Post your comments/questions or apply challenges for this week’s module here!
Academy of Engineering & Technology Freshman Physics
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Post your comments/questions or apply challenges for this week’s module here!
I did the first challenge, and my claim was that Multiplying the factors the charges were increased by will result in the factor the total force increased by. The constant in my experiment was the distance between the charges. My evidence is that when I multiplied one charge by 3 and the other by 2, the net force went from 0.5 N to 3 N. This shows that the force was multiplied by 6, which is the product of the factors the charges were multiplied by.
Challenge #2, where we are looking at the relationship between the electric force and the distance between the two objects in the simulation.
The electric force and the distance between the two objects are inversely, exponentially related. As one decreases the other increases at a rapid pace (in this case, as the distance goes down, the electric force goes up). This can be seen when we look at the data – the difference between 10 units away and 8 units away is only 0.6 N of force, but the difference between 4 units away and 2 units away is 22.9 N. The amount of force rapidly increased as the distance decreased.
Distance and force have an indirect relationship. As distance decreases, force exponentially increases.
q1 and q2 affected Felect by their magnitude. As q1 was doubled, halved, etc., the Force was doubled, halved, etc.. If both charges were altered, the Force was altered by the same factor as the sum of the factors acting on q1 and q2. For example, if q1 and q2 were both doubled, then the Force would be quadrupled. This is shown in the data, and one example of this would be the second the data element. As a control, we had the charge of q1 at
25*10-6 and the charge of q2 at 478*10-6.and the resulting force was 1.8N. The distance between the objects did not matter in this case since it was kept constant throughout the trials. When I doubled the charge of q1, the total force doubled as well. In another trial, I halved both the charges, the Force went down by a factor of 4.
The distance between the two objects, d, affects Felect. Using the 8 tile distance to start, we have a 1.8N force. When I divided the distance by 2, the force went up 4 times. Each time I halved the distance, the force went up 4 times. This shows that the Force is related to the distance squared. In the data, we see that when the distance was 1 square, the force was 116.5N and when the distance was 0.5 squares, the force was 466N. The distance was reduced by a factor of 2 and the force was increased by a factor of 4. This shows exponential growth.
Challenge #2: How is the electrostatic force (F elect) related to the separation distance (d) the two charges?
Since at 0 units, the force is also at 0, that means that the force is a product of the separation distance and something else.
Since the numbers of force get lower as the distance gets more, the constant multiplying the separation distance to get the force has to be a fraction or there has to be a negative exponent involved.
Dividing from 1 unit from 2 units, is about 0.26 N/units.
Dividing from 2 units from 3 units, is about 0.45 N/units.
Dividing from 3 units from 4 units, is about 0.57 N/units.
Since the numbers are different, this means that there is an exponent involved.
I think that the force is multiplied by some constant and the separation distance to some negative power.
When graphed, this is what it looks like.
This means that the F elect = 110d^-1.96.
I completed Challenge #2, which is about the relationship between the separation distance and the force.
As the separation distance (d) increases (I went by squares, so 2d means 2 squares between), the force (Felect) decreases. When the separation distance was 2d, the force was 5.7N (q1 charge was 25 * 10^-6 Coulombs, and q2 charge was 376 * 10^-6 Coulombs). When the separation distance increased to 6d, the force was 1.4N. When the separation distance went up to 10d, the force went down to 0.6N.
My claim was that the overall force is multiplied by the product of the factors acting on both of the charges. For example, in my case, the starting charge values were 25*10^-6 and 478*10^-6. When one of the charges was halved, while the other remained constant, the force was cut in half as well. And when both of the charges were halved, the electrostatic force decreased by a factor of 4. Additionally, when only one of the charges was doubled, the force was doubled as well.
The relationship between Felect and d is inverse because according to the data, as the distance between the two charges increases, the force between the two charges decreases and vice versa. In the first trial, the distance was 8 and the force was 1.8 N. In the fourth trial, the distance was 4 and the force was 7.2 N, showing that as the distance between two objects increases, the force between the two objects decreases. Furthermore, to prove that the relationship between the distance and force is an inverse relationship, when the distance between two objects was 8, the force was 1.8 N. When the distance was halved to 4, the force was increased by a factor of 4, leading to a force of 7.2 N. That is why the relationship between the distance and Felect is inverse.
The electric force and the distance of the objects seem to have an exponential relationship. As the distance between the objects decreases, the force between them decreases, making them inversely related as well. We can conclude that the force and the distance have an exponential relationship rather than a linear one from looking at the forces for the 4 unit distance (7.2 N), the 6 unit distance (3.3 N), and the 8 unit distance (1.8 N). While all distances are incremented by 2 units, when their forces are divided by their following force we can see that the quotients are all different. For example, 7.2N(4 units) divided by 3.3N(6 units) is 2.18, while 3.3N divided by 1.8N(8 units) is 1.83. This difference concludes that force and distance cannot be linearly related. Therefore, as the distance between the two objects decreases, the force between them exponentially increases.
