Why is the separation vector this (electrostatics):

[grandpa2390]

Homework Statement

As always, I wish I were allowed to upload a drawing ;)

This is not a problem, this is a derivation of sorts, so I don't have any numbers.

I have 3 points. one at the origin, a point charge q and a test charge Q

I have a vector from the origin to the point charge q with a magnitude of r' (there was a button for tutorial on typing math, I don't see it).
I have a vector from the origin to the test charge Q with a magnitude of r

the magnitude of the separation vector (vector connecting the point charge with the test charge) is given as $r - r'$

Homework Equations

$π = r-r'$
script r that is the vector connecting q and Q

The Attempt at a Solution

it isn't a problem, it is just a question. I would think t would be some trig function and we would need angles and so forth to calculate this magnitude, but it is just $r-r'$. If the angle between r and r' were 90 degree (not possible in this case, but I am just giving an example), I would expect the magnitude of π to be the square root of $\sqrt{r^2+r'^2}$

once again I am sorry for not using the math code, I can't find that button that gave the tutorial. I know a link to it is somewhere on this page (or used to be) but it is eluding me at the moment.

edit I just found the link to the LaTex tutorial (of course I would find it after posting), I am editing it right now. I knew it was right there, but perhaps it wasn't while I was creating the thread.

 

[Doc Al]

I wish I were allowed to upload a drawing

You can certainly upload a drawing.

it isn't a problem, it is just a question. I would think t would be some trig function and we would need angles and so forth to calculate this magnitude, but it is just $r-r'$. If the angle between r and r' were 90 degree (not possible in this case, but I am just giving an example), I would expect the magnitude of π to be the square root of $\sqrt{r^2+r'^2}$

OK.

Still not sure what the issue is. Realize that the separation is a vector $\vec{r} - \vec{r}'$.

 

[Doc Al]

The separation, which I'll call $\vec{s}$, is a vector that goes from q to Q. Thus: $\vec{r}' + \vec{s} = \vec{r}$.

Make sense?

 

[fresh_42]

The distances from the origin to the charges are the vectors $\vec{r}'$ and $\vec{r}$. As they start at the origin, their lengths are $r'=|\vec{r}'|$ and $r=|\vec{r}|$ which you denoted by the same letter and might therefore be a source of confusion. Now $\vec{\pi}= \vec{r}'-\vec{r}$ is also a vector. Its length is $|\vec{\pi}|=\sqrt{({r_1}'-{r_1})^2+({r_2}'-{r_2})^2}$ if the (planar) coordinates are $r_i$ and ${r'}_i$. The angle in between doesn't play a role here, it is simply a triangle. The fact they are charges start to play a role, if they are moving for then the electric field comes into play.

 

[grandpa2390]

The separation, which I'll call $\vec{s}$, is a vector that goes from q to Q. Thus: $\vec{r}' + \vec{s} = \vec{r}$.

Make sense?

I apologize, I know this is a dumb question (I said it, so no worries about the "there are no dumb questions bit"). But that doesn't make sense to me. As I said, in my mind that is like saying that on a right triangle, leg a but the hypotenuse, c, is equal to leg b. Obviously, I realize these two are not related, I just don't see how they aren't related.

The distances from the origin to the charges are the vectors $\vec{r}'$ and $\vec{r}$. As they start at the origin, their lengths are $r'=|\vec{r}'|$ and $r=|\vec{r}|$ which you denoted by the same letter and might therefore be a source of confusion. Now $\vec{\pi}= \vec{r}'-\vec{r}$ is also a vector. Its length is $|\vec{\pi}|=\sqrt{({r_1}'-{r_1})^2+({r_2}'-{r_2})^2}$ if the (planar) coordinates are $r_i$ and ${r'}_i$. The angle in between doesn't play a role here, it is simply a triangle. The fact they are charges start to play a role, if they are moving for then the electric field comes into play.

I am currently thinking about this. is the subscript 1 and 2 the x and y components of the vectors?

