Coulomb's law in three dimensions (xyz). HELP

AI Thread Summary
The discussion revolves around calculating the electric force exerted on charge q1 by charge q2 using Coulomb's law in three dimensions. The user initially struggles to express the force in vector form after calculating the distance between the charges. Guidance is provided to use the unit vector derived from the position difference between the charges, which simplifies the calculation. Ultimately, the correct vector form of the force is derived as F12 = (-3.05i - 1.02j + 4.07k) * 10^-3 N. The user acknowledges a misunderstanding of vector notation but successfully arrives at the correct answer with assistance.
jezejeze
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Homework Statement


A charge q1 = 5uC is placed at (x,y,z):(1,2,-1).
A charge q2 = -2uC is placed at (x2,y2,z2):(-2,1,3).

Determine the electric force (vectorized form) exerted on q1.

Homework Equations



F12= kQ1Q2 / r2

The Attempt at a Solution



At first I tried finding the distance between both charges:

d = sqrt{(x2-x)2+(y2-y)2+(z2-z)2 } = sqrt{26}

I then plugged that into Coulomb's formula, and found:

F12= 5,19*10-3 N

The thing is that I am not able to get the answer into vector form...

I tried using trig:

F1X= F12 cos THETA
F1Y= F12 sin THETA
F1Z= F12 * ?

I tried drawing it.. but it's a pain in 3D.

I logically know that the force is attractive. I know that the x displacement is 3 units, and the y displacement is 1 unit, and the z displacement is 4 units.

But i still can't give my answer in the form of:
F12 = (something \vec{i} +/- something \vec{j} +/- something \vec{k}) N.

If anyone could show me the light, it would be greatly appreciated!

Thanks!
 
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The full vector form of the law is
\displaystyle \vec{F} =\frac{KQ_1Q_2 ( \vec{x_1}-\vec{x_2})}{|\vec{x_1}-\vec{x_2}|^3}
where |\vec{x_1}-\vec{x_2}| is the distance between the points \vec{x_1} and \vec{x_1}.

RGV
 
Last edited:
Try this simple math:
Force in x direction = kq1q2/x^2 where x is separation in x direction. Similarly for y & z.
 
ashishsinghal said:
Try this simple math:
Force in x direction = kq1q2/x^2 where x is separation in x direction. Similarly for y & z.

:eek: What happens when, say, Δx is zero (and Δy and/or Δz are not)?

I'm afraid that the vector can't be decomposed that way. Although it might be nice if it could! :smile:
 
Oh Sorry, my mistake:cry:
Thanks gneill
 
So all I need to do is plug in the numbers in the formula .. and that's it ?! No angles and vector stuff... !

Thanks alot!

Ray Vickson said:
The full vector form of the law is
\displaystyle \vec{F} =\frac{KQ_1Q_2 ( \vec{x_1}-\vec{x_2})}{|\vec{x_1}-\vec{x_2}|^3}
where |\vec{x_1}-\vec{x_2}| is the distance between the points \vec{x_1} and \vec{x_1}.

RGV
 
jezejeze said:
So all I need to do is plug in the numbers in the formula .. and that's it ?! No angles and vector stuff... !

Thanks alot!

Well, the "vector stuff" is still there, but it's in a form that requires very little in the way of puzzling out directions, angles, components, and so forth. It's plug-and-play physics!:wink:
 
well after a while of trying with the formula stated by RGV in an earlier post, and still not coming to the correct answer.. ( I guess I was doing something wrong.. with a P&P formula..!).. I found a much more logical way!

Just need to use the distance between q1 and q2 which is sqrt{26}, use it in Coulomb's simple formula, and then multiply by the unit vector

vec{Q2Q1}:( 3/sqrt{26}{i} + 1/sqrt{26}{j} -4/sqrt{26}{k})

and then voilà, you get:

\vec{F} 12=(-3,05{i} - 1,02{j}+ 4,07{k}) * 10-3 N

Anyways, thanks for everyone's help!
 
Last edited:
jezejeze said:
well after a while of trying with the formula stated by RGV in an earlier post, and still not coming to the correct answer.. ( I guess I was doing something wrong.. with a P&P formula..!).. I found a much more logical way!

Just need to use the distance between q1 and q2 which is sqrt{26}, use it in Coulomb's simple formula, and then multiply by the unit vector

\vec{Q2Q1}:( 3\vec{i}+1\vec{j}-4\vec{k})

and then voilà, you get:

\vec{F} 12=(3,05 ; 1,02 ; 4,07) * 10-3 N

Anyways, thanks for everyone's help!

You're attempting to carry out what Ray's formula prescribes, but doing it in two parts. You've found the distance (well, the square of the distance): |x1 - x2|2 = 26m2 and used that in a scalar version of Coulomb's Law. Then you took that as a magnitude and (wanted to) multiply by a unit vector in the appropriate direction. But something looks fishy with your unit vector.

It should be:

u = \frac{ \vec{x_1} - \vec{x_2}}{|\vec{x_1} - \vec{x_2}|}

and that should yield something like (0.588, .196, -0.784)m as a unit vector.

Rewriting the overall calculation and making these "tasks" more obvious,

\vec{F} = K \frac{Q1 Q2}{|\vec{x_1} - \vec{x_2}|^2} \cdot \frac{ \vec{x_1} - \vec{x_2}}{|\vec{x_1} - \vec{x_2}|}

You should recheck your unit vector math.
 
  • #10
yeah

||Q2Q1|| = sqrt {32+12+(-4)2 } = sqrt{26}

so the unit vector is:

3/sqrt{26} = 0.588i
1/sqrt{26} = 0.196j
-4/sqrt{26}= -0.784k

I just misinterpreted the meaning of the '' | '' in the formula as being absolute values instead of the '' || '' of being the ''norme'', or magnitude you call it in English.

(I'm from Quebec, Canada, so everything is in french, and for us magnitude is written:

|| x1 - x2 ||, and absolute value is | x1 - x2 |. )

But anyways the answer coincides, so everything is a-ok! Thanks. And yes my vector math sucks!
 
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