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Tricky Electric Charge Problem

  1. Sep 22, 2007 #1
    1. The problem statement, all variables and given/known data

    38. Particles 2 and 4, of charge [itex]-e[/itex], are fixed in place on the y axis, at [itex]{y}_{2}[/itex] = -10.0 cm and [itex]{y}_{4}[/itex] = 5.00 cm. Particles 1 and 3, of charge [itex]-e[/itex], are placed on the x axis and can be moved along the x axis. Particle 5, of charge [itex]+e[/itex], is fixed at the origin. Initially particle 1 is at [itex]{x}_{1}[/itex] = -10.0 cm and particle 3 at [itex]{x}_{3}[/itex] = 10.0 cm.
    (a) To what x value must particle 1 be moved to rotate the direction of the net electrostatic force [itex]\vec{F}_{net}[/itex] on particle 5 by 30 degrees counterclockwise?
    (b) With particle 1 fixed at its new position, to what x value must you move particle 3 to rotate [itex]\vec{F}_{net}[/itex] back to its original direction?

    2. Relevant equations

    Coulomb's Law

    Vector Form:

    [tex]
    \vec{F}_{12} = \frac{k_{e}q_{1}q_{2}}{{r_{12}}^2}\hat{r}_{21}
    [/tex]

    Scalar Form:

    [tex]
    {F}_{12} = \frac{k_{e}{q_{1}}{q_{2}}}{{r_{12}}^2}
    [/tex]

    Magnitude Form:

    [tex]
    |\vec{F}_{12}| = \frac{k_{e}|q_{1}||q_{2}|}{{r_{12}}^2}
    [/tex]

    Superposition of Forces

    [tex]
    \sum{\vec{F}_{m}} = {\sum_{i=1}^{n}}{\vec{F}_{mn}}
    [/tex]

    3. The attempt at a solution

    So, there are five particles: [itex]{q}_{1}[/itex] and [itex]{q}_{3}[/itex] placed on the x-axis and [itex]{q}_{2}[/itex] and [itex]{q}_{4}[/itex] placed on the y-axis and [itex]{q}_{5}[/itex] placed at the origin.

    I note that there are two unknowns: the net force on particle 5, [itex]\sum\vec{F}_{5}[/itex] and the distance, [itex]{r}_{51}[/itex].

    Let,

    [tex]
    {r}_{51} = x
    [/tex]

    What I did was setup the [itex]x[/itex] and [itex]y[/itex] components for the net force and then solve each of the them for the (magnitude of the) net force and set them equal to each other and then solved for x.

    [tex]
    \sum\vec{F}_{5_{x}} = {{F}_{{51}_{x}}}{\hat{i}} + {{F}_{{53}_{x}}}{\hat{i}}
    [/tex]

    [tex]
    {\sum}{\vec{F}_{5_{x}}} = {{F}_{51}}{cos{\theta_{51}}}{\hat{i}} + {{F}_{53}}{cos{\theta_{53}}}{\hat{i}}
    [/tex]

    [tex]
    {\sum}{\vec{F}_{5_{x}}} = {\frac{{k_{e}}{q_{5}}{q_{1}}}{{{r}_{51}}^{2}}}{cos{\theta_{51}}}{\hat{i}} + {\frac{{k_{e}}{q_{5}}{q_{3}}}{{{r}_{53}}^{2}}}{cos{\theta_{53}}}{\hat{i}}
    [/tex]

    [tex]
    {\sum}{{F}_{5}}{cos{\theta}}{\hat{i}} = {\frac{{k_{e}}{q_{5}}{q_{1}}}{{{r}_{51}}^{2}}}{cos{\theta_{51}}}{\hat{i}} + {\frac{{k_{e}}{q_{5}}{q_{3}}}{{{r}_{53}}^{2}}}{cos{\theta_{53}}}{\hat{i}}
    [/tex]

    [itex]{\theta}[/itex] = 30 degrees

    [tex]
    {\sum}{{F}_{5}} = {\frac{{{k}_{e}}{{q}_{5}}}{{cos}{\theta}}}\left[{\frac{{q_{1}}}{{{r}_{51}}^{2}}}{cos{\theta_{51}}} + {\frac{{q_{3}}}{{{r}_{53}}^{2}}}{cos{\theta_{53}}}\right]
    [/tex]

