Coulomb's law vector question.

[mad]

Hello, I have a simple exercice I'm having trouble to solve.

So we have a) and b)

a) asked the force applied on a charge... and the answer was $$\vec{F_{res}.}= Qkq ( -0,222 \vec{i} -0,250\vec{j}) N$$
that's not a problem here.. that was the good answer. Now in b), they're asking where do they have to put a +2,5Q charge to nullify the force exerced.. so I thought yeah really simple, the force just has to be the inverse:

$$\vec{F_{12}}= Qkq ( 0,222 \vec{i} + 0,250\vec{j}) N$$

and since $$\vec{F_{12}}= \frac{2,5Qkq}{r^2_{21}} \vec{u_{21}}$$

then I just have to equal this last equation to the 2nd to last one, and I equal the $$\vec{i}$$ and $$\vec{j}$$ vectors in both ones to find out the vector of where the charge should be placed to nullify the force applied on it, but I'm having trouble finding it. Believe me, I have tried more than 10 times, and I can't find the vector. Can someone please help me on this one?

thanks a lot!

 

[quasar987]

$$\vec{F_{12}}= \frac{2,5Qkq}{r^2_{21}} \vec{u_{21}}$$

can also be written as

$$\vec{F_{12}}= \frac{2,5Qkq}{|\vec{r_{21}}|^3} \vec{r_{21}}$$

 

[dextercioby]

If you want help,then u have to help us too:the only way to do that is to post the ORIGINAL TEXT OF THE PROBLEM...

Daniel.

 

[dextercioby]

And BTW:if that N at the right of your first equation is what i think it is,then learn that VECTORS DO NOT HAVE "SI" UNITS...

Besides,even if it were a scalar,you should have used the traditional notation
$[unit]$

Daniel.

 

[K.J.Healey]

Please give a description of the first part. Like:
Is there another particle that causes this first force?
Is q the charge on the main particle? Is it positive?
If these are right then let's look at it as: (sorry I am not goot at latex)
charge 1 = main particle
charge 2 = causes force in part a
charge 3 is the 2.5Q charge

Force on 1 due to 2:
kqQ*(u12)/(r12)^2

Force on 1 due to 3:
kq(2.5Q)*(u13)/(r13)^2

They must sum to zero
cancel kqQ

(u12)/(r12)^2 = -2.5*(u13)/(r13)^2

since we know its on the other side, the unit vector would be the negative of the previous
u12=-u13
1/(r12)^2 = 2.5/(r13)^2

so r13=sqrt(2.5)*r12

The unit vector of the first is (-0.664i-0.748j) with the magnitude of 0.334 (u=r/|r|)

so r13=sqrt(2.5)*(0.334)=0.528

So the vector is magnitude*(0.664i+0.748j) = 0.351i+0.395j

Thats what I get, but I am not very confident in my skills.

 

[mad]

And BTW:if that N at the right of your first equation is what i think it is,then learn that VECTORS DO NOT HAVE "SI" UNITS...
Daniel.

A force has a unit, Newtons, that's why I posted the N.It has a force in a direction i and j.

Healey01, thanks for the help but that's not the answer.

The question was there's two charge that create a force (the first one I posted) on another charge. They then add a 2,5Q charge and ask where to put it in order to nullify the force applied on the charge..
Thanks for everyone that helped, and I hope someone can figure this out...

 

[mad]

$$\vec{F_{12}}= \frac{2,5Qkq}{r^2_{21}} \vec{u_{21}}$$

can also be written as

$$\vec{F_{12}}= \frac{2,5Qkq}{|\vec{r_{21}}|^3} \vec{r_{21}}$$

Yes, that's how I started the problem, but it gives me a wrong answer..

 

[dextercioby]

A force has a unit, Newtons, that's why I posted the N.It has a force in a direction i and j.

A SCALAR YES,but A VECTOR NO...
$$F_{x}=... [N]$$
and similar for F_{y} woud have been the correct formulas...

Daniel.

 

[mad]

A SCALAR YES,but A VECTOR NO...
$$F_{x}=... [N]$$
and similar for F_{y} woud have been the correct formulas...

Daniel.

I have been using vectors with units for over a year in my physics courses. Every teacher in the physics department teaches this. A force has components in i, j and k.. how can it not have units? But I don't want to start an argument here about this, that's what we use here. Thanks for pointing that out, but we always put units, vectors or not, in my college. So please let's go back to the question.. do you have any clues/tips on how to do it.. I've been trying for a long time now.

 

[dextercioby]

I'm asking you (for the second time) to post the original text of the problem...Else i cannot help,as i cannot lmake an idea about the charges' distribution and therefore i cannot compute any forces...

Daniel.

P.S.Ask a teacher at your school to measure (in Newtons) the vector $2\vec{i} + 3 \vec{j}$.

