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Solving for a variable in an equation that involves vectors

  1. Apr 9, 2014 #1
    1. The problem statement, all variables and given/known data
    I have the equation:
    [tex]\mathbf{F}_{2,1} = \frac{Q_1 Q_2}{4 \pi \varepsilon_0 {r_{2,1}}^2}\hat{r}_{2,1}[/tex] (standard electric force equation for 2 charges)

    I know the value of everything except Q2 and have to find it. The vectors each have 3 components.

    Normally in an algebraic equation, I would just solve for a variable by isolating it on one side of the = sign. But this equation involves vectors and I don't think there is a way to divide vectors. I could also subtract F2,1 from both sides which at least gets everything onto one side but I am still left with the vectors. I will effectively have three equations (one for each component of the vectors), but only one variable I have to solve for, right? Does the nature of this problem (it is physical) make it so that only certain vectors are even possible, so if I were to tweak F2,1 and not change the value of anything else, the problem would become impossible to solve because it would represent a physical impossibility?



    2. Relevant equations
    Nothing really


    3. The attempt at a solution
    I used only the x component of each vector and solved it using that equation and got the correct answer. But like I said before, if I tweaked only the y component of F2,1, the equation that used only the x components would be the same (so I'd get the same result) but the entire vector equation would not.
     
  2. jcsd
  3. Apr 9, 2014 #2

    ehild

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    r212 is not vector, but scalar. It is possible to divide with it. F2,1 is parallel with the unit vector ##\hat{r}_{2,1}## , a scalar multiple of it.
    What were the data? You should determine r212 from the coordinates.



    ehild
     
    Last edited: Apr 9, 2014
  4. Apr 9, 2014 #3
    That's true, but [itex]\hat{r}_{2,1}[/itex] is a vector, and I would want that to be on the other side as well. I just realized what I tried to do is impossible, because it would result in a scalar on one side of the = sign, and whatever you would (imaginatively) get if you divided two vectors, on the other side.

    I'm starting to believe that any problem like this would be set up very specifically, and that only certain values will work at all, allowing me to just use one component of the vectors.

    Edit: oops, I see you were still writing when I posted this.. sorry for jumping the gun so quickly, haha :P

    Edit 2:
    Oh dang, you're right. So you can't just tweak one component of F2,1.. doing that would prevent them from being parallel and would cause all kinds of issues. This is just a simple equation of a vector being equal to a magnitude times a direction unit vector. Sorry, it's been a little while since I've really worked with vectors in such a way, so I've been spending a lot of time on issues like this. Thanks again, very much!
     
    Last edited: Apr 9, 2014
  5. Apr 9, 2014 #4

    ehild

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    You can not put ##\hat{r}_{2,1}## on the other side. But you know that ##\vec{F}_{2,1}## is a scalar multiple of ##\hat{r}_{2,1}##, and that holds for all corresponding coordinates.
    It would be better to see the whole text of the problem.

    ehild
     
  6. Apr 9, 2014 #5
    Well, I have my question answered, but to be complete, here is the text of the problem. It's from an old Schaum's Outlines book from the early 1980's.

    Point charge Q1 = 300 µC, located at (1, −1, −3) m, experiences a force F2,1 = (8, −8, 4) N due to point charge Q2 at (3, −3, −2) m. Determine Q2.
     
  7. Apr 9, 2014 #6
    What you do is dot both sides of the equation by [itex]\hat{r}_{2,1}[/itex]. This will give you the scalar equation you desire. Do you know how to determine [itex]\hat{r}_{2,1}[/itex]?

    Chet
     
  8. Apr 9, 2014 #7
    Yep, that unit vector is just the vector r2,1 divided by its magnitude ([itex] \hat{r}_{2,1} = \frac{1}{r_{2,1}}\mathbf{r}_{2,1}[/itex], where [itex]r_{2,1} = ||\mathbf{r}_{2,1}|| = \sqrt{{r_{{2,1}_x}}^2 + {r_{{2,1}_y}}^2 + {r_{{2,1}_z}}^2}[/itex]). Sorry for the really weird notation inside the square root sign.

    That's a really neat trick, thanks for showing it to me. I never thought to use scalar products to get rid of the vectors, and of course the scalar product of any unit vector with itself is 1.
     
  9. Apr 10, 2014 #8

    ehild

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    The point charge Q1 experiences force from Q2:

    [tex]\vec F = k \frac {Q_1 Q_2}{(\vec r_1-\vec r_2)^2} \hat r_{12} [/tex]

    where ##\hat r_{12}=\frac{\vec r_1-\vec r_2} {|\vec r_1-\vec r_2|}##

    In your problem, r12=(1, −1, −3)-(3, −3, −2)=(-2, 2, -1). The magnitude is 3, so the components of ##\hat r_{12} ## are (-2/3, 2/3, -1/3)

    The force is F= (8, −8, 4) N. You see that the force is parallel to ##\hat r_{12} ##.

    You can write out the Coulomb Law in x,y,z components:

    [tex]F_x=8=\frac {kQ_1Q_2}{r_{12}^2}(-2/3)[/tex]
    [tex]F_y=-8=\frac {kQ_1Q_2}{r_{12}^2}(2/3)[/tex]
    [tex]F_z=4=\frac {kQ_1Q_2}{r_{12}^2}(-1/3)[/tex]

    You see that you get the same value for ##kQ1Q2/ {r_{12}^2}## from each equation, it is -12. kQ1Q2/9=-12. That is a scalar equation already.

    ehild
     
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