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Applications of Coulomb's Law

  1. Feb 3, 2016 #1

    CARNiVORE

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    1. The problem statement, all variables and given/known data
    In the figure particles 1 and 2 of charge q1 = q2 = +24.00 × 10-19 C are on a y axis at distance d = 18.0 cm from the origin. Particle 3 of charge q3 = +17.60 × 10-19 C is moved gradually along the x axis from x = 0 to x = +6.47 m. At what values of x will the magnitude of the electrostatic force on the third particle from the other two particles be (a) minimum and (b) maximum? What are the (c) minimum and (d) maximum magnitudes?

    2. Relevant equations
    E field strength at position r = kq/r^2
    E force = kq1q2/r^2

    3. The attempt at a solution
    I got the minimum, so (a) is complete - it's just 0 cm. I would like to start with (b).

    I am confused about (b). I imagine that the closer you get to 0 cm, the more powerful the force would be on q3, since the distance would be minimal, but not equal to 0 cm, since 0 cm is the location of the minimum. However, I think this is incorrect. Do I have to find the maximum using a derivative? If so, could you guys help me get started with this?

    Thanks so much in advance!
     
  2. jcsd
  3. Feb 3, 2016 #2

    gneill

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    The force is a vector quantity. You found a minimum at x = 0 because the forces from both q1 and q2 on q3 cancelled each other there. You'll need to write an expression for the magnitude of the net force at any location along the x-axis and, as you suggested, solve for a maximum (yes, differentiation is involved). Pay attention to the symmetry of the problem.
     
  4. Feb 3, 2016 #3

    CARNiVORE

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    Thanks for your response.
    I've determined that the net force on the third particle should be equal to the sum of the force of q1 on q3 and the force of q2 on q3. Thusly,

    ΣF = F13 + F23

    Then:

    ΣF = kq1q3/r13^2 + kq2q3/r23^2

    Where q1=q2=q and r13=r23. I then took the derivative with respect to r:

    dΣF/dr = -2kq3q/r^3 - 2kq3q/r^3
    dΣF/dr = -4kq3q/r^3

    From here, I set this function equal to zero in order to find the maximum. However:

    0 = -4kq3q/r^3

    There is no solution to this for r, so I believe I've made an error.
     
  5. Feb 3, 2016 #4

    gneill

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    The forces are vectors. They don't sum like scalars. You'll need to look at their components. Draw a diagram to see how symmetry can simplify the problem. (hint: some components cancel).
     
  6. Feb 3, 2016 #5

    CARNiVORE

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    My apologies; I meant to mention this in my last post:

    The y-components of F13 and F23 cancel each other out at all values of x. I actually do have a diagram:
    5f87e8cbfe15c76cdf9648db239ef475.png
    So, the only forces that are acting upon q3 are the x-components of F13 and F23. That means that I have to add a sinθ to the equation I mentioned before, right? But isn't θ changing in the same way that r is? If I'm right, that would mean that change must be accounted for in the derivative, but that obviously isn't possible.

    I feel like I'm missing something really obvious.
     
  7. Feb 3, 2016 #6

    gneill

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    Right, the y-components of the forces cancel. That leaves the x-components to deal with, implying the use of the cosine of the angle (not the sine) to select the component. Use similar triangles and known side lengths to form the cosine, thus avoiding the use of the angular form of the cosine.
     
  8. Feb 3, 2016 #7

    CARNiVORE

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    Another thing I forgot to mention is that even the x-component of each of these forces mirrors one another, so the problem is simplified even further.

    However, I feel like I'm going in circles, because I just made another attempt with an inconclusive result.

    ΣF = 2F13x

    because F13x=F23x, and

    ΣF = (2kq3q/r^2) * cosθ

    One thing that's giving me a bit of confusion is θ. I should be using this angle value, correct?
    ff341e222a4f5a7c2dce78ca9e1d8d55.png
     
  9. Feb 3, 2016 #8

    gneill

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    Yes. It's the same magnitude of angle that the force vectors make with the x-axis.

    upload_2016-2-3_23-19-40.png
    Lots of triangles and symmetry to take advantage of (and keep in mind that the whole image can be reflected about the y-axis, if the third charge is moved to the left of the origin).
     
  10. Feb 3, 2016 #9

    CARNiVORE

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    All right, great. So, I left off here:

    ΣF = (2kq3q/r^2) * cosθ

    cosθ is equal to adjacent over hypotenuse, so this means that cosθ=x/r, where x is the value we want to find, and r is the distance from q1 to q3. So,

    ΣF = (2kq3q/r^2) * (x/r) = 2kq3qx/r^3

    At this point, should I take the derivative with respect to r? In this case, x is also changing, so I think I'm missing a step.
     
  11. Feb 3, 2016 #10

    gneill

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    You don't want to deal with another variable. r is a function of x. What is r in terms of x? Think Pythagoras.
     
  12. Feb 3, 2016 #11

    CARNiVORE

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    Ah, that's right - by the Pythagorean Theorem, x2 = r2-d2. I don't think this makes things any simpler, though, so I believe I'm applying it incorrectly.

    Thank you so much for your time, by the way - this is taking way longer than I thought it would, and I really appreciate your assistance.
     
  13. Feb 3, 2016 #12

    gneill

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    Replace all occurrences of r accordingly. Write the cosine as a ratio of known sides.
    No problem. Once you've done one of these and found the tricks, the next ones will take much less time. The use of symmetry and recognizing the radius as a dependent variable are the key points.
     
  14. Feb 4, 2016 #13

    CARNiVORE

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    Hey, sorry for the late response - I decided yesterday night that I should take a break from this assignment and go to sleep. It's not due until Sunday, though, so I'm fine.

    Anyway, picking up where we left off:

    I substituted √x^2+d^2 for r, which left me with ΣF=2kq3qx/(x^2+d^2)3/2.

    I applied the quotient rule to this, which left me with what I can only describe as an abhorrent conflagration of letters and numbers. However, it seemed like I was on the right track, because I ended up with (√2)d = x when I substituted 0 for dΣF. However, this result of 25.something cm was apparently incorrect.

    At this point, I'm very frustrated, but honestly used to it. Do you happen to see an error in my reasoning, or should I show you more of my in-depth work?
     
  15. Feb 4, 2016 #14

    gneill

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    Your force expression looks good. Yes, differentiating it is not pretty. When you apply the quotient rule remember that you're going to be equating the result to zero, So only the numerator matters. Show what you've got so far.
     
  16. Feb 5, 2016 #15

    CARNiVORE

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    Woah, I did it! I got the correct answer, where x = d/√2 = 12.7279. I just had to check my math in the ugly differentiation - careless mistake.

    Part (c) asks what the minimum magnitude of electrostatic force is. Obviously, the answer is 0 N at x = 0 cm.

    Part (d) asks what the maximum magnitude is. Plugging x=12.7279 into the equation for net force, ΣF=2kq3qx/(x^2+d^2)3/2, gave me 6.0223E-25 as the maximum value. This was also correct.

    It was a pretty long journey (or at least, it felt like it), but I managed to pull through thanks to your assistance. I greatly appreciate your help, and I really can't stress that enough. Thanks, gneill!
     
  17. Feb 5, 2016 #16

    gneill

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    Well done! And your very welcome.
     
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