Download PH213 Chapter 27 Solutions

Chapter 27 Solutions PSS 27.2 The Electric Field of a Continuous Distribution of Charge
Description: Knight Problem-Solving Strategy 27.2 The Electric Field of a Continuous Distribution of
Charge is illustrated. (vector applet)
Learning Goal: To practice Problem-Solving Strategy 27.2 for continuous charge distribution
problems.
A straight wire of length
has a positive charge
distributed along its length. Find the magnitude of
the electric field due to the wire at a point located a distance
extending from the wire.
from one end of the wire along the line
The electric field of a continuous distribution of charge
Model the distribution as a simple shape, such as a line of charge or a disk of charge. Assume
the charge is uniformly distributed.
VISUALIZE: For the pictorial representation:
1. Draw a picture and establish a coordinate system.
2. Identify the point P at which you want to calculate the electric field.
PROBLEM-SOLVING STRATEGY 27.2
MODEL:
3.
Divide the total charge
4.
already know how to determine
. This is often, but not always, a division into point
charges.
Draw the electric field vector at P for one or two small pieces of charge. This will help you
identify distances and angles that need to be calculated.
Look for symmetries in the charge distribution that simplify the field. You may conclude that
5.
some components of
into small pieces of charge
are zero.
using shapes for which you
SOLVE:
The mathematical representation is
.

Use superposition to form an algebraic expression for each of the three components of
point P. (Note that one or more components may be zero.)

Let the

Replace the small charge


coordinate, such as
, that describes the shape of charge
. This is the critical step in
making the transition from a sum to an integral because you need a coordinate to serve as the
integration variable.
Express all angles and distances in terms of the coordinates.
Let the sum become an integral. The integration will be over the coordinate variable that is
at
coordinates of the point remain as variables.
with an equivalent expression involving a charge density and a
related to
. The integration limits for this variable will depend on your choice of
coordinate system. Carry out the integration, and simplify the result as much as possible.
ASSESS: Check that your result is consistent with any limits for which you know what the field should
be.
Model
No information is given on the cross section of the wire. For simplicity, assume that its diameter is
much smaller than the wire's length, and model the wire as a line of charge. Also, assume that the total
charge is uniformly distributed along the wire. Note that the point at which you want to calculate the
electric field is close to one of the ends of the wire, so you cannot make use of any symmetry to
simplify the problem. Instead, you will need to divide the total charge distributed along the wire into
many small segments and model each segment as a point-like charge.
Visualize
Part A
The diagram below is an incomplete pictorial representation of the situation described in this
problem: It shows the positively charged wire, a suitable coordinate system with the origin at the left
end of the wire, and the point P at which you need to calculate the electric field. The total charge
distributed along the wire has been divided into small segments, each of length
. For clarity, only the i-th segment at point
and charge
is shown in the diagram.
Complete this pictorial representation by drawing the electric field
Draw the vector starting at P.
at P due to segment i.
ANSWER:
View
Each small segment will contribute to the net electric field at P. By modeling each segment as a point
charge and using the principle of superposition, you can make a quick estimate of the direction of the
net field at P. To do that, it is helpful to draw the electric field due to one or two more charge
segments.
net field at P. To do that, it is helpful to draw the electric field due to one or two more charge
segments.
Part B
With reference to the coordinate system used in the previous part, which component, if any, of the
electric field due to the total charge is zero at P?
ANSWER:
both the x and y component
the x component
the y component
neither the x nor the y component
Since the point at which you want to calculate
lies on a straight line extending from the wire,
must be directed along the same line. Therefore, by choosing the x axis to be parallel to the wire, as
we did in Part A, we simplified the problem: Now the y component of the field at P is zero and the
magnitude of
is simply given by the magnitude of its x component.
Solve
Part C
Find
, the magnitude of the electric field at point P due to the total charge.
Hint
C.1
How to approach the problem
As explained in the strategy above, the mathematical expression for the net field is
. Because of the nature of the problem, which deals with a continuous
distribution of charge rather than a point-like distribution, the sum must be treated as an integral.
To set up the summation and find what to integrate, follow the steps listed in the strategy. After
dividing the total charge into small segments, and modeling each segment as a point charge, as we
did in Part A, construct a mathematical expression for
segment. Treat
, the x coordinate of the i-th segment, as a variable. Express the charge on the i-
th segment in terms of
. Then, let
calculus to evaluate the integral.
