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The water rises

  1. Apr 10, 2014 #1
    1. The problem statement, all variables and given/known data
    A plastic ball of 1cm diameter and 10^-8 Coulomb of charge is suspended by an insulating string. The lowest point of the ball is 1cm above a big container of saltwater. Indeed, the water's surface rises a bit. What is the water elevation height exactly below the ball?
    Ignore the water's surface tension and consider 1000kg/m³ the saltwater density

    2. Relevant equations
    Eletrostatic equations

    3. The attempt at a solution
    My english is quite rusty, I am sorry for that, but I would really appreciate the help xD
    In my solution, I considered the saltwater surface an infinite conductive plane, and so an image charge would be attracted by the plastic ball as the water rises. It was, basically the idea. From now, I considered the image charge as if it was made of saltwater but with the same plastic ball volume, indeed, the water will rise until the balance of forces (electric and weight):

    Let k = 9*10^9Nm²/C² be the electrical constant, d = 1000kg/m³ the density of salt water, g = 10m/s² the gravity, L the length between the ball's center and of its image, and V=4/3π(0,005)³ the plastic ball volume:
    The charge image will rise until x such that
    k(10^-8)²/x² = (dV)g
    (9*10^9)(10^-8)²/x² = (1000)(4/3π(0,005)³)(10)
    => x² = (9*10^9)(10^-8)²/(1000)(4/3π(0,005)³)(10)
    solving for x
    => x = 0,01311 m = 1,311 cm
    But, geometrically, as the charge image rises until x, namely, the distance L-x, the saltwater surface would rise y = (L-x)/2
    L = 0,5 + 1 +1 + 0,5 = 3cm
    => y = (3-1,311)/2 = 0,85cm

    But the answer seems to be...

    Were my considerations that wrong?
  2. jcsd
  3. Apr 10, 2014 #2
    Why are you using the volume of the sphere? it seems to play no role in the problem.
  4. Apr 10, 2014 #3
    Also, please refrain from plugging in any data until the very last step of your solution. That makes your solution much easier to understand.
  5. Apr 10, 2014 #4
    The volume of the sphere is irrelevant, as dauto says. Are you familiar with Gauss' Law? The charge outside the sphere is the same as if all the charge were at the centre, as far as all points outside the sphere are located. You need to find the effect on the water of the charge 10^-8 Coulomb being concentrated at the centre of the sphere...
  6. Apr 10, 2014 #5
    I know, this problem was driving me crazy. It makes no sense, but it seems to be a polish olympic problem, so I was trying to get an answer out of it... This question is wrong, right?
    Last edited: Apr 10, 2014
  7. Apr 11, 2014 #6
    No, the question is correct. It is possible to find the solution with the information provided.
  8. Apr 16, 2014 #7
    Then, how should I proceed?
  9. Apr 16, 2014 #8
    1st: Find the electric field at the surface field using the images method
    2nd: Find the surface charge density using Gauss' law on a "Pillbox" region enclosing the surface.
    3rd: Using the electric field and surface charge find the force acting on the surface. Beware of a factor of 1/2 that appears in such problem because half of the field is the field of the surface charge itself that cannot produce a force on itself.
    4th: Calculate the work done by that force on the rising water.
    5th: Subtract that work from the gravitational potential energy to find the total potential energy (electric plus gravitational).
    6th: Minimize that energy
    Last edited: Apr 16, 2014
  10. Apr 16, 2014 #9
    You may take a shortcut and use equilibrium between the electric force and gravity. That way you won't have to calculate the energy and minimize it.
  11. Apr 16, 2014 #10
    I solved the problem and found an answer 10 times smaller than the one provided in the spoiler box in the OP.
    h = 0.29 mm
  12. Apr 24, 2014 #11
    Thanks, guys. I tried to redo it yesterday and I got 0,56mm. Could you post your solution please, dauto?
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