Circuits coulombs law
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ECPI University, Newport News *
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Jan 9, 2024
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Unit 4 Lab 1 Coulomb’s Law and Circuits Lab
Part 1: Coulomb’s Law This part explores Coulomb’s Law, which provides the force two charges exert on each other.
F
=
(
9
×
10
9
)
(
q
1
) (
q
2
)
d
2
Remember that for this equation to work, the charge must be in Coulombs and the distance in meters, resulting in the force in Newtons.
Open the Coulomb’s Law simulation (link on Canvas Modules in Unit 4).
PhET Simulation Coulomb’s Law
Click the “Macro Scale” tab. Follow directions below on how to set up test cases
. Each student should have different settings and no one should use the values demonstrated by the instructor.
Trial 1: Pick a value for the left charge, either positive or negative, with any value for the charge except zero or one.
Set the simulation and record your value in
the table below in the trial 1 row. [The table is on the next page.] This instructor will use +6.
Pick a different charge
with the opposite type of charge
(positive or negative). This number should be different from the number chosen for the 1
st
charge and not zero or one.
Set the simulation and record your value in the table in the trial 1 row. The instructor will use –7.
Note: the unit for the charge in the simulation is designated µC
which stands for 1 microcoulomb
. The micro
prefix represents 1/1,000,000 or 1 millionth of a unit. To convert to Coulombs
, divide the number by 1 million or move the decimal six places to the left. For example, 2 µC
= 2/1,000,000 C
= 0.000 002 C
= 2 x 10
-6
C
.
Once you have chosen your charges and set the simulation, click and drag one of the charges toward or away from the other and observe how the magnitude of the Coulomb Force changes. Move the charges so that the distance between them is less than 5.0 cm. Use the ruler to measure the distance
between the charges. Note the ruler measures distances
in cm and changes in increments of 0.1 cm. Make sure
the ruler’s 0.0 point is aligned with the dotted line
marking the position of the center of the left charge.
The instructor will use 4.7 cm.
Indicate the magnitude of the force given by the simulation.
1
For Type of Force
indicate whether the force is attractive or repulsive. Points will be deducted if your values are not within the stated parameters above for Trial 1 and below for the other two trials. Ask if you have questions about whether your values are okay to use. Do not use the instructor’s values.
Trial
q
1
(
µC
)
q
2
(
µC
)
d (
cm
)
F
(
N
)
Type of force
1
-4
7
9
31.068
2
-2
5
7
18.342
3
-6
3
10
16.178
The simulation gives both the force on charge 1 by charge 2 and the force on charge 2 by charge 1. Observe these numbers to answer the following question.
Which of the following statements correctly describes the relationship between these
two forces?
a
The forces are not equal; the larger charge exerts the larger force on the smaller charge.
X
b
The forces have equal magnitudes. c
The forces are not equal; the smaller charge exerts the larger force on the larger charge.
Trial 2:
Keep the magnitude of the charges the same but change them so they are like charges (either both positive or both negative). Do not change the distance between the charges. Record your settings as Trial 2 and fill in the force and type of force.
How does the force change in trial 2 compared to trial 1? They are the same
Possible changes to note: If the magnitude of the force changes, indicate by what factor you would have to multiply the force in trial 1 to get the force in trial 2. If the type of force changes (attractive to repulsive or vice versa
) indicate that.
Trial 3:
Keep the charges the same as trial 2 but double the distance between them. Record your settings and force magnitude and type.
How does the force change in trial 3 compared to trial 2?
It is the same
Possible changes to note: If the magnitude of the force changes, indicate by what factor you would have to multiply the force in trial 2 to get the force in trial 3. If the type of force changes (attractive to repulsive or vice versa
) indicate that.
For trial 1
, convert the charge and distance from the non-standard units of the simulation to standard units of charge (
C
) and distance (
m
). Then do the calculation using the eCalc calculator or the Desmos calculator. eCalc:
https://www.eeweb.com/tools/calculator
). To keep the numbers manageable, use the EE key to enter the exponents of scientific notation power of ten. 2
Desmos:
https://www.desmos.com/scientific
. Put scientific notation numbers in parentheses to prevent problems with order of operations.
Charge 1 (C)
-0.000004
Charge 2 (C)
0.000007
Distance (m)
0.09
Force (N)
F
=
(
9
×
10
9
)
(
q
1
) (
q
2
)
d
2
=
(
9
×
10
9
)
(
×
10
❑
) (
×
10
❑
)
()
2
Use the snipping tool to cut and paste the image of the calculator
into the lab below the table. You can just snip the top of the calculator to show what was entered and the result. If your result does not equal the simulation force for trial 1, you’ve made a mistake. Check your work and ask for help if needed.
DO NOT Screen Shot the simulation.
Paste calculator screen shot here.
Continued on the next page.