Hint
C.2
, the electric field due to the i-th
become an integral and apply your knowledge of
Find an expression for the electric field due to segment i
Find the magnitude of the electric field
at point P due to a segment of charge
located at
on the x axis.
Hint
C.2.1
Find the distance between point P and a segment of charge at
Assume the origin of the x axis is at the left end of the wire and point P is located at a distance
from the other end of the wire.
is the length of the wire. What is the distance
between
on the x axis.
Hint
C.2.1
Find the distance between point P and a segment of charge at
Assume the origin of the x axis is at the left end of the wire and point P is located at a distance
from the other end of the wire.
is the length of the wire. What is the distance
point P and a segment of charge located at
Express your answer in terms of
,
between
on the x axis?
, and
.
ANSWER:
=
Express your answer in terms of some or all of the variables
constants
and
,
,
, and
, and the
.
ANSWER:
=
Hint
C.3
Find an expression for the charge of a small segment
Since you will integrate with respect to a coordinate, it is important to express the charge on a short
segment of wire,
, in terms of its length
Find a mathematical expression for
assuming the charge is uniformly distributed along the
wire. Recall that the total charge on the wire is
Hint
C.3.1
.
and the length of the wire is
.
Find the linear charge density
Assuming the wire, whose length is
, has a total charge
length, what is the linear charge density
of the wire?
Express your answer in terms of
.
and
uniformly distributed along its
ANSWER:
=
Now use the linear charge density to express the charge on a small segment of wire in terms of
its length as
.
ANSWER:
=
Now use the linear charge density to express the charge on a small segment of wire in terms of
its length as
.
Express your answer in terms of
and some or all of the variables
,
, and
.
ANSWER:
=
Hint
C.4
How to make the transition from a sum to an integral
At this point in the problem, you should have found an expression for the magnitude of the electric
field
due to a small charge segment of length
that your expression for
located at
along the x axis. Make sure
contains only the following quantities, besides the electrostatic
constant
: ,
,
,
, and
. If this is not the case, go back and make the
necessary substitutions (the previous hints may help you do that).
Assuming each charge segment has the same length
, you can construct an algebraic
expression for the magnitude of the net field using the principle of superposition, that is,
, where the sum contains as many terms as the number of charge segments. If you
assume that there is an infinite number of segments in the wire,
small and can be replaced with
. At this point, you can replace the sum with an integral and
drop the subscript i from the variable
Note that the remaining quantities
in the problem statement.
Hint
C.5
will become infinitesimally
, which will then become the integration variable
,
, and
.
are constants: They are known quantities given
Find the limits of integration
When you change the sum over charge segments into an integral, the variable of integration will be
the coordinate
ANSWER:
. What should the limits of integration be?
0 to
0 to
to
to
0 to
Hint
C.6
How to simplify the integral
You can simplify the integral by changing the variable of integration from
to
. Don't forget to adjust the limits of integration appropriately when you
make this change. The following formula will also be useful:
.
Express your answer in terms of some or all of the variables
and
,
, and
, and the constants
.
ANSWER:
=
Assess
Part D
Imagine that distance is much greater than the length of the wire. Intuitively, what should the
magnitude of the electric field at point P be in this case?
Hint
D.1
Envisioning the limiting case
In the limiting case
, the length of the wire is not relevant and the wire appears to be a
point charge in the distance. What is the magnitude of the electric field due to a point charge?
Express your answer in terms of some or all of the variables
The variable
and
, and the constants
and
should not appear in your answer.
ANSWER:
=
A short wire far away from point P creates a field very similar to the one created by a point charge.
.
ANSWER:
=
A short wire far away from point P creates a field very similar to the one created by a point charge.
Not surprisingly, in the limiting case
, the mathematical expression you derived for
reduces to that of the electric field due to a point charge.
To see this, note that one way of writing the answer you obtained in the Solve step is:
.
When
reduces to
,
becomes negligible compared to
and
. Thus, the denominator
, as we expected.
Visualizing Electric Fields
Description: Select the correct drawing of electric field lines for several situations and answer
questions about why other choices are incorrect. Then, these ideas are demonstrated with an applet.
Learning Goal: To understand the nature of electric fields and how to draw field lines.
