Coulomb's law vector question.

quasar987

Homework Helper
Gold Member
I still haven't received an answer for this one :P
What's wrong with the solution I gave in post #11 ?

I'm in my 2nd session, and my cote R from the first one is great, so if I can continue like that, maybe next year I can make an admission.
That's good. I think the trick is to build yourself a high cote R in the first 2 sessions while there are still very weak people in the classes for you to "feed on" . And for the last two sessions it's basically just a matter of maintaining it.

K.J.Healey

I get 1.8157i + 2.0446j as the force.(in newtons :p)

How I did it:
Your first way was right,
$$\vec{F_{12}}= Qkq ( 0,222 \vec{i} + 0,250\vec{j}) N$$
and
$$\vec{F_{12}}= \frac{2,5Qkq}{r^2_{21}} \vec{u_{21}}$$

So
$$( 0,222 \vec{i} + 0,250\vec{j}) = \frac{2,5}{r^2_{21}} \vec{u_{21}}$$

and
$$(0,222 \vec{i} + 0,250\vec{j}) = 0,3343(0,66399\vec{i} + 0,7477\vec{j}) = 0,3343\vec{u_{21}}$$

and so

$$0,3343=\frac{2,5}{r^2_{21}}$$
$$r_{21}=2,7345m$$

so multiply this magnitude onto the its unit vector and you have the answer above.

Healey01 said:
I get 1.8157i + 2.0446j as the force.(in newtons :p)

How I did it:
Your first way was right,
$$\vec{F_{12}}= Qkq ( 0,222 \vec{i} + 0,250\vec{j}) N$$
and
$$\vec{F_{12}}= \frac{2,5Qkq}{r^2_{21}} \vec{u_{21}}$$

So
$$( 0,222 \vec{i} + 0,250\vec{j}) = \frac{2,5}{r^2_{21}} \vec{u_{21}}$$

and
$$(0,222 \vec{i} + 0,250\vec{j}) = 0,3343(0,66399\vec{i} + 0,7477\vec{j}) = 0,3343\vec{u_{21}}$$

and so

$$0,3343=\frac{2,5}{r^2_{21}}$$
$$r_{21}=2,7345m$$

so multiply this magnitude onto the its unit vector and you have the answer above.

That is the right answer, except it's (-1,8 i - 2,05j) (no problem there)

I just want to know from where comes the $$0,3343(0,66399\vec{i} + 0,7477\vec{j}) = 0,3343\vec{u_{21}}$$
Thanks a lot for the help, can you just clarify this for me? :)

K.J.Healey

Well, the force is a vector. ALL vectors have a magnitude times a unit vector which describes its direction. Since we know the UNIT VECTORS have to be the same, i wanted to get that unit vector out of the main force vector, so i could cancel it out.
Unit vectors are defined as : VECTOR/MAGNITUDE
so i got the magnitude by sqrt(sum of squares) and divided the vector by it.
The magnitude of a unit vector is 1 (just for knowledge).

so

for vector A
|A|*A
------- == A
|A|

and
unit vector u
u = A/|A|
so
A=|A|*u

and u = A/|A| = A1i+A2j/Sqrt(x^2+y^2)

I hope that clears it up. Im jsut making the vector into a unitvector and a magnitude.

Thanks a lot Healey01, I understand . What I don't get is the teacher told me to equal the i and j vectors so it makes two different equations, and since I have 2 unknown values, I'd find them like that. Do you have any idea how to do that?

I tried
$$( 0,222 \vec{i} + 0,250\vec{j}) = \frac{2,5}{r^2_{21}} \vec{u_{21}}$$

which I substituted with x and y instead of i and j

$$( 0,222 \vec{i} + 0,250\vec{j}) = \frac{2,5}{ x^2 + y^2} x \vec{i} + y \vec{j}$$

but that doesnt make sense. I hope you understand what I was trying to do.

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K.J.Healey

Jsut take what you have, and split it up:
$$( 0,222 \vec{i} + 0,250\vec{j}) = \frac{2,5}{ x^2 + y^2} x \vec{i} + y \vec{j}$$

becomes
$$( 0,222 \vec{i}) = \frac{2,5}{ x^2 + y^2} x \vec{i}$$
and
$$( 0,250\vec{j}) = \frac{2,5}{ x^2 + y^2} y \vec{j}$$

Then its basically 2 equations with 2 unknowns.

I would then say solve the top one for y^2 and then plug it into the bottom. Its a lot of algebra from there but it would work. I honestly think splitting up vectors into the two parts gets rid of the point of even having vectors. But it DOES make things easier to computer in classes with dynamics and forces acting on objects. Sum the forces in each direction, set equal to m*a, youll see that a lot.

But you should always make sure you feel comfortable using something like full vector notation before you toss it aside for an easier way.

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When I solve for x and y , I get x= (+-) 3,16 and y= 1,27 ...
Anyway, I already understood your first method, it's a lot more easier.. but I dont know why it never works like you have written.
Thanks a lot for all your time and help, Healey01 :)

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