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Point Charge sliding on a semi-circle

  1. May 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Two point charges Q1 = 69.5 μC and Q2 = 19.9 μC are fixed on the two ends of the diameter of a semi-circle with a radius of r = 21.3 cm as shown in the figure.


    A massless point charge Q3 = 54.6 μC can slide on the semi-circle without any friction. At which value of the angle θ will the point charge Q3 be in equilibrum?

    2. Relevant equations
    Coulomb's Law : F = KQQ/R^2

    3. The attempt at a solution

    My understanding of this was to find the point at which there was a net Electrical field force of zero on Q that is on the semicircle. However, since all of these were positive charges, I was confused as to how this could stay on a semi-circular path. The attempt to cancel out the X-component Force Vectors (parallel to the diameter) I used the law of cosines to find the distance between Q3--Q1 and Q3--Q2. This distance was used to set the Forces equal to each other and hence have an equation with respect to theta.

    My final answer was theta = arccos((Q1^2-Q2^2)/(Q1^2+Q2^2). This was wrong.

    I am not looking for a direct answer, merely an explanation of the forces that are acting on Q3 and how they cancel to keep it in equilibrium.
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. May 26, 2010 #2

    Doc Al

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    Staff: Mentor

    Your image requires a log in and thus is not viewable.

    Since I can't see the diagram, I can only guess. I'd say that the net force on Q3 cannot have any component tangential to its path.
  4. May 26, 2010 #3
    I apologize for that, I forgot that my computer kept me logged in. Here is a new link:

    Essentially, it is two point charges on an x axis separated by a distance of 2R. A third point charge Q3 is located on a semicircular arc a distance of R away from the center of the two initial charges.
  5. May 26, 2010 #4


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    Try to find the location where the tangential component of the electric field on Q3 is zero (by tangential I mean tangential to the semicircle). :wink:

    As far as why there is equilibrium -- remember that the sliding charge is confined to the semicircle. It can't just fly off the semicircle all will-nilly. If it gets especially close to any one of the other point charges it is pushed away from it. But there should be one point in between (but still on the semicircle) where it can remain stationary. [btw., this problem makes a lot of assumptions with "massless" and "frictionless," but don't overthink them. Do something like you did before, but instead of trying to cancel out fields on the x-axis, try another some other axis (which might not be Cartesian) :wink:]

    [Edit: It appears Doc Al beat me to the advice. :smile:]
    Last edited: May 26, 2010
  6. May 26, 2010 #5

    Doc Al

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    OK, my comment remains. They can't mean the point where the electric force from Q1 and Q2 are zero, since they are both + charges. What I suspect they mean is that the track can exert a normal force (radial) on the Q3, but not a tangential force. Find the spot where the tangential component of the net force from Q1 and Q2 is zero.
  7. May 26, 2010 #6
    My latest attempt at solving this problem was to find the value on an x axis where the charge would feel no net force. I did this by setting (K*Q1*Q3)/(2R-x)^2=(K*Q2*Q3)/(x^2) and I got .148 meters. I then used the conversion to polar coordinates R*cos(theta)=x and I got 45.8 degrees. This is wrong however.

    Am I at least on the right track? I only have a couple tries left and I have done so many different approaches (except the right one) that I have basically done a review in trigonometry.
  8. May 26, 2010 #7

    Doc Al

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    Try this. For the position shown in the diagram, what's the tangential component of the force from each charge? First find the magnitude of the force, then the tangential component. (Express all distances needed in terms of r and theta.)
  9. May 26, 2010 #8
    After doing all of this before, I felt as though I was going in the wrong direction.

    I have two forces acting on Q3: Force of Q1 on Q3 and Force of Q2 on Q3.

    I know I need the tangential component which has always been easy to derive by multiply the force by some cos(angle). However, this angle is not given and therefore I derived one from the law of cosines along with the distance from Q1 and Q2 with respect to Q3. Here are the values I got:

    F1on3: (K*Q1*Q3)/(2*R^2(1+cos(theta)))
    F2on3: (K*Q2*Q3)/(2*R^2(1-cos(theta)))

    The distances in the denominator were derived from law of cosines.

    Now the tangential component is:

    F1on3*cos(alpha) : cos(alpha) = -(1+cos(theta))/SQRT(2(1+cos(theta)))
    F2on3*cos(beta) : cos(beta) = -(1-cos(theta))/SQRT(2(1-cos(theta)))

    I set F1on3*cos(alpha) = F2on3*cos(beta) and it simplifies to:

    cos(theta) = (Q1^2-Q2^2)/(Q1^2+Q2^2)

    However this is wrong.
  10. May 26, 2010 #9

    Doc Al

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    I don't understand how you got this result for cos(alpha). It should be simpler. Hint: In terms of theta, what is the angle that F1on3 makes with the the radius?
  11. May 26, 2010 #10


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    Seems alright to me so far...

    Is this a calculus based class?

    If so there may be another approach you can try instead of the above. I haven't actually gone through all the math for this problem yet, but using the electric potential approach should be easier than dealing with those awful unit vectors. This assumes you've learned about electric potential though.

    The strategy is to calculate the electric potential at the point where Q3 is. It depends on the charges of Q1 and Q2, and the scalar distances that you've already calculated above. But the great news is that this approach doesn't require any vectors -- meaning you don't have to convert those pesky unit vectors to tangential components -- i.e. you don't need to worry about tangents. :biggrin:

    Try to formulate an equation for the potential, V, at point Q3 (the actual charge on Q3 doesn't matter by the way [except for the fact that it's positive]). It should be a function of Q1, Q2, r, and θ. Then treating Q1, Q2, and r as constants, find the minimum of V as a function of θ (hint: take the partial derivative with respect to θ, and set it equal to zero). :wink:

    Again though, what I wrote above is just another possible approach. If you don't know calculus or electric potential this problem is still solvable. It just means you'll have to deal with those pesky unit vectors.

    [Edit: Okay, now I have worked out the math, and yes, the electric potential approach works nicely. I also did the problem your original way of trying to find the tangential electric field directly, but I had to use Wolfram Alpha to help with the trigonometry -- it was tough (keep in mind trig was never one of my favorite subjects, and I don't remember very many trig identities). If your class is calculus based, and if you have learned about electric potential, then I highly recommend finding the potential first, then converting back to the electric filed by finding the gradient (and setting it equal to zero to solve the problem). For me anyway, it was a lot easier.]
    Last edited: May 26, 2010
  12. May 26, 2010 #11
    I want to thank everyone for their help with this problem. I finally got the right answer.

    Looking at this problem now with a clearer mind, it is not that it is difficult, it just requires a review in angles and trigonometric formulas.

    I am in the calculus based course, but since we just started, I am assuming we are meant to struggle with vectors in the beginning before moving on to different approaches.
  13. May 26, 2011 #12
    Could you please give a detailed explanation of how you did it.
    I have 2 tries left on this problem.
  14. May 26, 2011 #13
    Agree, this problem is due in 2 hours. I've worked the equation to this form but i can't seem to get the right answer

    k = 8.99E9

    (k*q1*q2) / (2r^2+2r^2cos(theta)) * cos((pi/2)-(theta/2)) = (k*q2*q3) / (2r^2-2r^2cos(theta)) * cos(theta/2)

    i have no idea what i'm doing wrong at this point and i'm basically desperate haha.
    any help here would be hot

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