
2
= 3340 V
Assess: A positive ΔV confirms that the proton moves into a higher potential region.
29.30. Model: The net potential is the sum of the scalar potentials due to each charge.
Visualize:
Solve: Let the point on the y-axis where the electric potential is zero be at a distance y from the origin. At this
point, V1 + V2 = 0 V. This means
1
4πε 0
(
⎧⎪ q1 q2 ⎫⎪
⎨ + ⎬ = 0 V ⇒
⎪⎩ r2 r2 ⎪⎭
⇒ 3 16 cm
)
2
+ y2 = 4
−3.0 × 10−9 C
(
( −9 cm )
2
−9.0 cm
2
)
2
+ y2
+
4.0 × 10−9 C
(
16.0 cm
(
=0
+ y2
(
+ y 2 ⇒ 9 256 cm 2 + y 2 = 16 81 cm 2 + y 2
2
⇒ 7y = 1008 cm ⇒ y = ±12 cm.
)
)
2
)
29.45. Model: Energy is conserved. The proton’s potential energy inside the capacitor can be found from the
capacitor’s potential difference.
Visualize: Please refer to Figure P29.45.
Solve: (a) The electric potential at the midpoint of the capacitor is 250 V. This is because the potential inside a
parallel-plate capacitor is V = Es where s is the distance from the negative electron. The proton has charge q = e
and its potential energy at a point where the capacitor’s potential is V is U = eV. The proton will gain potential
energy
ΔU = eΔV = e(250 V) = 1.60 × 10 19 C (250 V) = 4.00 × 10 17 J
−
−
if it moves all the way to the positive plate. This increase in potential energy comes at the expense of kinetic
energy which is
K = 12 mv 2 =
1
2
(1.67 × 10
−27
)(
kg 200,000 m/s
)
2
= 3.34 × 10−17 J
This available kinetic energy is not enough to provide for the increase in potential energy if the proton is to reach
the positive plate. Thus the proton does not reach the plate because K < ΔU.
(b) The energy-conservation equation Kf + Uf = Ki + Ui is
1
2
(
mvf2 + qVf = 12 mvi2 + qVi ⇒ 12 mvf2 = 12 mvi2 + q Vi − Vf
2q
⇒ vf = v +
V − Vf =
m i
(
2
i
)
( 2.0 × 10
5
)
2
m/s +
(
)
)(
2 1.60 × 10−19 C 250 V − 0 V
1.67 × 10−27 kg
) = 2.96 × 10
5
m/s
29.48. Model: Mechanical energy is conserved.
Visualize:
Solve: The initial energy of the electron is the same as the energy at the turning point, when its speed is zero.
1
m v 2 + U1i + U 2i = U1f + U 2f
2 e i
⎛ 1
⎞
⎛ 1 e −e
e −e
1
⇒ me vi 2 + 2 ⎜
⎟ = 2 ⎜
−11
⎜⎝ 4πε 0 5.5 × 10 m ⎟⎠
⎜⎝ 4πε 0
2
r
Ei = Ef ⇒
( )( )
( )( ) ⎞
⎟
⎟⎠
)(
)
1.6 × 10−19 C
2
1
⇒ 9.11 × 10−31 kg 1.5 × 106 m/s − 2 9.0 × 109 Nm 2 /C2
2
5.5 × 10−11 m
(
(
)(
= −2 9.0 × 10
9
)
(1.6 × 10
Nm /C )
2
2
−19
(
C
)
2
r
⇒ 1.025 × 10−18 J − 8.38 × 10−18 J = −
4.61 × 10−28 Nm 2
r
⇒ r = 6.27 × 10−11 m = 0.0627 nm
2
(
Since r 2 = 0.055 nm
)
2
+ y2 , y =
2
(0.0627 nm ) − (0.055 nm )
2
= 0.030 nm.
Assess: The electron moves a distance outward that is less than the distance to each of the protons.
29.56. Model: Energy is conserved.
Visualize:
The alpha particle is initially at rest (vi alpha = 0 m/s) at the surface of the thorium nucleus. The potential energy of
the alpha particle is U i alpha . After the decay, the alpha particle is far away from the thorium nucleus, Uf alpha = 0 J,
and moving with speed vf alpha .
Solve: Initially, the alpha particle has potential energy and no kinetic energy. As the alpha particle is detected in
the laboratory, the alpha particle has kinetic energy but no potential energy. Energy is conserved, so Kf alpha + Uf
=
alpha
Ki alpha + Ui alpha. This equation is
1
2
⇒ vf alpha =
(
)
mvf2alpha + 0 J = 0 J +
2
1 360e
=
4πε 0 mri
( )( )
1 2e 90e
4πε 0
ri
(9.0 × 10 N m /C ) 360(1.60 × 10 C)
4 (1.67 × 10 kg ) ( 7.5 × 10 m )
9
2
2
−19
−27
−15
2
= 4.1 × 107 m/s
29.64. Solve: A charged particle placed inside a uniformly charged spherical shell experiences no electric
force. That is, E = 0 V/m inside the shell. We know from Section 29.5 that a difference in potential between two
points or two plates is the source of an electric field. Since the potential on the surface of the shell is
V = Q 4πε 0 R, the potential inside must be the same. This ensures that the potential difference is zero and hence
the electric field is zero inside the shell. The potential at the center of the spherical shell is thus the same as at the
surface. That is, Vcenter = Vsurface = Q 4πε 0 R.
