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Thread: The water rises

  1. #1 The water rises 
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    Aug 2014
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    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/m the saltwater density.

    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*10^9Nm/C be the electrical constant, d = 1000kg/m the density of salt water, g = 10m/s the gravity, L the length between the ball's center and of its image, and V=4/3π(0,005) the plastic ball volume:
    The charge image will rise until x such that
    k(10^-8)/x = (dV)g
    (9*10^9)(10^-8)/x = (1000)(4/3π(0,005))(10)
    => x = (9*10^9)(10^-8)/(1000)(4/3π(0,005))(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...
    Spoiler Alert, click show to read: 
    0.29mm
    Reply With Quote  
     

  2. #2  
    Senior Member
    Join Date
    Jul 2014
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    Quote Originally Posted by Dinheiro View Post
    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/m the saltwater density.

    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*10^9Nm/C be the electrical constant, d = 1000kg/m the density of salt water, g = 10m/s the gravity, L the length between the ball's center and of its image, and V=4/3π(0,005) the plastic ball volume:
    The charge image will rise until x such that
    k(10^-8)/x = (dV)g
    (9*10^9)(10^-8)/x = (1000)(4/3π(0,005))(10)
    => x = (9*10^9)(10^-8)/(1000)(4/3π(0,005))(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...
    Spoiler Alert, click show to read: 
    0.29mm
    I think V is also depend on x
    Reply With Quote  
     

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