Incredibly basic Coulomb's law question - am I being silly?

In summary, the conversation discusses using the parallelogram method to add vectors and the importance of considering the magnitude of the charges when doing so. The individual is revising for an exam and is unsure if their answer is correct. They are reminded to follow the homework template and not to waste time with graphical addition of vectors on an exam.
  • #1
smileandbehappy
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0
member reminded that problem statement should be text and not a graphic
I've attached my workings... But is this correct? I can't see the fault with it. However I can't see how there will be no y-component to the force... Can someone a bit smarter than me tell me if I'm being really dim.

17387568979_224c324095_b.jpg
https://flic.kr/p/sutQu4][/PLAIN]
 
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  • #2
smileandbehappy said:
However I can't see how there will be no y-component to the force...
What's the y-component of the force from B on C? Compare that to the y-component of the force from A on C.

(I'm having trouble reading your diagram.)
 
  • #3
Doc Al said:
What's the y-component of the force from B on C? Compare that to the y-component of the force from A on C.

(I'm having trouble reading your diagram.)

Apologies about the poor iphone photo/my diagram drawing skills. If you click on the link you can then zoom in and see it better. I am no good at typing things out - and it seems like a waste of time to do so to ask a question, so I just photo my work. But I do realize it's a pain.

I have done this and found that the y-component for Fca and Fcb cancel each other out. Which is what I have a problem with. I have tried to draw out the what the ovreral force should look like... However I think I should get a y-component from the overall force... Noting here I am just using the paralellogram method and am not taking into account the magnitude of the charges...

Hold on - are you saying I should look at the relative sizes of the charges and compare them to the y component using the angle... If so I've seen this done before however I have never done it myself. I always just use the algebra and maths to give me an answer. I usually get the right answer to drop out - however sadly you are always a little unsure if you have it right or not.
 
  • #4
smileandbehappy said:
I have done this and found that the y-component for Fca and Fcb cancel each other out.
Right.

smileandbehappy said:
Which is what I have a problem with.

smileandbehappy said:
I have tried to draw out the what the ovreral force should look like... However I think I should get a y-component from the overall force... Noting here I am just using the paralellogram method and am not taking into account the magnitude of the charges...
If you used the parallelogram method properly to add those vectors, you would get a net force on C that would have no y-component. You must take into account the magnitude of the charges as they determine the magnitude of the forces!

smileandbehappy said:
Hold on - are you saying I should look at the relative sizes of the charges and compare them to the y component using the angle...
You certainly must consider the relative charges to properly scale your forces when drawing your parallelogram.
 
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  • #5
Doc Al said:
Right.

If you used the parallelogram method properly to add those vectors, you would get a net force on C that would have no y-component. You must take into account the magnitude of the charges as they determine the magnitude of the forces!

You certainly must consider the relative charges to properly scale your forces when drawing your parallelogram.

Thanks - I will do that in the future. I am currently revising for an exam where I only have around 5 minutes so answer both parts of that question including reading and checking time - so I can't do too much. But i do appreciate if I am going to do something I should do it properly.

For the avoidance of doubt - and because I'm a bit dim. Is the answer correct? Thanks Sam
 
  • #6
smileandbehappy said:
I am no good at typing things out - and it seems like a waste of time to do so to ask a question, so I just photo my work.
Posting your problem as an image and not following the homework template is against our rules. (See: How to Ask for Homework Help)

(My bad for not pointing this out sooner.)

smileandbehappy said:
Thanks - I will do that in the future. I am currently revising for an exam where I only have around 5 minutes so answer both parts of that question including reading and checking time - so I can't do too much.
On an exam, I would not waste time with graphical addition of vectors.

smileandbehappy said:
For the avoidance of doubt - and because I'm a bit dim. Is the answer correct?
Type out your answer and I'll check it out.
 

FAQ: Incredibly basic Coulomb's law question - am I being silly?

What is Coulomb's law?

Coulomb's law is a fundamental law in physics that describes the relationship between electric charges. It states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

How is Coulomb's law calculated?

Coulomb's law can be calculated using the equation F = k(q1q2)/r2, where F is the force between the two charges, k is the Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.

What is the unit of measurement for electric charge?

The unit of measurement for electric charge is the coulomb (C). One coulomb is equal to the amount of charge that passes through a point in one second when there is a constant current of one ampere (A).

How does distance affect the force in Coulomb's law?

The force between two charged objects is inversely proportional to the square of the distance between them. This means that as the distance between the charges increases, the force decreases. Similarly, as the distance decreases, the force increases.

Is Coulomb's law applicable to all types of charges?

Yes, Coulomb's law is applicable to all types of charges, including positive and negative charges. The law remains the same regardless of the type of charges involved, as long as they are stationary relative to each other.

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