29.68. Model: The net potential is the sum of potentials from all the charges.
Visualize: Please refer to Figure P29.68. Point P at which we want the net potential due to the linear electric
quadrupole is far away compared to the separation s, that is, y >> s.
Solve: The net potential at P is
Vnet = V1 + V2 + V3 =
=
q1
q3
1
1 q2
1
+
+
4πε 0 y − s 4πε 0 y 4πε 0 y + s
( ) (
) ( )
−2q
+q ⎤
1 ⎡ +q
q
⎢
⎥ =
+
+
4πε 0 ⎢ y − s
y
y + s ⎥ 4πε 0
⎣
⎦
At distances y >> s ,
⎡ y 2 + ys − 2 y 2 + 2s 2 + y 2 − ys ⎤
q
2s 2
⎢
⎥ =
⎢
⎥ 4πε 0 y y 2 − s 2
y y 2 − s2
⎣
⎦
(
)
(
)
(
)
2
1 2qs
1 Q
=
3
4πε 0 y
4πε 0 y 3
Vnet =
where Q = 2qs2 is called the electric quadrupole moment of the charge distribution.
Assess: This charge distribution is in fact a combination of two dipoles. As seen above their effects do not completely
cancel.
(
)
2
29.72. Model: The disk has a uniform surface charge density η = Q A = Q π Rout
− Rin2 .
Visualize: Please refer to Figure 29.31. Orient the disk in the xy-plane, with point P at distance z. Divide the
disk into rings of equal width Δr. Ring i has radius ri and charge ΔQi.
Solve: Using the result of Example 29.11, we write the potential at distance z of ring i as
Vi =
1
4πε 0
ΔQi
2
ri + z
⇒ V = ∑Vi =
2
i
ΔQi
1
∑
4πε 0 i r 2 + z 2
i
Noting that ΔQi = ηΔAi = η2πriΔr,
V =
=
r Δr
1
η
η 2π ∑ i
=
2
2
4πε 0
2ε 0
i
ri + z
Rout
∫
Rin
r dr
r2 + z2
=
Rout
η ⎡ 2
η ⎡ 2
2 ⎤
r
+
z
=
R + z 2 − Rin2 + z 2 ⎤⎥
⎢
⎥
⎣
⎦
⎦
2ε 0
2ε 0 ⎣⎢ out
Rin
Q
⎡ R 2 + z 2 − R 2 + z 2 ⎤
out
in
2
⎦⎥
2πε 0 Rout
− Rin2 ⎣⎢
(
)
In the limit Rin → 0 m,
V =
Q
⎡ R 2 + z 2 − z ⎤
2 ⎢
⎥⎦
⎣ out
2πε 0 Rout
This is the same result obtained for a disk of charge in Example 29.12.
29.83. Model: Assume the wire is a line of charge with uniform linear charge density. The electric potential at
the point is the sum of contributions from all charges present.
Visualize: Please refer to Figure CP29.83. Let the origin be the center of our coordinate system. Divide the
charged wire into two straight sections and the central semicircle. Divide each section into N small segments,
each of length Δx and with charge Δq. Segment i, located at position xi, contributes a small amount of potential Vi
at the semicircle center.
Solve: For the right hand straight section, the contribution of the ith segment is
Vi =
Δq
Δq
λΔx
=
=
4πε 0 ri 4πε 0 xi 4πε 0 xi
where Δq = λΔx. The Vi are now summed and the sum is converted to an integral giving
Vstraight section =
λ
4πε 0
3R
∫
R
3R
dx
λ
λ
⎡ ln x ⎤ =
=
ln 3
⎣
⎦
R
x 4πε 0 R
4πε 0 R
()
()
The integration limits are set by the physical location of the straight section. By symmetry, the left hand straight
section adds the same amount to the total potential at the semicircle center.
The potential due to the ith segment of the semicircular section is
Vi =
(
)
λ R Δθ
Δq
Δq
λ
=
=
=
Δθ
4πε 0 ri 4πε 0 R
4πε 0 R
4πε 0 R
where we have used the arc length Δx = R Δθ . The Vi are now summed and the sum converted to an integral
giving
Vsemicircle =
π
λ
λπ
λ
dθ =
=
∫
0
4πε 0 R
4πε 0 R 4ε 0 R
The total potential is the sum of the potentials due to the three sections:
⎛ λ
⎞
λ
λ ⎛
2ln(3) ⎞
V = Vsemicircle + 2Vstraight section =
+ 2 ⎜
ln 3 ⎟ =
⎜⎝ 1 + π ⎟⎠
4ε 0 R
4
πε
R
4
ε
R
⎝
⎠
0
0
()