Converting Coulomb's force to vector form in 3D

[jezejeze]

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:

F_1X= F₁₂ cos THETA
F_1Y= F₁₂ sin THETA
F_1Z= 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!

 

[Ray Vickson]

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

 

[ashishsinghal]

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

 

[gneill]

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

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!

 

[ashishsinghal]

Oh Sorry, my mistake
Thanks gneill

 

[jezejeze]

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

Thanks a lot!

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

 

[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 a lot!

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!

 

[jezejeze]

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!

 

[gneill]

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.

 

[jezejeze]

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!