    [tex]
    {\sum}{{F}_{5}} = {\frac{{{k}_{e}}{{q}_{5}}}{{cos}{\theta}}}\left[{\frac{{q_{1}}}{{\left(x\right)}^{2}}}{cos{\theta_{51}}} + {\frac{{q_{3}}}{{{r}_{53}}^{2}}}{cos{\theta_{53}}}\right]
    [/tex]

    [tex]
    {\sum}{{F}_{5}} = {\frac{{{k}_{e}}{{q}_{5}}}{{cos}{\theta}}}\left[{\frac{{q_{1}}}{{x}^{2}}}{cos{\theta_{51}}} + {\frac{{q_{3}}}{{{r}_{53}}^{2}}}{cos{\theta_{53}}}\right]
    [/tex]

    Doing the same for the [itex]y[/itex] components leads to,

    [tex]
    \sum\vec{F}_{5_{y}} = {{F}_{{52}_{y}}}{\hat{j}} + {{F}_{{53}_{4}}}{\hat{j}}
    [/tex]

    [tex]
    {\sum}{{F}_{5}}{sin{\theta}}{\hat{j}} = {\frac{{k_{e}}{q_{5}}{q_{2}}}{{{r}_{52}}^{2}}}{sin{\theta_{52}}}{\hat{j}} + {\frac{{k_{e}}{q_{5}}{q_{4}}}{{{r}_{54}}^{2}}}{sin{\theta_{54}}}{\hat{j}}
    [/tex]

    [itex]{\theta}[/itex] = 30 degrees

    [tex]
    {\sum}{{F}_{5}} = {\frac{{{k}_{e}}{{q}_{5}}}{{sin}{\theta}}}\left[{\frac{{q_{2}}}{{{r}_{52}}^{2}}}{sin{\theta_{52}}} + {\frac{{q_{4}}}{{{r}_{54}}^{2}}}{sin{\theta_{54}}}\right]
    [/tex]

    Letting,

    [tex]
    {\sum}{{F}_{5}} = {\sum}{{F}_{5}}
    [/tex]

    [tex]
    \left({\frac{{{k}_{e}}{{q}_{5}}}{{cos}{\theta}}}\left[{\frac{{q_{1}}}{{x}^{2}}}{cos{\theta_{51}}} + {\frac{{q_{3}}}{{{r}_{53}}^{2}}}{cos{\theta_{53}}}\right]\right) = \left({\frac{{{k}_{e}}{{q}_{5}}}{{sin}{\theta}}}\left[{\frac{{q_{2}}}{{{r}_{52}}^{2}}}{sin{\theta_{52}}} + {\frac{{q_{4}}}{{{r}_{54}}^{2}}}{sin{\theta_{54}}}\right]\right)
    [/tex]

    Solving for x and simplifying leads to,

    [tex]
    {x} = {\sqrt{\frac{{{q}_{1}}}{{\frac{{cot}{\theta}}{{cos}{\theta}_{51}}\left[\frac{{q}_{2}}{{{r}_{52}}^{2}}{{sin}{\theta}_{52}} + \frac{{q}_{4}}{{{r}_{54}}^{2}}{{sin}{\theta}_{54}}\right] - {\frac{{q}_{3}}{{{r}_{53}}^{2}}{{cos}{\theta}_{53}}}}}}}
    [/tex]

    sig. fig. [itex]\equiv[/itex] 3

    Where,

    [itex]{\theta}_{}[/itex] = 30 degrees
    [itex]{\theta}_{51}[/itex] = 180 degrees
    [itex]{\theta}_{52}[/itex] = 270 degrees
    [itex]{\theta}_{53}[/itex] = 0 degrees
    [itex]{\theta}_{54}[/itex] = 90 degrees

    Using the above equation I get,

    x = [itex]\sqrt{-0.0016139...}[/itex]

    Which is ofcourse wrong because it is imaginary.

    So, how do I solve this because I thought for sure my approach was right and I don't think I made any mistakes.

    Any help is appreciated.

    Thanks,

    -PFStudent
     
    Last edited: Sep 22, 2007
  2. jcsd
  3. Sep 22, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Your approach looks OK, but your overly and unnecessarily formal approach makes checking your work rather painful.
    Why not just write this last as:

    [tex]{\sum}{{F}_{5}}{cos{\theta}}{\hat{i}} = (-\frac{ke^2}{x^2} + \frac{ke^2}{{{r}_{53}}^2}) {\hat{i}}[/tex]

    And similarly for the y-component.
     
