Electrostatic Forces in Equilateral Triangles

In summary: It's a bit wose than that. There is also an extra closing parenthesis at the end, making the whole thing look completely wrong, rather than just a minor omission.In summary, the student attempted to solve an equation for the forces on the vertices of an equilateral triangle, but made a mistake in adding the vectors. The teacher considered the work to be incorrect, marking it down as such.
  • #1
Mohamed Walid
4
0

Homework Statement


I've encountered a question about electrostatic forces on vertices of an equilateral triangle and I believe that I solved it correctly but my Physics teacher has marked it as incorrect. Am I correct? Amy clue why my Physics teacher marked it wrong? This is the question and the Physics teacher's marking of my answer. I attached the entire thing.

Homework Equations


F = KQ1Q2/r^2
Vectors/Trigonometry

The Attempt at a Solution


The attempt of the solution is attached down with the teacher's marking.
 

Attachments

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  • #2
Looks like you put a parenthesis in the wrong place, changing the meaning drastically.
 
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  • #3
haruspex said:
Looks like you put a parenthesis in the wrong place, changing the meaning drastically.
I understand what you mean. I missed one closing paranthesis on the paper, but I typed it in correctly in the calculator and my final answer of Fnet is correct. The thing is that the teacher deducted 4 marks out of 5, this means that my entire work is virtually incorrect, not only this step (parenthesis).
 
  • #4
Mohamed Walid said:
I missed one closing paranthesis on the paper
It's a bit wose than that. There is also an extra closing parenthesis at the end, making the whole thing look completely wrong, rather than just a minor omission.
Mohamed Walid said:
my final answer of Fnet is correct.
It is? Looks a bit on the low side to me.
 
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  • #5
You worked out the magnitude of the two forces, good; but you made your mistake in adding the vectors.
... there are a number of ways to add the vectors.
You should check your reasoning by adding the vectors head-to-tail and see what happens.

The final equation looks like it is trying to be:
##F^2 = \big(F_1 + F_3\cos(60)\big)^2 + F_3^2\sin^2(60)##

You seem to have a ##F_3=6\times10^{-13}## in the term that got circled, but ##F_3=6\times10^{-3}##N elsewhere. Could that be it?

The way to think about your long answers, though, is this: imagine you have to mark schoolwork, it's your job ... you have marked 200 of them, it's 2am, and someone hot is waiting for you in bed, "come to me honey..."; but you can't because you have just one more question to mark... and it is the one in the pic you sent. How much effort are you going to put into figuring out what this student was thinking?

Make it easier to award the mark than withhold it.
 
Last edited:
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  • #6
According to the teacher, this is supposed to be the correct answer. But how?
 

Attachments

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  • #7
haruspex said:
It's a bit wose than that. There is also an extra closing parenthesis at the end, making the whole thing look completely wrong, rather than just a minor omission.

It is? Looks a bit on the low side to me.
I understand bro. I closed that missing parenthesis that made it look wrong when I put the equation in the calculator to find Fnet. So it must not significantly be the error that made me grant only 1 mark out of 5. I think that the teacher solves the thing in a wrong way as I posted in the previous reply.
 
  • #8
Mohamed Walid said:
I think that the teacher solves the thing in a wrong way as I posted in the previous reply.
The person who got all 5 points worked it correctly except for evaluating the square root at the end.

I do think your teacher was overly severe in grading your work.
 

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  • #9
Simon Bridge said:
You worked out the magnitude of the two forces, good; but you made your mistake in adding the vectors.
... there are a number of ways to add the vectors.
You should check your reasoning by adding the vectors head-to-tail and see what happens.

The final equation looks like it is trying to be:
##F^2 = \big(F_1 + F_3\cos(60)\big)^2 + F_3^2\sin^2(60)##

You seem to have a ##F_3=6\times10^{-13}## in the term that got circled, but ##F_3=6\times10^{-3}##N elsewhere. Could that be it?

The way to think about your long answers, though, is this: imagine you have to mark schoolwork, it's your job ... you have marked 200 of them, it's 2am, and someone hot is waiting for you in bed, "come to me honey..."; but you can't because you have just one more question to mark... and it is the one in the pic you sent. How much effort are you going to put into figuring out what this student was thinking?

Make it easier to award the mark than withhold it.

Makes sense.
 
  • #10
Simon Bridge said:
##F^2 = \big(F_1 + F_3\cos(60)\big)^2 + F_3^2\sin^2(60)##

You seem to have a ##F_3=6\times10^{-13}## in the term that got circled, but ##F_3=6\times10^{-3}##N elsewhere. Could that be it?
Yep - that is probably it all right.

The correct paper you show us uses the same basic reasoning as you, just broken into more stages.

When I do the calc using the ##6\times 10^{-13}## figure, I get your answer. If I use ##6\times 10^{-3}=0.006## instead, then I get F=0.016N

However: I'd say it was more like the last straw ... you already made your teacher work to try to figure out what you meant, but then it looked like you had the wrong value anyway so... X.

The marking in the "correct" paper shows that there are no marks awarded for the numerical answer at the end.
0.05N is clearly incorrect - the biggest those forces can add to is if they are in-line for: 0.012N+0.006N=0.018N < 0.05N.
 
  • #11
Mohamed Walid said:
Makes sense.
As Simon says, there are other ways to add the vectors. I would have considered the triangle formed by the two forces (in a noise-to-tail depiction) and their resultant. The cosine rule gives F2=F212+F232-2F21F23cos(120).
 
  • #12
haruspex said:
As Simon says, there are other ways to add the vectors. I would have considered the triangle formed by the two forces (in a noise-to-tail depiction) and their resultant. The cosine rule gives F2=F212+F232-2F21F23cos(120).
His version is the cosine rule using the external angle ;)
 
  • #13
Simon Bridge said:
His version is the cosine rule using the external angle ;)
It produces the same equation necessarily, but it looks to me that the OP method was to resolve one of the forces into components parallel to and perpendicular to the other.
 
  • #14
haruspex said:
It produces the same equation necessarily, but it looks to me that the OP method was to resolve one of the forces into components parallel to and perpendicular to the other.
That's what I got too.

The other example arbitrarily assigned x and y axes and resolved components wrt those... much more work.
 

Related to Electrostatic Forces in Equilateral Triangles

1. What is an equilateral triangle?

An equilateral triangle is a triangle with all three sides of equal length. This means that all three angles in the triangle are also equal, measuring 60 degrees each.

2. How do electrostatic forces work in equilateral triangles?

Electrostatic forces in equilateral triangles are determined by the charges present on the vertices of the triangle. Opposite charges attract each other, while like charges repel. This creates a force that acts along the lines connecting the charges, known as the Coulomb force.

3. How are the magnitude and direction of electrostatic forces calculated in equilateral triangles?

The magnitude of electrostatic forces can be calculated using Coulomb's law, which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. The direction of the force is determined by the orientation of the charges and follows the line connecting them.

4. Can electrostatic forces in equilateral triangles be balanced?

Yes, electrostatic forces in equilateral triangles can be balanced if the charges on the vertices are carefully chosen. For example, if two equal and opposite charges are placed on two vertices, the third vertex will experience a net force of zero.

5. What are some real-life applications of electrostatic forces in equilateral triangles?

Electrostatic forces in equilateral triangles have various applications in fields such as physics, engineering, and chemistry. Some examples include the use of electrostatic force to manipulate particles in microfluidic devices, the functioning of electronic circuits, and the formation of chemical bonds between atoms in molecules.

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