3
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Part 2: Circuits
This part requires each student create his or her own simulated circuit using the PhET Circuit Construction Kit Virtual Lab (link also provided in the assignment instructions area). PhET Circuit Construction Kit: DC
After opening the simulation, click on the Lab
tab. Create the following circuit:
Before After
Continued on next page.
4
Leave default electrons setting.
Check boxes for Labels and Values.
Drag components from the box along the left side.
Rotate by clicking and dragging on one dotted circle.
Wires connect automatically when close to another circe.
To change the value of a battery or resistor (light bulb), click on the component and use the slider to adjust.
Part A: Ohm’s Law Each student will use data designated below. Highlight the data you are using.
Student
Voltage
Resistor
Amin, Zahedullah
8 V
2 Ω
Anderson, Shane
12 V
3 Ω
Breeding, Matthew
16 V
4 Ω
Brooks, Brandon
8 V
3 Ω
Cotto, Sebastien
12 V
4 Ω
Ellison, Julius
16 V
5 Ω
Ferry, David
8 V
4 Ω
Genovese, Jacob
12 V
5 Ω
Leon, Bailey
16 V
6 Ω
Melvin, Blake
8 V
5 Ω
Myers, Morgan
12 V
6 Ω
Puttkammer, Travis
16 V
7 Ω
Short, Nick
8 V
6 Ω
Simmons, Tommy
12 V
7 Ω
Thompson, Davis
16 V
8 Ω
Velasco, An
8 V
7 Ω
Wayson, Cody
12 V
2 Ω
Williamson, Tad
16 V
2 Ω
8 V
8 Ω
12 V
8 Ω
16 V
3 Ω
Instructor
10 V
4 Ω
Once the circuit is conducting current, make sure the simulation is set to show electrons
. Current should be blue circles with a negative sign in each.
In what direction is the electron current flowing in the circuit?
X
a
From the negative terminal (black end) through the resistor ending at the positive terminal (copper end)
b
From the positive terminal (copper end) through the resistor ending at the
negative terminal (black end)
c
The current alternates between one direction and the other.
Put an X in the yellow box to the left of your choice.
Change the setting so it shows conventional current. Current should be red arrows.
5
Now which way does the current flow?
a
From the negative terminal (black end) through the resistor ending at the positive terminal (copper end)
X
b
From the positive terminal (copper end) through the resistor ending at the
negative terminal (black end)
c
The current alternates between one direction and the other.
Substitute the voltage (
V
) and the resistance (
R
) into Ohm’s Law to demonstrate the
current reading is correct. Your simulation readings should confirm this.
value
Unit (V, Ω, or A)
Current
=
I
=
V
R
=
❑
❑
4
A
Now double the voltage. Write the new voltage here
Component
Quantity
Value
Unit
(V, Ω, or A)
Battery
Voltage
16
V
Resistor (light bulb)
Resistan
ce
4
Ω
Record the reading on your current meter:
Current
4
A
When the voltage is doubled, the current is a
½ of what it was.
b
¼ of what it was.
c
4 times what it was
X
d
2 times what it was.
Do not change the voltage of the battery from new value (listed above). Continued on next page.
6
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Double the resistance. Record your values.
Component
Quantity
Value
Unit
(V, Ω, or A)
Battery
Voltage
16
V
Resistor (light bulb)
Resistan
ce
8
Ω
Record the reading on your current meter:
Current
2
A
When the resistance is doubled, the current is X
a
½ of what it was.
b
¼ of what it was.
c
4 times what it was
d
2 times what it was.
Keep the voltage and Resistance the same and calculate the power of the current
.
Calculation
Value
Unit
𝑃𝑜𝑤𝑒𝑟
=
𝑉𝐼
=
(
16
) (
2
)
=
¿
32
W
Cut the resistance of the light bulb in half and recalculate power.
Component
Quantity
Value
Unit
(V, Ω, or A)
Battery
Voltage
16
V
Resistor (light bulb)
Resistan
ce
4
Ω
Record the reading on your current meter:
Current
4
A
Calculation
Value
Unit
𝑃𝑜𝑤𝑒𝑟
=
𝑉𝐼
=
(
16
) (
4
)
=
¿
64
W
The lower resistance light bulb ________________ than the bulb with the higher resistance.
X
a
was brighter
b
was the same brightness
c
was dimmer
Part B: Adding Resisters in Series
For this part, reset the circuit to how it was at the start of Part A
.
Component
Quantity
Value
Unit
(V, Ω, or A)
Battery
Voltage
16
V
7
Resistor (light bulb)
Resistan
ce
4
Ω
Record the reading on your current meter:
Current
4
A
Break the circuit between the resistor and the ammeter (bottom right) and insert a
light bulb
so the current flows through the first light bulb and then the second and then the ammeter.
Change the resistance of the new bulb to equal the first bulb.
