Ashley1nOnly said:
F=(KQ1Q2)/r^2 • (r-hat)
Why are we multiplying by r-hat?
I understand that the equation helps us calculate the force between two charges.
A hat sign above a variable indicates a unit vector: the magnitude of the vector is equal to 1.
In this case, ##\hat r = \vec r/|\vec r|## ... ie, ##\hat r## points in the direction of ##\vec r## but has length 1.
The cartesian unit vectors i, j, k, should really be ##\hat\imath##, ##\hat\jmath## and ##\hat k## and can also be called ##\hat x, \hat y, \hat z##.
Your text uses bold-face to represent a general vector so ##|{\bf v}| = |\vec v| = v## and the magnitude can be any real number.
The point of using it in equations like above is to indicate the preferred positive direction.
However, as you wrote it, the equation is misleading ... it does not properly define the vector ##\vec r##.
usually we would want to put ##q_1## at position ##\vec r_1## and ##q_2## at position ##\vec r_2## and say that the force on charge 2 due to charge 1 is:
$$\vec F_{21} = \frac{kq_1q_2}{|\vec r_{21}|^3}\vec r_{21}: \vec r_{ij} = \vec r_i - \vec r_j$$ ... notice the cube in the denominator?
If you use ##\hat r_{21} = \vec r_{21}/|\vec r_{21}|## you get a unit vector pointing from 1 to 2 (check I got that right.)
The book says that r=|r(bold)|
##r = |{\bf r}|## ... right: the bold-face indicates an arbitrary vector type, we like to use an arrow over the top of the letter.
I know that this is the magnitude sqrt(rx^2+ry^2+rz^2)
##r = \sqrt{r_x^2+r_y^2+r_z^2}## yes, well done.
That is because ##\vec r = r_x\hat\imath + r_y\hat\jmath + r_z\hat k##
rx= r with respect to the X position and so on.
But we can neglect z since the charges are not in the k direction
... not strictly correct: but you can anticipate the result to make your maths easier to do.
Since there are two charges, and you want to know the force on charge 2 due to the presence of charge 1, it makes sense to choose your coordinates system so the origin is on charge 1, and the x-axis goes through charge 2 (unless you get told to do it otherwise) ... then ##\vec r_1=(0,0,0)## and ##\vec r_2 = (x,0,0) = x\hat\imath##
Put that into the more complete equation I gave you above and see what happens.
(notice, in this case the x direction and the ##\vec r_2## direction are the same direction, so ##\hat r = \hat\imath## and ##r=x##)
[/quote]Because it says so ... that bit of math is telling you that the label "r" is going to be used to represent the magnitude of the vector ##\bf r##.