The distance between charges and the electrostatic force are inversely related. When the distance between the charges is 10 units, the force is only 0.8 Newtons. However, when the distance is 8 units, the force is 1.2 Newtons. Also, as the distance got incremented by 2, the difference between forces got smaller. For example, the difference between 2 and 4 units is 4.1 Newtons, while the difference between 8 and 10 units is 0.4 Newtons. This shows that the relationship is exponential.
I did challenge #1 where the distance stayed the same. The magnitude of the two charges multiplied equals the magnitude of the F_elect. An example is if you have one charge with a magnitude of 10 and the other with a magnitude of 200 then the F_elect is 0.3. If you multiply either by 3, say 10 to 30, then the F_elect is now 0.9, Then if you multiple the other by 2, from 200 to 400, the F_elect is now 1.8, because 3 x 2 = 6, and 6 x 0.3 = 1.8.
I did the first challenge, and I found that the magnitude of q1 and q2 (how much charge each of them have) affect F(elect) mostly directly. When, for example, q1 was doubled from 10*10^-6 Coulombs to 20*10^-6 (q2 stayed constant at 200*10^-6), the force also doubled from 0.3 N to 0.6 N. This shows that when calculating F(elect), q1 and q2 are directly related to it, which means they are multiplied by a constant to get F(elect). Since distance was kept constant, that was not a factor that would have affected the final outcome. It also did not matter which charge was changed– when q1 was kept the same, and q2 was doubled, F(elect) was also doubled. Therefore, if either of the objects’ charge is multiplied by a constant, then F(elect) will also be multiplied by that same constant.
What I learned from this module is that the amount of charges on two objects has a direct relationship with the amount of force between them. This can be proven by how 5*10^-6 on q1 and 200*10^-6 on q2 has only a force of 0.2 N, but 50*10^-6 on q1 and 600*10^-6 on q2 has a force of 4.5N. Another thing I have learned is that the amount of distance between two charged objects and the force between them is inverse. This can be proven by how at 1 inch apart, the force was 29.4 N, and at 5 inches apart, the force was only 5.7 N. A question I have about this unit is how can it be decided which object picks up electrons or protons? Like in the balloon rubbing against the sweater example. Why does the balloon pick up the electrons? Why not the sweater?
I did Challenge #2.
The force and separation distance seem to be inversely related to one another. As separation distance increases, the force decreases. At 1 unit between the two charges, the force was 12.9 N. At 5 units, however, the force was a mere 2.4 N. It is also important to note that the amount of force lost decreases while separation distance increases. From 1 to 2 units of separation, there was a force loss of 5.7 N. From 4 to 5 units, there was a force loss of 0.8 N.
Challenge #1:
When you double any of the charges the force doubles, however if you multiply q1 * q2 it doesn’t equal Force, therefore you must multiply it by 2 * 10^8 to get the force. The constants in this experiment was the second object’s (q2) charge.
(challenge 2) The lesser the separation distance, the greater the force. The experiment shows that as the two objects draw nearer to each other, the force between them increases. When the objects are 9 units apart, the force between them is 1.2 N, but once they’re 7 units apart, the force between them is 2 N.
d affects Felect. At an 8 tile distance, we can see that the force is 1.8N, and when I halved the distance, we can see that the Force went to 7.4N. After halving it again, it went to 30.1N, and after the last half, it was at 109.5N. Each time I halved the distance, the force went up approximately 4 times. This shows that the force is related to the distance squared, which also shows that d is related to Felect exponentially.
1.Make a claim describing the relationship between Felect and q1 and q2. Support your claim with evidence (references to the data) and reasoning.
It seems doubling the charge of the negatively charged hotspot force roughly doubled the force, except for the first test in which adding -2*10^6 to the second value would cause it to be double the force so I assume rounding is the result of the somewhat inconsistent results.
2.Make a claim describing the relationship between Felect and d. Support your claim with evidence (references to the data) and reasoning.
As distance is a squared quantity in Coulomb’s law it makes sense that as you halve distance the force is quadrupled due to distance being squared and dividing the force.
I did Challenge #2. From this module, I discovered that the charge on q1 and q2 are directly proportional to the electrostatic force (Felect). This is because every time I increased the charges of q1 and q2, the electrostatic force (Felect) increased. For example, when q1 = 15*10-6 and q2 = 400 * 10-6, Felect = 0.9, and when q1 = 25 * 10-6 and q2 = 600 * 10-6, Felect = 2.3. The scientific reasoning behind this is that in the formula, Felect = k * (q1q2) / r2, q1 and q2 are in the numerator and are being multiplied to (k/r2), so the greater their values are, the greater the product of (k/r2) and (q1 * q2), and, as a result, the greater the electrostatic force (Felect) will be.
