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Electric Forces, Balloon Problem

  1. Jan 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Two balloons A and B are filled with He gas. They have identical radii of 10 cm. The balloons are made with rubber so that they can be charged by rubbing against hair. They are held together with insulating strings, whose mass and charge can be ignored for this problem. You can also ignore the mass of the balloons.

    a. The magnitude of the charge on Balloon A is 80 nC and that on Balloon B is 40 nC. The strings attached to the balloons make angles θ1 and θ2 with the vertical. Which angle is larger?

    b. Density of He gas is 0.17 kg/m3 and density of air is 1.20 kg/m3. Calculate the angular separation θ1 + θ2 at equilibrium. Hint: assume the angles are small. Let L, the length of the string be 1.0m.

    3. The attempt at a solution

    So basically, for part a. I found that if you draw a FBD of all the forces acting on the two balloons, all the forces are the same. (Vertical - Buoyant force, gravity, component of tension
    Horizontal - electric force, component of tension). Therefore, the angles should be the same,
    θ[itex]_{1}[/itex] = θ[itex]_{2}[/itex].

    Then for part b, I take the equations for the forces acting on a balloon in the x direction and y direction (with the above mentioned forces). Then substituting one equation into another,
    I get a pretty messy equation. But since we assume the angles are small, I let
    sinθ[itex]_{1}[/itex] ≈ θ[itex]_{1}[/itex] and tanθ[itex]_{1}[/itex] ≈ θ[itex]_{1}[/itex].

    Now assuming part b is correct, we know θ[itex]_{1}[/itex] + θ[itex]_{2}[/itex] = 2θ.
    Then we simply just rearrange the equations and solve for θ. I obtain a fairly reasonable value of
    ≈3.2 deg. for theta.

    Although everything seems to make sense to me, I'm not entirely confident with this answer.
    If I made a mistake along the way, if you could please let me know its much appreciated.
  2. jcsd
  3. Jan 18, 2014 #2
    That answer cannot be correct. At 3.2 degrees and 1 m string, the distance between the centers of the balloons is 12 cm, which is impossible since their radii are 10 cm each.
  4. Jan 18, 2014 #3


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    Agree with thetas being equal.
    Don't agree with the 3.2 deg. Can you show us where it comes from?
  5. Jan 18, 2014 #4
    So for part b, I took the FBD of a balloon.

    X component of net force:

    [itex]F_{net, x}[/itex] = Tsinθ[itex]_{1}[/itex] - [itex]\frac{k\left|q_{a}q_{b}\right|}{r^{2}}[/itex] = 0 ....(1)

    Y component of net force:

    [itex]F_{net, y}[/itex] = [itex]V_{balloon}\rho_{air}g - V_{balloon}\rho_{helium}g - Tcosθ_{1}[/itex] = 0 ......(2)

    Then the horizontal distance between the two balloons, r is given by [itex]r = L(sinθ_{1} + sinθ_{2})[/itex]

    Substituting into equation 2 into 1 we have:

    [itex]V_{balloon}g(\rho_{air}-\rho_{helium}) = \frac{k\left|q_{a}q_{b}\right|cosθ_{1}}{L^{2}sin\theta_{1}(sin\theta_{1} + sin\theta_{2})^{2}}[/itex]

    Then using small angle approximations sinθ[itex]_{1}[/itex] ≈ θ[itex]_{1}[/itex] and tanθ[itex]_{1}[/itex] ≈ θ[itex]_{1}[/itex], and since θ[itex]_{1}[/itex] = θ[itex]_{2}[/itex], the above equation simplifies considerably.
    Rearranging we can solve for θ.
    This is how I got that value for θ, if something is incorrect please let me know.
  6. Jan 18, 2014 #5
    This is not correct, because the radii of the balloons are not taken into account. Other than that, you are on the right track.
  7. Jan 18, 2014 #6
    So then r = (L + 0.10)(sin[itex]\theta_{1} + sin\theta_{2})[/itex]

    (the question did say to treat the balloons as point charges located at the centre of the balloon when calculating the coulomb force, so that would make sense)

    With this, I obtain 2.98 deg which isn't really that different, and still doesn't make sense considering the radii are 0.10 m.
  8. Jan 18, 2014 #7


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    So we are starting to wonder how the strings are attached.
    Could it be we have to start over with a different assumption? The picture we (I) had thus far was two strings attached to a single point. Could it be two strings attached to two points 20 cm apart ?
  9. Jan 18, 2014 #8
    No the strings are attached to a single point.
    It looks something like this:
    Θ -------- Θ
    . ----- .
    . --- .
    .- .

    where the dotted lines represent the respective strings and Θ represents the balloon (don't bother with the dashes).
    The angle between the left string and vertical is [itex]θ_{1}[/itex], and the angle between the vertical and right string is θ[itex]_{2}[/itex]. I hope this is clear.
    Edit: The left string should be slanted not vertical.
    Last edited: Jan 18, 2014
  10. Jan 19, 2014 #9
    Using your formula, I obtain 4.73 degrees for the half-angle, which yields 0.09 m for the half-distance between the centers. Sill short of the minimum 10 cm required.
  11. Jan 19, 2014 #10
    Any ideas to what I'm doing wrong?

    Also note, I think r = L(sinθ1+sinθ2), because L is the length of the string extended to the centre of the balloon.
  12. Jan 19, 2014 #11
    I suggest that you double check all your steps when you calculate the answer. Your method seems good to me - as long as you are told to treat the balloons as point charges.
  13. Jan 19, 2014 #12
    Yeah I've been doing that for the past two days. Still no luck though. Everything seems sound, just keeps yielding odd answer.
  14. Jan 19, 2014 #13
    This is probably time that you spoke with your instructor. Explain your method, and explain why you are not happy with the result.
  15. Jan 19, 2014 #14
    I was thinking that since we are assuming the balloons to be point charges maybe then the answer makes sense in that context. I think its the assumption that the balloons are point charges that puts us in hot water.

    But since we calculated electrostatic force between point charges instead of balloons, then maybe its not justifiable to claim afterwards the point charges to be balloons with radii 0.10m, and thus on this basis claiming the answer we obtained assuming the balloons as point charges is not valid is not well justified.
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