Component
Quantity
Value
Unit (V, Ω, or A)
Battery
Voltage
16
V
2 resistors in series, with the same resistance
Record the reading on your current meter:
Current
2
A
What has happened to the current in the circuit?
X
a
It is half of what it was.
b
It is ¼ of what it was.
c
It is 2 times what it was.
d
It is 4 times what it was.
Use Ohm’s Law for calculating resistance to calculate the total resistance of the circuit.
value
Unit (V, Ω, or A)
Resistance
=
R
=
V
I
=
❑
❑
=
¿
8
Ω
Double the second light bulb’s resistance
(R
2
) and record the two resistance values below. Component
Quantity
Value
Unit (V, Ω, or 8
R1, same resistance as in Part 1
A)
Battery
Voltage
16
V
Resistor 1 (top) = R
1
R
1
4
Ω
Resistor 2 (bottom) = R
2
R
2
8
Ω
Record the reading on your current meter:
Current
1.33
A
Click and drag the voltmeter from the right onto the simulation.
Place the red probe between the top resistor and the positive (copper) terminal of the battery.
Place the black probe on the other side of the top resistor. See diagram for how this
should look (except your voltage reading will be different).
Record the voltage across Resistor 1 (top). Call this V
1
. Record it below.
Move the probes down to read the voltage across the bottom resistor. Call this V
2
.
Quantity
Value
Unit (V, Ω, or A)
V
1
=
Voltage dropacrosstop resistor
=
¿
5.33
V
V
2
=
Voltage dropacross bottomresistor
=
¿
10.67
V
Total Drop across both resistors =
16
V
This total should be the same as your battery voltage.
Part C: Adding Resistors in Parallel
For this part, reset the circuit to how it was at the start of Part A
.
Component
Quantity
Value
Unit
(V, Ω, or A)
Battery
Voltage
16
V
Resistor (light bulb)
Resistan
ce
4
V
Record the reading on your current meter:
Current
4
A
Now add a resistor in parallel, setting the second light bulb’s resistance equal to double the first
. Add a switch between the two resistors as shown below.
9
If your voltage reads a negative number, reverse the black and red probe positions.
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Your circuit should look like this: Value
Unit (V, Ω, or A)
Current with switch open
4
A
Current with switch closed
6
A
When the switch is closed, the first light bulb
a
gets dimmer
b
gets brighter
X
c
stays the same brightness
Add an Ammeter to each leg of the parallel circuit to measure the current through each light bulb. The circuit should look something like this:
Record the current through the first (left) resistor. Call this I
1
.
Record the current through the second resistor with the switch closed. Call this I
2
.
Record the total current going back to the battery. Call this I
total
.
10
Make sure the right side resistor has twice the resistance of the left side resistor.
Switch in the open position; no current flows in this position.
Value
Unit (V, Ω, or A)
I
1
=
Current through
¿
=
¿
4
A
I
2
=
Current through
¿
resistor
=
¿
2
A
Total current going back to battery
6
A
I
total
should equal I
1
+ I
2
Open and close the switch so the right light bulb goes out and comes back on again.
What happens to the current through the left light bulb (I
1
)?
X
a
It stays the same.
b
It doubles.
c
It is cut in half.
Calculate the total Resistance of the two resisters in parallel. value
Unit (V, Ω, or A)
Resistance
=
R
=
V
I
total
=
❑
❑
=
¿
2.67
Ω
Conclusions:
Adding resistors in parallel causes the total resistance of the circuit to
X
a
go up.
b
stay the same
c
go down
Adding resistors in parallel causes the current coming out of the battery to
X
a
go up.
b
stay the same
c
go down
As resistors are added in parallel the current through a resistor already carrying current
11
X
a
goes up.
b
stay the same
c
goes down
As light bulbs are added in parallel, the bulbs already carrying current
a
get brighter
X
b
stay the same brightness
c
get dimmer
Listen for the instructions. The instructor may demonstrate the next part if time is limited; watch the demonstration and then answer each question. If there is not enough time even for a demonstration, the next part will be skipped.
Disconnect the battery from the circuit before making changes described below. When you have all the components in place, then connect the battery back, leaving switch at the bottom right open.
Change the resistance of each resister (light bulb) to 2.5 Ohm. Change the voltage of the battery to 13 V.
Add a fuse next to the battery. Change the fuse rating to 12 A.
(To find the fuse, scroll down the components on the left.)
The circuit should look like this.
Close the switch so you add current through the third light bulb to the circuit. When the third parallel resistor is added, what happens to the circuit?
The fuse breaks because the current goes above 12 amps
12
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Open the switch. Remove the fuse and complete the circuit without it. Close the switch without the fuse.
What happens to the circuit when the third parallel resistor is added?
The current goes above the voltage, causing the battery to go on fire 13
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