Electric field lines are a tool used to visualize electric fields. A field line is drawn beginning at a
positive charge and ending at a negative charge. Field lines may also appear from the edge of a picture
or disappear at the edge of the picture. Such lines are said to begin or end at infinity. The field lines are
directed so that the electric field at any point is tangent to the field line at that point.
The figure
shows two
different ways to visualize an electric field. On the left, vectors are drawn at various points to show the
direction and magnitude of the electric field. On the right, electric field lines depict the same situation.
Notice that, as stated above, the electric field lines are drawn such that their tangents point in the same
direction as the electric field vectors on the left. Because of the nature of electric fields, field lines never
cross. Also, the vectors shrink as you move away from the charge, and the electric field lines spread out
as you move away from the charge. The spacing between electric field lines indicates the strength of the
electric field, just as the length of vectors indicates the strength of the electric field. The greater the
spacing between field lines, the weaker the electric field. Although the advantage of field lines over
field vectors may not be apparent in the case of a single charge, electric field lines present a much less
cluttered and more intuitive picture of more complicated charge arrangements.
Part A
Which of the following figures
correctly depicts the
field lines from an infinite uniformly negatively charged sheet? Note that the sheet is being viewed
edge-on in all pictures.
Hint A.1
Description of the field
Recall that the field around an infinite charged sheet is always perpendicular to the sheet and that
the field strength does not change, regardless of distance from the sheet.
ANSWER:
A
B
C
D
Part B
In the diagram from part A
, what is wrong with
figure B? (Pick only those statements that apply to figure B.)
Check all that apply.
ANSWER:
Field lines cannot cross each other.
The field lines should be parallel because of the sheet's symmetry.
The field lines should spread apart as they leave the sheet to indicate the
weakening of the field with distance.
The field lines should always end on negative charges or at infinity.
Part C
Which of the following figures
shows the correct
electric field lines for an electric dipole?
ANSWER:
A
B
C
shows the correct
electric field lines for an electric dipole?
ANSWER:
A
B
C
D
This applet shows two charges. You can alter the charge on each independently or alter the distance
between them. You should try to get a feeling for how altering the charges or the distance affects the
field lines.
Part D
In the diagram from part C
, what is wrong with
figure D? (Pick only those statements that apply to figure D.)
Check all that apply.
, what is wrong with
figure D? (Pick only those statements that apply to figure D.)
Check all that apply.
ANSWER:
Field lines cannot cross each other.
The field lines should turn sharply as you move from one charge to the
other.
The field lines should be smooth curves.
The field lines should always end on negative charges or at infinity.
In even relatively
simple setups as in the figure, electric field lines are quite helpful for understanding the field
qualitatively (understanding the general direction in which a certain charge will move from a specific
position, identifying locations where the field is roughly zero or where the field points a specific
direction, etc.). A good figure with electric field lines can help you to organize your thoughts as well
as check your calculations to see whether they make sense.
Part E
In the figure
, the
electric field lines are shown for a system of two point charges,
and
following could represent the magnitudes and signs of
?
In the following, take
and
. Which of the
to be a positive quantity.
ANSWER:
,
,
,
,
,
Very far from the two charges, the system looks like a single charge with value
. At large enough distances, the field lines will be indistinguishable from the
field lines due to a single point charge
.
Charged Ring
Description: Find the electric field from a uniformly charged ring (qualitative and quantitative parts) at
points along its axis. Then use this to find the frequency of small oscillations of a oppositely charged
object placed on the axis.
Consider a uniformly charged ring in the xy plane, centered at the origin. The ring has radius
positive charge
and
distributed evenly along its circumference.
Part A
What is the direction of the electric field at any point on the z axis?
Hint A.1
How to approach the problem
Approach 1
In what direction is the field due to a point on the ring? Add to this the field from a point on the
opposite side of the ring. In what direction is the net field? What if you did this for every pair of
points on opposite sides of the ring?
Approach 2
Consider a general electric field at a point on the z axis, i.e., one that has a z component as well as a
component in the xy plane. Now imagine that you make a copy of the ring and rotate this copy
about its axis. As a result of the rotation, the component of the electric field in the xy plane will
rotate also. Now you ask a friend to look at both rings. Your friend wouldn't be able to tell them
apart, because the ring that is rotated looks just like the one that isn't. However, they have the
component of the electric field in the xy plane pointing in different directions! This apparent
contradiction can be resolved if this component of the field has a particular value. What is this
value?