(I reposted my comment with the carrot symbol for exponents to make it easier to read and understand.)
I did Challenge #2. From this module, I discovered that the charge on q1 and q2 are directly proportional to the electrostatic force ( F(elect) ). This is because every time I increased the charges of q1 and q2, the electrostatic force ( F(elect) ) increased. For example, when q1 = 15*10^-6 and q2 = 400 * 10^-6, F(elect) = 0.9 N, and when q1 = 25 * 10^-6 and q2 = 600 * 10^-6, F(elect) = 2.3 N. The scientific reasoning behind this is that in the formula, F(elect) = k * (q1 * q2) / r2, q1 and q2 are in the numerator and are being multiplied to (k/r2), so the greater their values are, the greater the product of (k/r2) and (q1 * q2), and, as a result, the greater the electrostatic force ( F(elect) ) will be.
I did challenge #1:
My constants were 25*10^-6 Coulumbs for the q1 and 478*10^-6 Coulumbs for q2.
If the magnitude of charge of one of the particles is doubled, then the force is also doubled. The first row of data was unaltered, so we can compare the other’s to it. In row 2 of the data, I doubled the magnitude of the charge of q_1 which doubled the force from the first row (from 1.8 N to 3.5 N). If we look at the formula from Coulumb’s Law (k * (q_1 * q_2)/(r^2)), it appears that the relationship between the magnitude of the charge and the force value is a direct relationship (as proven above by the data) since both charges have a power of 1.
When separation distance is decreased, electrostatic force increases. This can be depicted by the results of the interactive lab I conducted. As you can see from the results, when there was a greater separation distance, we got smaller electrostatic forces between the two charges compared to smaller separation distances. Our largest separation distance was 14 d, which gave a value of 0.6 N in terms of electrostatic force. However, the electrostatic force was much more, 3.3 N, when we decreased the distance to only 6 d.
I did challenge 2.
As the distance decreases, the force increases exponentially. For example, 10 units has a force of 1.2 and 5 units has a force of 4.5, but 2.5 units has a force of 12.4. Clearly it isn’t increasing by a certain amount or being multiplied by a certain amount, and it is increasing rapidly, so it is increasing exponentially.
The constants were the position of the ruler, the magnitude of the charges, and the position of one of the charges.
Challenge 2:
Using the data gathered, it can be determined that the force and separation distance of two charges have an inverse, exponential relationship. As the distance decreases, the force increases at an exponential rate. The forces of 4 and 5 units have a difference of only 0.8 N, but the force at 1 unit is greater than the force at 2 units by 5.8 N.
Challenge #2
The relationship between Force and distance is negative, meaning that as the distance decreases, the force increases dramatically. For example, at 8 blocks away the objects contained 1.2 Newtons of force, but when moved to 3 blocks away (decreasing the distance by 5 blocks), they have 4.7 Newtons of force. It can also be interpreted that not only does force increase as distance is decreased, it does so at a curve. Between 8 and 6 blocks (a change in distance of 2) the change in force is a mere .6 Newtons, but between 4 and 2 (with the same change in distance), the change in force is a drastic 3.7 Newtons.
Challenge#2: Determine the relation between electrostatic force and distance of the charges involved.
Constants: q1 magnitude of charge = 25 * 10^-6 coulombs; q2 magnitude of charge = 478 * 10^-6 coulombs.
Claim: The relationship between separation distance and force is inversely related. According to the data, greater distance yields less force between the charges and vice versa. For example, 10 squares yields 0.8N of force, but 2 squares yields 7.2N of force.
I did the lab for challenge 2. My claim was that as the distance between both objects increased, the electrostatic force between them decreased. For example, when the distance between them was 2 squares, the force was 29.1 Newtons. But when the distance was 3 squares, the force was only 12.4 Newtons. And by the time the distance was 6 squares, the force was 4.7 Newtons.
Challenge #1:
The relationship between the force and q1 and q2 is a positive linear. As the charge of q1 increases by a constant of 10 * 10^-6, the force is also increasing at a constant of 0.6 newtons. The constants in this experiment were the distance between q1 and q2, the charge of q2, and the interval of change for q1.
Challenge #2:
As separation distance decreases, the electrostatic force increases and vice versa. At 10 d, the force was 0.1 N and at 2 d, the force was at 2.5 N proving that separation distance and electrostatic force have an inverse relationship. They also have an exponential relationship which means that the differences between the forces increased as separation distance decreased. The difference in force between 8 d and 6 d is only 0.2 N, whereas the difference between 4 d and 2 d is 1.9 N.
