# Coulomb's Law

## Homework Statement

[PLAIN]http://img547.imageshack.us/img547/7932/14207376.png [Broken]

[PLAIN]http://img684.imageshack.us/img684/719/75458442.png [Broken]

[PLAIN]http://img88.imageshack.us/img88/4883/55224430.png [Broken]

Assume $$q_{1}=q_{2}=q_{3}$$ and that all charges are positive.

## The Attempt at a Solution

*if someone could, please tell me the proper code for vectors, because I am having trouble

For the first of the problem

$$\vec{E_{1}} = \vec{E_{21}}$$

Since it sort of just "sits in space", I put q_{2} on the origin.

So $$\vec{E_{21}} = <0, k\frac{q_2}{d^2}>$$ and the magnitude should simply be $$k\frac{q_2}{d^2}$$.

For the second part

$$\vec{E_{1}} = \vec{E_{21}} + \vec{E_{31}}$$

$$\vec{E_{31}} = k\frac{q_{3}}{d^2}<-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}>$$

Since $$\vec{E_{21}} = <0, k\frac{q_2}{d^2}>$$

Then the sum would be $$\vec{E_{1}}= \frac{k}{d^2}<-q_{3}\frac{\sqrt{2}}{2}, q_{3}\frac{\sqrt{2}}{2} + q_{2}>$$

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tiny-tim
Homework Helper
hi flyingpig!

(it's easiest to use bold letters for vectors )

i] the field is kq/d2, but the force is kqq/d2

ii] i think you're using 45° for the angle, it's 60°

iii] don't forget the question says you can assume the qs are all the same!

hi flyingpig!

(it's easiest to use bold letters for vectors )

i] the field is kq/d2, but the force is kqq/d2

ii] i think you're using 45° for the angle, it's 60°

iii] don't forget the question says you can assume the qs are all the same!
Oh right...what am I doing!!!? I was reading "electrostatic" and it immediately turned into field.

But what if I want to use arrows? I actually find it more difficult to tell if it is bold.

$$\vec{F_{1}} = \vec{F_{21}}$$

$$\vec{F_{21}}= k\frac{q^2}{d^2}<0,1>$$

That's part one

For part two

$$\vec{F_{1}} = \vec{F_{21}} + \vec{F_{31}}$$

$$\vec{F_{31}}= k\frac{q^2}{d^2}<-\frac{\sqrt{3}}{2}, \frac{1}{2}>$$

$$\vec{F_{21}}= k\frac{q^2}{d^2}<0,1>$$

So the sum of $$\vec{F_{21}} + \vec{F_{31}}$$ is then

$$k\frac{q^2}{d^2}<-\frac{\sqrt{3}}{2}, \frac{3}{2}>$$

So now, my question is, should I turn this into a unit vector?

Oh right...what am I doing!!!? I was reading "electrostatic" and it immediately turned into field.
Try to be polite next time.

______________________________________________________

No, why would turn this into a unit vector?
Unit vector has mag of 1
it will mean that that your force also has a unit vector of one!!!

Try to be polite next time.

______________________________________________________

No, why would turn this into a unit vector?
Unit vector has mag of 1
it will mean that that your force also has a unit vector of one!!!
I wasn't being rude =(! Not sure where you caught that from.

But I thought the unit vector will only take care of the direction of my force, not the force itself? Is it okay to leave my answer above like that?

Did I get it right!??

Thanks!

And direction can still be founded with this result. unit vectors just provide direction without changing mag of something!!!

10(i) + 10(j) has same direction as its unit vector (1/√2)(i) + (1/√2)(j)

And direction can still be founded with this result. unit vectors just provide direction without changing mag of something!!!

10(i) + 10(j) has same direction as its unit vector (1/√2)(i) + (1/√2)(j)
I noticed something, this problem just want to assume that I place q2 at the origin, why?

Also, I thought the concept of unit vector is like multiplying and dividing by one, does it really matter?

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I noticed something, this problem just want to assume that I place q2 at the origin, why?
Well this problem doesnot depend on the choice of origin
Also, as long as orientation of XYZ is same direction will also remain same ... but mag of force always remain same no matter where is the origin

tiny-tim
Homework Helper
Try to be polite next time.
cupid.callin, whatever are you talking about?
… So now, my question is, should I turn this into a unit vector?
Sort of.

The question asks for "the magnitude and direction".

So you need to find the magnitude anyway …

once you've done that, what's left is the unit vector of the direction!
I noticed something, this problem just want to assume that I place q2 at the origin, why?