P.P.S.Take the table of units (SI-MKgs) and search for FORCE.Tell me if they put the vector on the "oblect"...

 

[quasar987]

Ok, my first post was a little reckless.

Here's a way to do it...

Draw a cartesian coordinate with the charge q at the center and draw also the vector force. Find the angle it makes with the x-axis using the sine law or something.

You know the charge 2,5Q will have to be put somewhere along the line along which the actual force act. So you can use the scalar version of coulombs law to find the distance r along this line where you must but the 2,5Q charge. And use the angle found previously to find the components. Sclalar version of coulomb's law:

$$|\vec{F}| = \frac{k(2,5Q)q}{r^2}$$

I hope that's clear enough.

 

[mad]

I'm asking you (for the second time) to post the original text of the problem...Else i cannot help,as i cannot lmake an idea about the charges' distribution and therefore i cannot compute any forces...

Daniel.

P.S.Ask a teacher at your school to measure (in Newtons) the vector $2\vec{i} + 3 \vec{j}$.

P.P.S.Take the table of units (SI-MKgs) and search for FORCE.Tell me if they put the vector on the "oblect"...

Sorry for the question, I posted the one they asked but forgot to add that they're ponctual charges. It's in french, that's why I translated it. But basically, the two charges create a resultant force of the one I wrote in my post. They then ask where to put a ponctual 2,5Q charge to nullfy the force applied on the charge. So that's why the force has to be the inverse of the other one. You can just take on from there, I just need to equal the 2 equations I posted to find the 2,5Q charge's position vector.

I'll ask the teacher, that is if he doesn't kill me because he's nuts about units.

Take a position vector R= (3i + 2j) m . It has to have units, or else what is 3 in i and 2 in j ?.. anyways, thanks

 

[MathStudent]

sorry to interfere, but for

$$|\vec{F}| = \frac{k(2,5Q)q}{r^2}$$

I was just wondering do the commas represent decimals?
if not, then what?

Thanks,
MS

 

[mad]

sorry to interfere, but for

$$|\vec{F}| = \frac{k(2,5Q)q}{r^2}$$

do the commas represent decimals?
if not, then what?

Thanks,
MS

yes, decimals, as in 2 and a half Q

 

[dextercioby]

sorry to interfere, but for

$$|\vec{F}| = \frac{k(2,5Q)q}{r^2}$$

do the commas represent decimals?
if not, then what?

Thanks,
MS

Yes in his and (incidentally on mine) side of the world the comma is a decimal and the decimal point indicates 3 consecutive zero-s in a number which are at the left of the comma...
$$1000\frac{3}{10} =1,000.3=1.000,3$$

Daniel.

 

[MathStudent]

Yes in his and (incidentally on mine) side of the world the comma is a decimal and the decimal point indicates 3 consecutive zero-s in a number which are at the left of the comma...
$$1000\frac{3}{10} =1,000.3=1.000,3$$

Daniel.

Thanks for the explanation. I should have figured it was just a crazy "Euro-thing"

 

[quasar987]

mad,

I'm guessing your textbook is by Harris Benson?

 

[mad]

mad,

I'm guessing your textbook is by Harris Benson?

Exactly :)

And I'm not in Europe ;) in Montreal , Canada

 

[quasar987]

That's where I'm from too. Actually South shore of montreal. I used to go to Cegep Edouard-Montpetit; where do you go?

 

[MathStudent]

Exactly :)

And I'm not in Europe ;) in Montreal , Canada

My bad,,, I should have said crazy "wannabe Euro-thing"
j/k

 

[mad]

That's where I'm from too. Actually South shore of montreal. I used to go to Cegep Edouard-Montpetit; where do you go?

Bois-de-Boulogne. What program are you in now? I want to enter UdeM in medecine

 

[quasar987]

Bois-de-Boulogne.. this rings a bell.. is there a physics teacher named Yves Charbonneau there?

I'm presently studying at UdeM in math-physics.

 

[mad]

Bois-de-Boulogne.. this rings a bell.. is there a physics teacher named Yves Charbonneau there?

I'm presently studying at UdeM in math-physics.

I think the only Charbonneau there is named Pierre Charbonneau..

 

[quasar987]

Oh.. he must work elsewhere then.

Good luck getting the grades to enter in medecine, and feel free to post some more if you encounter difficulties!

 

[mad]

Oh.. he must work elsewhere then.

Good luck getting the grades to enter in medecine, and feel free to post some more if you encounter difficulties!

I still haven't received an answer for this one :P . I'll check tomorrow with my teacher. 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. :) Thanks bro

 

[quasar987]

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.

 

[mad]

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. I am just making the vector into a unitvector and a magnitude.

 

[mad]

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 doesn't make sense. I hope you understand what I was trying to do.

 

[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.

 

[mad]

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 don't know why it never works like you have written.
Thanks a lot for all your time and help, Healey01 :)