Does a similar argument hold for the z component of the field?
ANSWER:
parallel to the x axis
parallel to the y axis
parallel to the z axis
in a circle parallel to the xy plane
parallel to the z axis
in a circle parallel to the xy plane
Part B
What is the magnitude of the electric field along the positive z axis?
Hint
B.1
Formula for the electric field
You can always use Coulomb's law,
, to find the electric field (the Coulomb force per unit
charge) due to a point charge. Given the force, the electric field say at
due to
is
.
In the situation below, you should use Coulomb's law to find the contribution
field at the point
from a piece of charge
integrate over the ring to find the value of
charge
in your answer, where
, and
.
ANSWER:
=
away. Then, you can
. Consider an infinitesimal piece of the ring with
. Use Coulomb's law to write the magnitude of the infinitesimal
positive z axis due to the charge
Use
on the ring at a distance
to the electric
at a point on the
shown in the figure.
. You may also use some or all of the variables
,
Use
in your answer, where
, and
. You may also use some or all of the variables
,
.
ANSWER:
=
Hint
B.2
Simplifying with symmetry
By symmetry, the net field must point along the z axis, away from the ring, because the horizontal
component of each contribution of magnitude
component of a similar contribution of magnitude
is exactly canceled by the horizontal
from the other side of the ring. Therefore,
all we care about is the z component of each such contribution. What is the component
electric field caused by the charge on an infinitesimally small portion of the ring
direction?
Express your answer in terms of
=
in the z
, the infinitesimally small contribution to the electric field;
, the coordinate of the point on the z axis; and
ANSWER:
of the
, the radius of the ring.
ANSWER:
=
Hint
B.3
Integrating around the ring
If you combine your results from the first two hints, you will have an expression for
vertical component of the field due to the infinitesimal charge
, the
. The total field is
.
If you are not comfortable integrating
charge
over the ring, change to a spatial variable. Since the total
is distributed evenly about the ring, convince yourself that
.
Use
in your answer, where
.
ANSWER:
=
Notice that this expression is valid for both positive and negative charges as well as for points located
on the positive and negative z axis. If the charge is positive, the electric field should point outward.
For points on the positive z axis, the field points in the positive z direction, which is outward from the
origin. For points on the negative z axis, the field points in the negative z direction, which is also
outward from the origin. If the charge is negative, the electric field should point toward the origin.
For points on the positive z axis, the negative sign from the charge causes the electric field to point in
the negative z direction, which points toward the origin. For points on the negative z axis, the negative
sign from the z coordinate and the negative sign from the charge cancel, and the field points in the
positive z direction, which also points toward the origin. Therefore, even though we obtained the
above result for postive
and
, the algebraic expression is valid for any signs of the parameters.
As a check, it is good to see that if
approximately
is much greater than
the magnitude of
is
, independent of the size of the ring: The field due to the ring is almost the same
as that due to a point charge
at the origin.
Part C
Imagine a small metal ball of mass
and negative charge
. The ball is released from rest at
the point
and constrained to move along the z axis, with no damping. If
what will be the ball's subsequent trajectory?
ANSWER:
,
repelled from the origin
attracted toward the origin and coming to rest
oscillating along the z axis between
and
circling around the z axis at
Part D
The ball will oscillate along the z axis between
What will be the angular frequency
and
of these oscillations? Use the approximation
simplify your calculation; that is, assume that
Hint
D.1
in simple harmonic motion.
to
.
Simple harmonic motion
Recall the nature of simple harmonic motion of an object attached to a spring. Newton's second law
for the system states that
, leading to oscillation at a frequency of
(here, the prime on the symbol representing the spring constant is to distinguish it from
). The solution to this differential equation is a sinusoidal function of time with
angular frequency
find the
Hint
D.2
What is
. Write an analogous equation for the ball near the charged ring in order to
term.
Find the force on the charge
, the z component of the force on the ball on the ball at the point
approximation
Hint
D.2.1
.
A formula for the force on a charge in an electric field
The formula for the force
on a charge
in an electric field
is
? Use the
approximation
Hint
D.2.1
.
