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Homework Help: Columb force question

  1. Aug 29, 2006 #1
    a small .4g sphere carries a charge of 3 X 10^-10 and is attached to one end of a silk fiber 8cm long. The other end of the fiber is attached to a large vertical insulating sheet that has surface charge density equal to 25 X 10 -6 C/m^2. The sphere on the end of the fiber is repelled outward and finally reaches a state of equilibrium. Find the angle that the fiber makes with the vertical sheet.

    Okay here is what I have done. I know I need to get the tan of theta. What I did is I calculated out the 2 forces. There is gravity which is Tcos(theta) = mg and there is the colomb force which is Tsin(Theta) which is qE. From there I am getting confused on what I should do next. Any help is much appreciated.
  2. jcsd
  3. Aug 29, 2006 #2


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    Good start. You realize that this is a static force analysis problem.

    Let's assume that the insulating sheet is on the right and the sphere is being repelled to the left. Breaking it down into x and y components:

    - Electric force pushing to the left (Felectric)
    - String tension pulling to the right (Fstring)

    - Gravity pulling down (Fgrav)
    - String tension pulling up (Fstring)

    How many unknowns do we have and can we solve with the number of equations available? We know that we have two equations, one in x: Fnet_x = 0, and similiarly in y: Fnet_y = 0. There are two unknowns, the angle theta and the string tension T. You're on the right track ^^.

    P.S. You'll also need to know that the electric field due to a large sheet charge is uniform everywhere and equal to [tex]E = \frac{\sigma}{2\epsilon_0}[/tex] and of course, the Colomb force is then simply Fe = qE as you have mentioned.
  4. Aug 29, 2006 #3

    Andrew Mason

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    [tex]T\cos\theta = mg[/tex]

    [tex]T\sin\theta = qE[/tex]

    [tex]tan\theta = \sin\theta/\cos\theta[/tex]

    Seems to me you have everything you need.

  5. Aug 29, 2006 #4
    okay so if I am correct I would take the atan of my answer and get about 6.2 degrees is this correct?
  6. Aug 29, 2006 #5

    Andrew Mason

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    You will have to show us how you got E.

  7. Aug 29, 2006 #6
    I got E by [tex]\sigma[/tex] = 25 * 10^-6 C/m^2 then I took that and divided by 2e0 with e0 being 8.85 *10^-12 I took that answer(1.4 * 10^6) and multiplied it by the charge of the ball which is 3 X 10^10 C which gave a force of 4.23 X 10^-4.
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