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What is the reason for r-hat?

  1. Sep 12, 2016 #1
    1. The problem statement, all variables and given/known data
    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.

    The book says that r=|r(bold)|
    I know that this is the magnitude sqrt(rx^2+ry^2+rz^2)
    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
    Why does r=|r(bold)|?

    If the charges has the same charge then it's just
    F=(KQ^2)/r^2 • (r-hat)
    Why are we multiplying by r-hat?


    Then I am also given another equation

    F=(KQ1Q2) •(x1-x2)/(|x1-x2|^3)

    Why did we get rid of r^2 and multiply by the unit vector ?


    2. Relevant equations
    None

    3. The attempt at a solution
    I just have question to better understand. Sorry if this is a stupid question.
     
  2. jcsd
  3. Sep 12, 2016 #2

    Simon Bridge

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    A hat sign above a variable indicates a unit vector: the magnatude 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.)

    ##r = |{\bf r}|## ... right: the bold-face indicates an arbitrary vector type, we like to use an arrow over the top of the letter.
    ##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##
    ... 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 magnatude of the vector ##\bf r##.
     
  4. Sep 12, 2016 #3
    This is an amazing explanation.
    Why was the bottom cubed?
    When doing Xi-Xj, how do you know which charge position to put first. I believe that if you are looking for the force in charge 2 with respect to charge one that Xi would be charge 1 and Xj would be charge 2

    If you are looking for the force on charge 2 with respect to charge 1 that Xi would be 2 and Xj would be 1. Is my thinking right?
     
  5. Sep 12, 2016 #4

    jtbell

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    Note the definition ##\hat r_{21} = \vec r_{21} / | \vec r_{21}|##. Substitute it into $$\vec F_{21} = \frac {k q_1 q_2} {|\vec r_{21}|^2} \hat r_{21}$$
     
  6. Sep 12, 2016 #5
    Got it. Thanks so much to you all.
     
  7. Sep 13, 2016 #6

    Simon Bridge

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    The maths is a language. Draw a picture and then describe the picture using the maths. I don't actually remember these equations, I just remembered that it was an inverse square law and like charges repel. If you end up with like charges moving towards each other, you did it wrong.

    ... well done :)


     
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