A formula for the force on a charge in an electric field
The formula for the force
on a charge
in an electric field
is
.
Therfore, in particular,
.
You have already found
in Part B. Use that expression in the equation above to find an
expression for the z component of the force
to use the approximation given.
Express your answer in terms of
,
,
on the ball at the point
,
, and
. Don't forget
.
ANSWER:
=
Compare this to the generic force equation for an oscillator to find the oscillation frequency.
Express your answer in terms of given charges, dimensions, and constants.
ANSWER:
=
Electric Field Vector Drawing
Description: Simple conceptual question about determining the electric field vector in a region of space
from the change in the path of a charged particle. (vector applet)
Each of the four parts of this problem depicts a motion diagram for a charged particle moving through a
region of uniform electric field. For each part, draw a vector representing the direction of the electric
field.
Part A
Hint
A.1
Relationship between electric field and electric force
The relationship between the electric force that acts on a particle and the electric field at the
location of the particle is
.
This formula indicates that the force and the electric field point in the same direction for a
positively charged particle, and in opposite directions for a negatively charged particle.
Hint
A.2
Determining the direction of the electric field
location of the particle is
.
This formula indicates that the force and the electric field point in the same direction for a
positively charged particle, and in opposite directions for a negatively charged particle.
Hint
A.2
Determining the direction of the electric field
The acceleration of the particle can be determined from the change in its velocity. By Newton’s
2nd law, the force acting on the particle is parallel to its acceleration. Finally, since this is a
positively charged particle, the electric field is parallel to the force. Putting this all together results
in an electric field that is parallel to the particle’s acceleration.
Draw a vector representing the direction of the electric field. The orientation of the vector will be
graded. The location and length of the vector will not be graded.
ANSWER:
View
The motion diagram shows that the particle's acceleration points to the right. Because the particle has
positive charge, the electric field should point to the right.
Part B
Hint
B.1
Relationship between electric field and electric force
The relationship between the electric force that acts on a particle and the electric field at the
location of the particle is
.
This formula indicates that the force and the electric field point in the same direction for a
positively charged particle, and in opposite directions for a negatively charged particle.
Hint
B.2
Determining the direction of the electric field
The acceleration of the particle can be determined from the change in its velocity. By Newton’s
2nd law, the force acting on the particle is parallel to its acceleration. Finally, since this is a
negatively charged particle, the electric field is directed opposite to the force. Putting this all
together results in an electric field that is directed opposite to the particle’s acceleration.
Draw a vector representing the direction of the electric field. The orientation of the vector will be
graded. The location and length of the vector will not be graded.
ANSWER:
View
The motion diagram shows that the particle's acceleration points to the right. Because the particle has
negative charge, the electric field should point to the left.
ANSWER:
View
The motion diagram shows that the particle's acceleration points to the right. Because the particle has
negative charge, the electric field should point to the left.
Part C
Hint
C.1
Relationship between electric field and electric force
The relationship between the electric force that acts on a particle and the electric field at the
location of the particle is
.
This formula indicates that the force and the electric field point in the same direction for a
positively charged particle, and in opposite directions for a negatively charged particle.
Hint
C.2
Determining the direction of the electric field
The acceleration of the particle can be determined from the change in its velocity. By Newton’s
2nd law, the force acting on the particle is parallel to its acceleration. Finally, since this is a
positively charged particle, the electric field is parallel to the force. Putting this all together results
in an electric field that is parallel to the particle’s acceleration.
Because the electric field is uniform, you can find the direction of the particle's acceleration by
subtracting any two consecutive velocity vectors graphically. If
velocities, you can subtract
from
by placing
and
at the tip of
are any two consecutive
.
is the
vector that starts at the tail of
and ends at the tip of
.
To find the direction of the particle's acceleration graphically, use two unlabeled vectors to
represent
and
. Pick any two vectors
and
that would make your
subtraction easier; you can verify your result by subtracting any other pair of consecutive vectors.
Draw a vector representing the direction of the electric field. The orientation of the vector will be
graded. The location and length of the vector will not be graded.
ANSWER:
View
Part D
Hint
D.1
Relationship between electric field and electric force
The relationship between the electric force that acts on a particle and the electric field at the
location of the particle is
.
This formula indicates that the force and the electric field point in the same direction for a
positively charged particle, and in opposite directions for a negatively charged particle.
