PH213 – Chapter 29 Solutions - WOU Homepage - …schoenfw/Old Courses/PH213 Spring 2012/Solutions/PH213...If the space outside the cell is defined as zero electric potential, then

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 80 . A schematic of this situation is shown in the figure.

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 (the

charge on a single proton). Rather than substituting a numerical value for , it is often more

convenient to use the constant as a unit. Thus, a proton located at a potential of 100 has energy

, 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, .)

ANSWER:

40

+40

80

+80

0

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


.





or


and recalling that . Thus,

.)

ANSWER: 40

+40

80

+80

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


.





or



.)

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

+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


.





or



.)

Hint D.2 Algebraic sign of the work


. 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 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 and and six points (a through f) at various distances from the two charges.

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 paths also involves a positive change in electric potential (i.e., electric potential that increases along the path)?

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 , what was its initial speed ? Hint A.1 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 of the particle with charge . Hint A.3.1 Formula for the electric potential energy

The formula for the electric potential energy between charges and separated by

distance is

,

where .


distance is

,

where .

Express your answer in terms of , , and appropriate constants. Use instead of . The numeric coefficient should be a decimal with three significant figures. ANSWER:

=

Hint A.4

Find the final potential energy

Find the potential energy of the particle with charge at the center of the original square. Hint A.4.1 Formula for the electric potential energy


distance is

,

where .


=

ANSWER:

=

Hint A.5

Find the kinetic energy

Using conservation of energy, find the initial kinetic energy of the particle with charge . Hint A.5.1 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.


=

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 instead of . The numeric coefficient should be a decimal with three significant figures. ANSWER:

=

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 sits in the xy plane, with its center at the origin

of the coordinate axes. The annulus has an inner radius 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 at a point on the

axis of the annulus a distance from its center. Hint 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

,

where is the distance from a point on the annulus to the point at which the potential is to be

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 is the area of a thin annular slice of thickness and radius .

Hint B.2 Find the area of an annular slice

What is , the area of a thin annular slice of thickness and radius ?

Express your answer in terms of and . ANSWER:

=

Now do the integral over . Don't forget that is a function of . You will need to use a variable substitution.

Hint B.3

Doing the integral

Set . What is ?

Express your answer in terms of and ? ANSWER:

=

Substitute for and in terms of and to do the integral. Find the new limits of integration carefully.

Substitute for and in terms of and to do the integral. Find the new limits of integration carefully.

Hint B.4 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 can be

obtained from the expression above by setting and . 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 throughout the droplet.

The droplet then breaks into identical spherical droplets, each of which has electric potential

throughout its volume. The small droplets are far enough apart form one another that they do not interact significantly. Part A

Find , the ratio of , the electric potential throughout the initial drop, to , the electric potential throughout one of the smaller drops. Hint A.1 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 throughout the volume of a spherical

conducting droplet with charge and radius is given by . Write expressions for

both electric potentials and . Express all the quantities involved in terms of the

charge and radius of the big drop and , and then compute the ratio.

Hint A.2 Find the charge on the small droplets

Keeping conservation of charge in mind, find the charge on each small droplet.

Express your answer in terms of , the charge on the big droplet, and . ANSWER:

=

Hint A.3 Find the radius of a small droplet

What is the radius of each small droplet? Hint A.3.1 Consider volume

ANSWER:

=

Hint A.3 Find the radius of a small droplet

What is the radius of each small droplet? Hint A.3.1 Consider volume

The volume of each small spherical droplet is that of the large droplet. In general, the volume

of a sphere of radius is equal to .

Express your answer in terms of , 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 , and the outer

cylinder has an equal but opposite charge per unit length, . Find a generic expression for the

electric field inside the Geiger counter as a function of the radial distance from the inner wire. Hint A.2.1 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

cylinder has an equal but opposite charge per unit length, . Find a generic expression for the

electric field inside the Geiger counter as a function of the radial distance from the inner wire. Hint A.2.1 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 , such that , where is the radius of the inner wire and 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 on the central wire will be used in the calculations. From Gauss's law, then,

, 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:

=

Hint A.3

Find an expression for the potential difference

Consider a Geiger counter whose central wire has radius and outer cylinder has radius , and

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:

=

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 in terms of electric field and the radial distance .

Express your answer in terms of some or all the variables , , , , and any appropriate constants. 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 of the electron when it is 10.0 from charge 1? Hint A.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 at the midpoint between the two stationary positive charges? Hint A.2.1 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.


ANSWER:

=

For a collection of point charges, the total potential at any point is the sum of the potentials due to each charge.