Also, I thought the concept of unit vector is like multiplying and dividing by one, does it really matter?
hmmm … not sure what you mean by either of those.

The question asks for "the magnitude and direction".

So you need to find the magnitude anyway …

once you've done that, what's left is the unit vector of the direction!
But I thought taking the unit vector according to cep will yield a magnitude of 1N

hmmm … not sure what you mean by either of those.
Sorry I wasn't sure what my question is now that I read over again.

once you've done that, what's left is the unit vector of the direction!
Both unit vector and vector itself gives same direction.
If answer is written with mag <not original vector> then only you need to give unit vector, or even better just give answer in degrees!!!

If he's giving answer in original vector, why waste time in finding unit vector also???

tiny-tim
Homework Helper
But I thought taking the unit vector according to cep will yield a magnitude of 1N
now i'm really confused

you got kq2/x2 < -√3/2, 3/2 > …

where does 1N come into that?

you need to convert < -√3/2, 3/2 > to a magnitude times a unit vector
Sorry I wasn't sure what my question is now that I read over again.
he he

you got kq2/x2 < -√3/2, 3/2 > …

where does 1N come into that?

you need to convert < -√3/2, 3/2 > to a magnitude times a unit vector
you agree that unit vector of < -√3/2, 3/2 > will have mag of 1, right?

So by this you mean that net force is just kq2/x2

< -√3/2, 3/2 > has nothing to do in the mag?

now i'm really confused

you got kq2/x2 < -√3/2, 3/2 > …

where does 1N come into that?

you need to convert < -√3/2, 3/2 > to a magnitude times a unit vector

he he

I got it from this

No, why would turn this into a unit vector?
Unit vector has mag of 1
it will mean that that your force also has a unit vector of one!!!
But once I get it to the unit vector, wouldn't my force be one like he said? Or is that actually not surprising since the distance is is d after all?

But once I get it to the unit vector, wouldn't my force be one like he said? Or is that actually not surprising since the distance is is d after all?
Which distance is d?

tiny-tim
Homework Helper
you agree that unit vector of < -√3/2, 3/2 > will have mag of 1, right?

So by this you mean that net force is just kq2/x2

< -√3/2, 3/2 > has nothing to do in the mag?
cupid.callin, are you drunk?

But once I get it to the unit vector, wouldn't my force be one like he said? Or is that actually not surprising since the distance is is d after all?
express < -√3/2, 3/2 > as a magnitude times a unit vector, say M <a,b> where <a,b> is the unit vector

then the magnitude of the force is M times kq2/x2, and it is in the direction of <a,b>

Which distance is d?
We don't know! It could be anything and it shouldn't matter

cupid.callin, are you drunk?

express < -√3/2, 3/2 > as a magnitude times a unit vector, say M <a,b> where <a,b> is the unit vector

then the magnitude of the force is M times kq2/x2, and it is in the direction of <a,b>
Don't you mean 1/M???

$$k\frac{q^2}{\sqrt{3}{d^2}}<-\frac{\sqrt{3}}{2}, \frac{3}{2}>$$

But then I would get back at $$k\frac{q^2}{d^2}$$

tiny-tim
Homework Helper
hi flyingpig!

(just got up :zzz: …)

no …

for example if the vector was 3 <4,0>,

that would be 12 <1,0> …

magnitude 12, in the direction of unit vector <1,0>

Then what am I thinking of? For 3<4,0>

M = 12

So 1/12<12,0>?

tiny-tim
Homework Helper
1/12 of <12,0> is the unit vector <1,0> …

are you thinking of a way to make unit vectors?

1/12 of <12,0> is the unit vector <1,0> …

are you thinking of a way to make unit vectors?
Isn't that the goal?

tiny-tim
Homework Helper
half the goal …

the goal (in the original question) was to find the magnitude and the direction …

ie a magnitude and a unit vector

half the goal …

the goal (in the original question) was to find the magnitude and the direction …

ie a magnitude and a unit vector
$$k\frac{q^2}{\sqrt{3}{d^2}}<-\frac{\sqrt{3}}{2}, \frac{3}{2}>$$ <=== Unit vector

$$k\frac{\sqrt{3}q^2}{d^2}$$ <=== magnitude.

tiny-tim
$$k\frac{q^2}{\sqrt{3}{d^2}}<-\frac{\sqrt{3}}{2}, \frac{3}{2}>$$ <=== Unit vector
$$k\frac{\sqrt{3}q^2}{d^2}$$ <=== magnitude.