Challenge #1
There is a direct relationship between the charges and the force. As you multiply a charge by something, the force is multiplied by the same amount. It does not matter what charge you multiply.
The relationship between the force and q1 and q2 is proportional. As the magnitude of the charges increases so does the force. This is shown by the data table above where the magnitude of q2 is being increased at a constant rate of 100 Coulombs and the force is also consistently rising by 0.7-0.8 Newtons for every hundred. This shows that there is a positive relationship between these two values.
I did challenge one and my claim is that, as you multiply the charges by a certain constant, the force is also multiplied by the same constant. For example, when I multiplied the 1st charge by 1.5 and the second charge by 2. The electrostatic force was multiplied by 3.
I did Challenge 2, where the goal is to find the relationship between the distance of the charges and the force. My claim is that the relationship between the distance and the force is exponential and inverse. As the distance increases, the force increases. But the factor by which the force increases is not constant. If you divide the force at 8 units (1.8 N) by the force at 10 units (1.2 N), the result is 1.5. But if you divide the force at 6 units (3.3 N) by the force at 8 units, the result is 1.83. This pattern continues throughout the rest of the data, which proves that the force increases exponentially with distance.
Challenge 1:
As q1 and q2 increase, Felect increases. In the experiment,, Felect would increase when I would increase q1 and q2.
I did both challenges.
Challenge 1: The relationship between Felect and q1 and q2 appears to be positive because when one charge goes down so does Felect. The relationship also appears to be linear, since when one charge is halved, the force is halved as well. This is shown in the 3rd and 5th trial, where q1 and q2 were halved and as a result the force was halved as well. (1.8N to 0.9N).
Challenge 2:
According to the graph in the google spread sheet linked below, the relationship between Separation Distance (in squares) and Force (Felect) is exponential. The relationship is also indirect, as distance decreases force increases.
You can see the graph and data on this sheet document: https://docs.google.com/spreadsheets/d/1xrCD3vg3FnnlgyxCFMkmwjYulsR54myiPf-BwaRFtRY/edit?usp=sharing
I completed Challenge #1. The higher the charge on either q1 or q2, the higher the force gets. It’s a positive correlation between the combined charges and the force. By doubling the first charge and keeping the second, I got a force of 0.3 Newtons. By keeping the first charge and doubling the second, I got the same force mentioned before. It doesn’t matter how large the charge is, force scale directly with the charges.
The constant was the distance between the charges.
Challenge 2:
The relation between d and Felect is exponentially inverse. When d is 6 units, the force is around 2 N. At 5 units, the force doubles to around 4 N. Shortening the distance by one each time will double the force.
Challenge #2
I had a constant charge of 25*10^-6 and 200*10^-6. I found that the relationship between distance and force is an inverse relationship.
As the distance decreased, the force increased. For example, when the distance was 5 squares apart, the force was equal to 4N. But when I decreased the distance to 3 squares apart, the force became 10.8N. In the same way, when I increased the distance, the force decreased. Once the distance became 7 squares apart, the force became 2N.
Challenge number 1: Felect is directly related to q1, as shown by the relationship between trials 2 (5*10^-6, 400*10^-6, 0.3 N) and 4 (10*10^-6, 400*10^-6, 0.6 N), or trials 4 and 6 (20*10^-6, 400*10^-6, 1.2 N). q2 is also related to Felect, but has less of an impact
I did challenge #2, and a constant for me was the number of blocks I decreased the separation distance by each time. When the separation distance between electrons is greater, the magnitude of the electrostatic force will in turn also decrease. I find this to be true according to my data; when the objects are 10 squares away, the force between them is 1.2 N, and when the distance is 6 squares away, the force between the objects is 3.2 N.
Challenge #2 – the constant in the challenge was the increment that changed the separation distance of the two objects(q1, q2), which was one square. I believe that the separation between the two objects(q1 and q2) is directly proportional to the electrostatic force(Felect). One reason this claim is true is that when the objects are very close, such as 1 square apart, the force is very high, but once I move it over by one square, the force loses almost half of its magnitude. That means that the electrostatic force gets higher and higher until super close, where the force skyrockets. On the inverse side, as the separation distance gets higher, the electrostatic force (Felect) lowers by diminishing amounts. (losing 1.3 N to 0.2 N)
The relationship between F(elect) and magnitude of charge on both objects is linear, since when one object’s magnitude was halved, so was F(elect). When both were halved, F(elect) was quartered, and the opposite would happen when they were multiplied. My constants were the distance between the two objects and their positive/negative charges.
This is for Challenge #1.