=

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 associated with a test charge is related to the potential by

the equation . 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 of the electron at the midpoint between the two stationary positive charges. Hint A.4.1 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 from charge 1. Hint A.5.1 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 from charge 1. Express your answer numerically in volts. 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 by applying conservation of energy:

. 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 = U0 + 0 J( ) − U0 + eEd( ) = −eEd

This decrease in potential energy appears as an increase in the proton’s kinetic energy:

ΔK = Kf − Ki =

12 mpvf p

2 − 12 mpvi p

2 = 12 mpvf p

2

Applying the law of conservation of mechanical energy ΔK + ΔUp = 0 J, we have

12 mvf p

2 + −eEd( ) = 0 J ⇒ vf p2 =

2eEdmp

When the proton is replaced with a helium ion and the same experiment is repeated,

vf ion

2 =2eEdmion

Dividing the two equations,

vf ion =

mp

mion

vf p =14 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 =1q

Ki − Kf( ) = 1e( )

12 mvi

2 − 0 J( )

⇒ ΔV =mvi

2

2e=

1.67 × 10−27 kg( ) 800,000 m/s( )2

2 1.60 × 10−19 C( )= 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

14πε0

q1

r2

+q2

r2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= 0 V ⇒

−3.0 × 10−9 C

−9.0 cm( )2+ y2

+4.0 × 10−9 C

16.0 cm( )2+ y2

= 0

⇒ 3 16 cm( )2

+ y2 = 4 −9 cm( )2+ y2 ⇒ 9 256 cm2 + y2( ) = 16 81 cm2 + y2( )

⇒ 7y2 = 1008 cm2 ⇒ y = ±12 cm.

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 = 1

2 mv2 = 12 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

12 mvf

2 + qVf =12 mvi

2 + qVi ⇒ 1

2 mvf2 = 1

2 mvi2 + q Vi −Vf( )

⇒ vf = vi

2 +2qm

Vi −Vf( ) = 2.0 × 105 m/s( )2+

2 1.60 × 10−19 C( ) 250 V − 0 V( )1.67 × 10−27 kg

= 2.96 × 105 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.

Ei = Ef ⇒12

mevi2 +U1i +U2i =U1f +U2f

⇒12

mevi2 + 2 1

4πε0

e( ) −e( )5.5× 10−11 m

⎛

⎝⎜⎜

⎞

⎠⎟⎟= 2 1

4πε0

e( ) −e( )r

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⇒12

9.11× 10−31 kg( ) 1.5× 106 m/s( )2− 2 9.0 × 109 Nm2 /C2( )

1.6 × 10−19 C( )2

5.5× 10−11 m

= −2 9.0 × 109 Nm2 /C2( )1.6 × 10−19 C( )2

r

⇒ 1.025× 10−18 J − 8.38 × 10−18 J = − 4.61× 10−28 Nm2

r

⇒ r = 6.27 × 10−11 m = 0.0627 nm

Since r 2 = 0.055 nm( )2

+ y2 , y = 0.0627 nm( )2

− 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

12 mvf alpha

2 + 0 J = 0 J + 14πε0

2e( ) 90e( )ri

⇒ vf alpha =1

4πε0

360e2( )mri

=9.0 × 109 N m2 /C2( )360 1.60 × 10−19 C( )2

4 1.67 × 10−27 kg( ) 7.5× 10−15 m( )= 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 4πε0 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 4πε0 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 =1

4πε0

q1

y − s+

14πε0

q2

y+

14πε0

q3

y + s

=1

4πε0

+q( )y − s

+−2q( )

y+

+q( )y + s

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

q4πε0

y2 + ys − 2y2 + 2s2 + y2 − ysy y2 − s2( )

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥=

q4πε0

2s2

y y2 − s2( )

At distances y >> s ,

Vnet =

14πε0

2qs2( )y3 =

14πε0

Qy3

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.

29.72. Model: The disk has a uniform surface charge density η = Q A = Q π Rout

2 − 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

4πε0

ΔQi

ri2 + z2

⇒

V = Vii∑ =

14πε0

ΔQi

ri2 + z2i

∑

Noting that ΔQi = ηΔAi = η2πriΔr,

V =1

4πε0

η2πriΔr

ri2 + z2i

∑ =η

2ε0

r dr

r 2 + z2Rin

Rout

∫ =η

2ε0

r 2 + z2⎡⎣⎢

⎤⎦⎥Rin

Rout

=η

2ε0

Rout2 + z2 − Rin

2 + z2⎡⎣⎢

⎤⎦⎥

=Q

2πε0 Rout2 − Rin

2( )Rout

2 + z2 − Rin2 + z2⎡

⎣⎢⎤⎦⎥

In the limit Rin → 0 m,

V =

Q2πε0 Rout

2 Rout2 + z2 − z⎡

⎣⎢⎤⎦⎥

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 =

Δq4πε0ri

=Δq

4πε0xi

=λΔx

4πε0xi

where Δq = λΔx. The Vi are now summed and the sum is converted to an integral giving

Vstraight section =

λ4πε0

dxxR

3R

∫ =λ

4πε0 Rln x( )⎡⎣ ⎤⎦R

3R=

λ4πε0 R

ln 3( )

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 =

Δq4πε0ri

=Δq

4πε0 R=λ R Δθ( )

4πε0 R=

λ4πε0 R

Δθ

where we have used the arc length Δx = R Δθ. The Vi are now summed and the sum converted to an integral giving

Vsemicircle =

λ4πε0 R

dθ0

π

∫ =λπ

4πε0 R=

λ4ε0 R

The total potential is the sum of the potentials due to the three sections:

V = Vsemicircle + 2Vstraight section =

λ4ε0 R

+ 2 λ4πε0 R

ln 3( )⎛

⎝⎜

⎞

⎠⎟ =

λ4ε0 R

1+ 2ln(3)π

⎛

⎝⎜⎞

⎠⎟

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PH213 – Chapter 29 Solutions - WOU Homepage - …schoenfw/Old Courses/PH213 Spring 2012/Solutions/PH213...If the space outside the cell is defined as zero electric potential, then