# Electrostatic Forces in Equilateral Triangles

1. Jan 12, 2017

### Mohamed Walid

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
F = KQ1Q2/r^2
Vectors/Trigonometry

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

#### Attached Files:

• ###### IMG_3511.JPG
File size:
29.9 KB
Views:
93
2. Jan 12, 2017

### haruspex

Looks like you put a parenthesis in the wrong place, changing the meaning drastically.

3. Jan 12, 2017

### Mohamed Walid

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. Jan 12, 2017

### haruspex

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.

5. Jan 12, 2017

### Simon Bridge

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.

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 in to figuring out what this student was thinking?

Make it easier to award the mark than withhold it.

Last edited: Jan 12, 2017
6. Jan 12, 2017

### Mohamed Walid

According to the teacher, this is supposed to be the correct answer. But how?

#### Attached Files:

• ###### IMG_3515.JPG
File size:
21.4 KB
Views:
23
7. Jan 12, 2017

### Mohamed Walid

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. Jan 12, 2017

### TSny

The person who got all 5 points worked it correctly except for evaluating the square root at the end.

#### Attached Files:

• ###### Charges.png
File size:
82.2 KB
Views:
29
9. Jan 12, 2017

### Mohamed Walid

Makes sense.

10. Jan 12, 2017

### Simon Bridge

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. Jan 12, 2017

### haruspex

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. Jan 12, 2017

### Simon Bridge

His version is the cosine rule using the external angle ;)

13. Jan 12, 2017

### haruspex

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. Jan 12, 2017

### Simon Bridge

That's what I got too.

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