Hint
D.2
Determining the direction of the electric field
The acceleration of the particle can be determined from the change in the illustrated velocity
vectors. By Newton’s 2nd law, the force acting on the particle is parallel to its acceleration. Finally,
since this is a negatively charged particle, the electric field is directed opposite to the electric force.
Putting this all together results in an electric field that is directed opposite to the particle’s
acceleration.
Because the electric field is uniform, you can find the direction of the particle's acceleration by
subtracting any two consecutive velocity vectors graphically. If
velocities, you can subtract
from
by placing
and
at the tip of
are any two consecutive
.
is the vector
that starts at the tail of
and ends at the tip of
.
To find the direction of the particle's acceleration graphically, use two unlabeled vectors to
represent
and
. Pick any two vectors
and
that would make your
subtraction easier; you can verify your result by subtracting any other pair of consecutive vectors.
Draw a vector representing the direction of the electric field. The orientation of the vector will be
graded. The location and length of the vector will not be graded.
ANSWER:
View
What About Finite Sheets?
Description: The electric field near two oppositely charged finite sheets is calculated and compared to
the result for infinite sheets. Then, these ideas are demonstrated with an applet.
Frequently in physics, one makes simplifying approximations. A common one in electricity is the notion
of infinite charged sheets. This approximation is useful when a problem deals with points whose
distance from a finite charged sheet is small compared to the size of the sheet. In this problem, you will
look at the electric field from two finite sheets and compare it to the results for infinite sheets to get a
better idea of when this approximation is valid.
Consider two thin disks, of negligible thickness, of radius
that the x axis runs through the center of each disk.
oriented perpendicular to the x axis such
The disk centered at
has positive charge density
, and the disk centered at
has negative charge density
, where the charge density is charge per unit area.
Part A
What is the magnitude
Hint
A.1
of the electric field at the point on the x axis with x coordinate
?
How to approach the problem
When calculating the electric field from more than one charge or a continuous charge distribution,
one makes use of the superposition principle, which states that the electric field from multiple
charges equals the vector sum of the fields from each individual charge. To find the electric field at
a point, add the field due to each disk at that point. Be careful of whether the field magnitudes
should add or subtract. The easiest way to be sure is to draw a figure with the two disks and arrows
for the electric field direction on each side of each disk. If the arrows for the two disks point the
same way, then the magnitudes add. If they point in opposite directions, the magnitudes subtract.
Hint
A.2
The magnitude of the electric field due to a single disk
The magnitude
is
of the electric field along the x axis for a charged disk centered at
,
where
is the radius of the disk,
is the charge density on the disk,
is the permittivity of free
space, and is the x coordinate. Be careful in determining the direction in which the electric field
due to each disk points.
Hint
A.3
Determine the general form of the electric field between the disks
space, and is the x coordinate. Be careful in determining the direction in which the electric field
due to each disk points.
Hint
A.3
Determine the general form of the electric field between the disks
Which of the following equations represents the magnitude
disks?
Hint A.3.1
of the electric field between the two
Determine whether the magnitudes should add or subtract
Which of the following statements properly describes the directions of the electric fields due to
each disk in the region between the two disks?
ANSWER:
The electric fields due to both disks point to the left.
The electric fields due to both disks point to the right.
The electric field due to the positive disk points to the left, while the
electric field due to the negative disk points to the right.
The electric field due to the positive disk points to the right, while the
electric field due to the negative disk points to the left.
Since the two fields point in the same direction, their magnitudes add.
ANSWER:
Express your answer in terms of
,
,
, and the permittivity of free space
.
ANSWER:
=
Notice that as
approaches
, this expression approaches
, the result for two infinite
sheets. Also, note that the minimum value of the electric field, which corresponds in this case to the
greatest deviation from the result for two infinite sheets, occurs halfway between the disks (i.e., at
).
Part B
For what value of the ratio
field at the point
sheets?
Hint
B.1
of plate radius to separation between the plates does the electric
on the x axis differ by 1 percent from the result
for infinite
Percent difference
Recall that the percent difference between two numbers, in this case the electric field
A and the electric field
from Part
due to two infinite sheets, is given by
.
Express your answer to two significant figures.