 

[grandpa2390]

The distances from the origin to the charges are the vectors $\vec{r}'$ and $\vec{r}$. As they start at the origin, their lengths are $r'=|\vec{r}'|$ and $r=|\vec{r}|$ which you denoted by the same letter and might therefore be a source of confusion. Now $\vec{\pi}= \vec{r}'-\vec{r}$ is also a vector. Its length is $|\vec{\pi}|=\sqrt{({r_1}'-{r_1})^2+({r_2}'-{r_2})^2}$ if the (planar) coordinates are $r_i$ and ${r'}_i$. The angle in between doesn't play a role here, it is simply a triangle. The fact they are charges start to play a role, if they are moving for then the electric field comes into play.

I think I have it. This is Physics one stuff. It has just been awhile.

 

[grandpa2390]

@fresh_42 @Doc Al

I get the $|\vec{\pi}|=\sqrt{({r_1}'-{r_1})^2+({r_2}'-{r_2})^2}$
but I don't see how it simplified to that formula. :(

I guess I could just memorize it. but that is not me, it bothers me... I get this. It is what I expect. But it then makes me ask what the separation vector $π$ is.
I just realized that the notation in the book (and from the professor) confused me. bold π is the notation for the vector. π not in bold is the magnitude. and $\hat π$ is the direction.

Once again, dumb question since I should know the answer to by now, but it has been awhile.
so what does π mean? It is the obviously one vector minus the other. a vector is a magnitude and direction. I don't know. It is just weird to me at the moment. Like it is just a number that represents the vector, and to get more useful information, you have to have other info to extract magnitude/direction from it.

 

[fresh_42]

$\vec{\pi} = \vec{r}' - \vec{r}$ which was your definition above. Now let $\vec{r}'=({r_1}',{r_2}',\ldots ,{r_n}')$ and $\vec{r}=({r_1},{r_2},\ldots ,{r_n})$ be the respective coordinates. Then $\vec{\pi}=({r_1}'-r_1,{r_2}'-r_2,\ldots ,{r_n}'-r_n)$ and its length is
$$ |\vec{\pi}|=|({r_1}'-r_1,{r_2}'-r_2,\ldots ,{r_n}'-r_n)| = \sqrt{({r_1}'-r_1)^2+({r_2}'-r_2)^2+ \ldots +({r_n}'-r_n)^2} $$
For a plane with $n=2$ you get Pythagoras. $\vec{\pi}$ is the vector between the two charges and you can move it into the origin without changing its length or direction. If you do so, then you get the formula above where $\vec{\pi}$ is the hypotenuse of a right triangle and its catheti are of length $|{r_1}'-r_1|$ and $|{r_2}'-r_2|$.

 

[grandpa2390]

$\vec{\pi} = \vec{r}' - \vec{r}$ which was your definition above. Now let $\vec{r}'=({r_1}',{r_2}',\ldots ,{r_n}')$ and $\vec{r}=({r_1},{r_2},\ldots ,{r_n})$ be the respective coordinates. Then $\vec{\pi}=({r_1}'-r_1,{r_2}'-r_2,\ldots ,{r_n}'-r_n)$ and its length is
$$ |\vec{\pi}|=|({r_1}'-r_1,{r_2}'-r_2,\ldots ,{r_n}'-r_n)| = \sqrt{({r_1}'-r_1)^2+({r_2}'-r_2)^2+ \ldots +({r_n}'-r_n)^2} $$
For a plane with $n=2$ you get Pythagoras. $\vec{\pi}$ is the vector between the two charges and you can move it into the origin without changing its length or direction. If you do so, then you get the formula above where $\vec{\pi}$ is the hypotenuse of a right triangle and its catheti are of length $|{r_1}'-r_1|$ and $|{r_2}'-r_2|$.

I get that now. I was wondering what a vector is besides magnitude and direction. all I remember is coordinates (or is it the magnitudes) of each component in angle brackets. $\langle 5, 4, 4 \rangle$ and then I know how to use that with calculus formulas to find other things.edit: a vector is written as an equation with components. doh that thing above is the magnitude of the vector (right?)

edit: a vector is written with two separate parts, a direction and a magnitude? an equation with different components and the magnitude in components.

the equation (or some other notation) would give the direction, and then after that you would write that notation (or an angle, or any other way) to notate the magnitude.