  4. Sep 23, 2007 #3

    dynamicsolo

    User Avatar
    Homework Helper

    I just love how a lot of the textbooks make a point of showing that formal vectorial definition for Coulomb's Law, as if any practicing physicist really solves electrostatic problems using it exactly as written. (It's good enough to feed to a computer, but that doesn't mean *you* have to eat it...)

    For these discrete charge problems, it pays to draw a picture first. Use the "opposites-attract, likes-repel" rule to figure out which way each of the force vectors on particle 5 (the one at the origin) points. Before even starting a calculation, what can you say about the components of Fnet(initial)?

    Now, all the surrounding charges are identical (-e) and the central charge is +e and all the distances are identical (10 cm) except y4 (which is half that). *Don't* compute a value for the magnitude of the forces: just call the force magnitude between, say, particle 5 and particle 2, F. What is the force magnitude between 4 and 5, *relative* to F?

    OK, now we're ready for part (a). To make that Fnet change, by moving particle 1, so that its direction changes by 30 degrees CCW, what would need to be true about the relative magnitudes of Fnet's components then? How would that be related to x1? (Which forces change and which don't?) That should give you a route to solving for the new position of particle 1.

    For part (b), we want Fnet to be returned to its original direction. What overall condition must apply to the configuration of the charges in order for that to happen? So where must particle 3 be moved to?

    Note that the problem doesn't ask you for the actual *values* of the forces. So much of the charge arrangement remains the same among the parts of the problem that it is simpler to deal with the questions by using a reference magnitude that *you* define. (You can always check your answers later by computing all the force components if you wish...)
     
    Last edited: Sep 23, 2007
  5. Sep 23, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    wrong angle

    I think I found the source of the problem--you are using the wrong value for theta:

    [tex]\tan\theta = F_y/F_x[/tex]

    where [itex]\theta[/itex] should be 90+30=120 degrees, not 30 degrees.
     
  6. Sep 23, 2007 #5
    Hey,

    Yea...while I do realize this when working through my physics problems, I also recognize that I should get the same answer taking the formal rigorous approach (though it may be unnecessary). Overall, I feel taking such an approach (the formal rigorous one) strengthens my ability to deal with increasingly complex problems as I will see when I take the next higher physics courses (Classical Mechanics and Classical Electricity & Magnetism).

    So, sorry about the symbolic messiness... :redface:

    Thanks, I will try that right now.

    However, just to be clear the first attempt, is it not (symbolically) correct?

    Hmm….I tried that as well and got the following answer,

    x = 0.0488173693… m

    However, in looking over the solution to this problem in the solution’s manual (since this is an even problem the solution is not in the back of the book) I believe this may be an error on the book’s part.

    Here is their answer,

    x = 6.05 cm

    Here is their solution,

    [​IMG]

    What I do not understand is why my approach is wrong, after looking over my work meticulously I just do not see any error in my approach.

    Is the solution's manual perhaps incorrect?

    In the mean time I will rework thoroughly my solution and simplify it.

    Thanks,

    -PFStudent
     
    Last edited: Sep 24, 2007
  7. Sep 23, 2007 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Well... no one will accuse you of being lazy, that's for sure! :smile:

    You made two errors: one algebraic, which I'll point out below, the other using the incorrect angle as I have already pointed out.



    Using the method that I described, which you should realize is exactly what you were doing, I get the same answer as your textbook. (So will you when you correct your errors.)

    In your final expression:
    That [itex]\cos\theta_{53}[/itex] should be [itex]\cos\theta_{53}/\cos\theta_{51}[/itex].
     
  8. Sep 24, 2007 #7
    lol, thanks! :smile:

    Ahh...I see, thanks for the correction!

    So my final solution is,

    [tex]
    {x} = {\sqrt{\frac{{{q}_{1}}}{{\frac{{cot}{\theta}}{{cos}{\theta}_{51}}\left[\frac{{q}_{2}}{{{r}_{52}}^{2}}{{sin}{\theta}_{52}} + \frac{{q}_{4}}{{{r}_{54}}^{2}}{{sin}{\theta}_{54}}\right]-\frac{{q}_{3}{cos}{\theta}_{53}}{{{{r}_{53}}^{2}}{cos}{\theta}_{51}}}}}}
    [/tex]

    Thanks for the help Doc Al!

    Thanks,

    -PFStudent
     
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