2022W 136-2 01 Coulomb
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Northwestern University *
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Course
136
Subject
Mathematics
Date
May 23, 2024
Type
docx
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Lab 1: A Law of Electrostatic Force
In this lab, you will use an online simulation to investigate how the charges of two objects and the distance between them effect the electrostatic force between them. After this lab, you should be able to describe the functional dependence of the force on charge and radius and draw simple conclusions from this understanding. You should also be able to linearize data and understand how to investigate functional dependence of a physical quantity by systematically changing related physical quantities.
Introduction
You are probably aware that like charges repel, while unlike charges attract. You may also have
learned the quantitative formula for the attractive or repulsive force. These facts, like all physical
laws, were established through rigorous experiments. In this lab we will use a simulation to
experimentally derive the equations for electrostatic force.
Electric charge is measured in units of Coulombs
, abbreviated as C. In practice, a single
Coulomb is a lot of charge, so in this lab we will mostly be dealing with µC (microcoulomb, =
1
×
10
−
6
C).
Familiarization and Setup
First, open the simulation
1
and click on the simulation. Play with the simulation for a little
while to familiarize yourself with its controls. We will use the “Macro Scale” option.
a)
The vertical dotted lines through the centers of the two charges, q
1
and q
2
indicate
the locations of the center of the charges. The ruler itself is also movable.
b)
The legends above the two charges give the force of interaction. The “force
values” checkbox at the bottom right enables this option.
c)
The controls at the bottom allow you to change the value of the two charges.
1
https://phet.colorado.edu/sims/html/coulombs-law/latest/coulombs-law_en.html
Page | 1
Force versus Charge
First, determine how the electrostatic force depends on the electric charge. This simulation
allows us to vary the charge on the two objects (
q
1
and q
2
) as well as the x-positions of those
charges (
x
1
and x
2
). To do this, we will set x
1
, x
2
, and q
2
to arbitrary values and leave them alone
while varying only q
1
. 1.1 (5pts) Record your values for x
1
, x
2
, r
and q
2
below. (
r
=
x
2
−
x
1
, the distance between the
two charges.) A positive q
2
and x
1
=
0
will be most convenient, but any values will work.
x
1
(cm)
x
2
(cm)
r
=
x
2
−
x
1
(cm)
q
2
(
C)
4.00
7.00
3.00
8
C
1.2 (5 pts)
Now, vary q
1
as shown in Table 1 and complete the table. If the force is attractive
(arrow on q
2
points toward q
1
), enter a negative number. If it is repulsive, enter a positive
number.
Table 1: Electrostatic force versus electric charge data from PhET simulation.
Charge 1 (
C)
Force on Charge 2 (N)
-9
-639.115
-6
-479.336
-3
-239.668
0
0.000
3
239.668
6
479.336
9
639.115
1.3 (20 pts)
We will guess that force and q
1
have a linear relationship
: that is, the force obeys
an equation of the form
F
=
k
1
q
1
(1)
for some as-yet-unknown value of the constant k
1
. Using the LineFit.xslx file included in the
Canvas assignment, plot the force (vertical axis) vs. the charge (horizontal axis) and fit the data
to this straight-line equation (set the intercept to zero). Save the graph as an image file, and insert
the image here. Do not
just copy and paste the graph into this document – if you do that and
submit your work as a .docx file, your graph may not show up in the submitted file! Record the
slope k
1
(determined by the straight line through the data) here. Don’t forget to state the units of
Page | 2
k
1
. (Note: in a future lab we will talk about uncertainty
in your slope, but for now you do not
need to record that. Just record the slope itself.)
You now know how the force depends on q
1
. In principle, you should also verify how the force
depends on q
2
. But intuition suggests that “electric charge is electric charge.” Why should the
electrostatic force depend differently on q
2
than on q
1
? This implies that the force depends
linearly on both charges, as:
F
=
k
12
q
1
q
2
(2)
Page | 3
\
f Slope:
in units of
74.183
N/
C
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[Side note: of course, it is possible that the F vs. q
1
graph is not a perfectly straight line after all,
and that the relation between the force and charges is entirely different. This is particularly
likely if our range of charges covers only a small part of what is possible in nature, so that our
linear fit is only an approximation. For example, we didn’t test values larger than 9 µC or closer
to zero than 3 µC, so it’s possible that something strange happens there. Ideally, we would
verify the linear dependence with additional data, but you do not need to do this now.]
Page | 4
Force versus Distance
2.1 (5pts) Next, determine whether (and how) the electrostatic force depends upon the distance
r
between the two charges. Move the first charge all the way to the left, x
1
=
0
. Select arbitrary,
non-zero values for q
1
and q
2
and record them here.
q
1
(
C)
q
2
(
C)
-4
8
2.2 (10pts)
Vary the position of Charge 2 and complete Table 2 with 7 data points. Use all of
the ruler and try to spread out your points uniformly. Also, for each position, calculate the value
of the constant k
12
from the previous section using the known q
1
and q
1
and measured F
.