ANSWER:
=
As mentioned above, this is the point on the x axis where the deviation from the result for two infinite
sheets is greatest. A common component of electrical circuits called a capacitor is usually made from
two thin charged sheets that are separated by a small distance. In such a capacitor, the ratio
is
far greater than 50. Based on your result, you can see that the infinite sheet approximation is quite
good for a capacitor.
This applet shows the electric field lines from a pair of finite plates (viewed edge-on). You can adjust
the surface charge density. You can also move the test charge around and increase or decrease its
charge to see what sort of force it would experience. Notice that the deviation from uniform electric
field only becomes noticeable near the edges of the capacitor plates.
Operation of an Inkjet Printer
Description: Find the charge that an ink droplet must be given to be deflected by a uniform electric
field between two parallel plates.
In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of
paper from a rapidly moving nozzle. The pattern on the paper is controlled by an electrostatic valve that
determines at each nozzle position whether ink is squirted onto the paper or not.
The ink drops have a mass
= 1.00×10−11
toward the paper at velocity
= 25.0
drop a positive charge
. The drops pass through a charging unit that gives each
by causing it to lose some electrons. The drops then pass between parallel
deflecting plates of length
magnitude
each and leave the nozzle and travel horizontally
= 7.60×104
= 1.50
, where there is a uniform vertical electric field with
.
Part A
If a drop is to be deflected a distance
= 0.250
deflection plate, what magnitude of charge
the ink drop is 1000
Hint
A.1
by the time it reaches the end of the
must be given to the drop? Assume that the density of
, and ignore the effects of gravity.
How to approach the problem
First, find the amount of time spent by the drop between the plates and use the result to find the
acceleration needed for the drop to have a vertical deflection of 0.250
. Next, using Newton's
2nd law, find the force needed for this vertical deflection. Finally, calculate the charge by setting
the electric force equal to this required force.
Hint
A.2
Calculate the time spent between the plates
How much time does it take for the ink drop to travel horizontally from the start to the end of the
deflection plates?
Hint
Relating horizontal distance and velocity
Hint
A.2
Calculate the time spent between the plates
How much time does it take for the ink drop to travel horizontally from the start to the end of the
deflection plates?
Hint
A.2.1
Relating horizontal distance and velocity
The ink drop has no acceleration in the horizontal direction. Therefore, the distance traveled by
the drop is equal to the product of the horizontal velocity and the time traveled. Use this relation
to calculate the time the ink drop spends between the plates.
Express your answer in seconds.
ANSWER:
=
Hint
A.3
Calculate the vertical acceleration
Calculate the acceleration needed for the ink drop to be deflected vertically by 0.250
during its trip between the deflection plates.
Hint
A.3.1
How to calculate the vertical acceleration
Since the ink drop initially has zero velocity in the vertical direction, the vertical deflection
0.250
is related to the vertical acceleration by
. Once you know the
amount of time the ink droplet spends between the plates, you can use this equation and solve for
the vertical acceleration
.
Express your answer in meters per second squared.
ANSWER:
=
Hint
A.4
Calculate the force that must be acting on the drop using Newton's law
Calculate the force
Hint
A.4.1
needed to deflect the ink drop vertically by a distance of 0.250
How to approach the problem
Recall Newton's 2nd law:
. Once you know the droplet's mass and its vertical
acceleration, you can compute the force needed to deflect the ink droplet vertically.
Express your answer in newtons.
ANSWER:
=
.
Recall Newton's 2nd law:
. Once you know the droplet's mass and its vertical
acceleration, you can compute the force needed to deflect the ink droplet vertically.
Express your answer in newtons.
ANSWER:
=
Now find the charge the ink droplet must have so that it will experience an electric force of this
magnitude when in the presence of an electric field of magnitude 7.60×104
Hint
A.5
.
Relating the electric force and the electric field
By definition, the magnitude of the electric field is the absolute value of the ratio of the magnitude
of the force acting on a charged particle in the field to the magnitude of the charge of the particle:
.
Express your answer numerically in coulombs.
ANSWER:
=
Here is something to think about: Is it reasonable to ignore the effect of gravity on the droplet in our
calculations? For an average inkjet printer, the magnitude of the acceleration due to the electric field
will be over ten times larger than the magnitude of the acceleration due to gravity. However, gravity
will still cause a small deflection of the droplet and hence should not be ignored if the accuracy of the
placement of the ink droplet is particularly important.