Table 2: Electrostatic force versus distance between electric charges for simulated data.
r
=
x
2
−
x
1
(cm)
Force (N)
k
12
=
F
/
q
1
q
2
(N/
C
2
)
2
-719.004
22.469
3
-319.557
9.986
4
-179.751
5.617
5
-115.041
3.595
6
-79.889
2.497
8
-44.938
1.404
10
-28.760
0.899
2.3 (20 pts)
In the previous section, we thought that k
12
was a constant because it didn’t
depend on q
1
or q
2
. However, to explain the data in Table 2, k
12
clearly must depend on r
. We
will assume that the dependence is a power law:
k
12
=
Ar
B
(3)
for unknown constants A
and B
. B
is an exponent so it will not have units, but A
will have
whatever units are necessary to balance Equation (3).
To determine values for A
and B
, we would normally use a computer program to fit Equation (3)
to our data. However, not all data analysis software can fit arbitrary non-linear equations such as
Equation (3). Also, when we start talking about fitting uncertainty later in the course, it is more
complicated to think about that for non-linear fittings. So, we’d really like a trick to let us fit
Page | 5
Equation (3) as though it were linear. Luckily, this trick exists! Consider taking a logarithm of
both sides of the equation, as:
log
(
k
12
)
=
log
(
Ar
B
)
¿
log
(
A
)
+
B
log
(
r
)
(4)
If we let y
=
log
(
k
12
)
and x
=
log
(
r
)
, then Equation (4) looks like y
=
b
+
mx
. So log
(
k
12
)
plotted as a function of log
(
r
)
should be a straight line with slope B
and y
-intercept log
(
A
)
. This trick of changing what you plot so that the result should be linear is called linearizing
the data.
Plot y
=
log
(
k
12
)
vs x
=
log
(
r
)
, and fit this graph with a straight line
(non-zero intercept). Save the graph as an image file and insert it below. Note: logarithms don’t have units, so the axes of this plot will not have units. Similarly, log
(
A
)
has no units.
2.3 continued:
Record the values of A
and B
here, and given their units (if any). Note: record
A
, not
log
(
A
)
, and make sure you know whether you’re using log base 10 or base e!
Page | 6
B = -2.00
A = 89.95 Ncm
2
/
C
2
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Results
The electrostatic force law is known to be
F
=
1
4
π ϵ
0
q
1
q
2
r
2
(5)
ϵ
0
is a fundamental physical constant, called the “vacuum permittivity” or the “permittivity of
free space.” Other scientists have carefully measured
ϵ
0
to
have a value of
8.8542
×
10
−
12
C
2
/
N m
2
. We have now collected enough data to determine this natural constant
for ourselves.
3.1 (10 pts) In the first part of this lab, you determined k
1
, defined by F
=
k
1
q
1
. Thus
k
1
=
1
4
π ϵ
0
q
2
r
2
(6)
with q
2
and r
having the fixed values you selected in 1.1
above. Use the value of k
1
you wrote
down in 1.3
above to calculate ϵ
0
and enter it here. Please express ϵ
0
in units of C
2
/
N m
2
, which
may require converting from cm or µC.
3.2 (10 pts)
Comparing Equation (5) with our expressions F
=
k
12
q
1
q
2
and k
12
=
A r
B
from
Equations (2) and (3), we get
A
=
1
4
π ϵ
0
(7)
Use the value of A
you entered in 2.3 to calculate ϵ
0
and enter it here. Again, please express ϵ
0
in units of C
2
/
N m
2
, which may require converting from cm or µC.
Page | 7
ϵ
0
= 9.54 * 10
-12
C
2
/Nm
2
ϵ
0
= 8.85 * 10
-12
C
2
/Nm
2
3.3 (5 pts)
Are the two estimates of ϵ
0
close to each other? Are they close to the established
value for ϵ
0
? We have not yet discussed a formal way to decide whether two values are close—
that is the topic of a future lab—so for now, investigate the percent difference. The percent
difference between two values v
1
and v
2
is their difference over their average: v
1
−
v
2
(
v
¿¿
1
+
v
2
)/
2
¿
For this lab, we will say that if the two values are within a few percent, they are in decent
agreement.
If you were writing a report on this exercise, you would want to report your measured value for
the physical constant, ϵ
0
, in your Conclusions. Additionally, you would state whether you have
verified that your data supports Equation (5) as the electrostatic force of interaction between two
electric charges.
History and Summary
These results were first published (with a few variations) by Charles-Augustin de Coulomb in
1785. Unlike the simulation, Coulomb could move his charges around in three dimensions, and
noted that the force (a vector) is always directed along the line connecting the centers of the two
charges. Equation (5) relates only the magnitude (not the direction) of this force. Feel free to play with the “Atomic Scale” simulation for a more realistic set of achievable charge
distributions.
Page | 8
Using value from 3.1: 0.0746 or 7.46% - not very accurate
Using value from 3.2: 4.74 * 10
-4
or 0.474% - more accurate