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Charge density-related question

  1. Feb 24, 2010 #1
    1. The irregularly shaped area of charge in the figure has surface charge density n_i. Each dimension (x and y) of the area is reduced by a factor of 3.57.

    An electron is very far from the area. What is the ratio F_f / F_i of the electric force on the electron after the area is reduced to the force before the area was reduced?




    2. n=Q/A, electric field equation



    3. So the ration of the charge densities is n_f/n_i = 12.57. Force is E*q, so dividing the force of the final field by that of the first field, I get q_f/q_i. But what is the charge q of each of these shapes? Since the electron is "very far" is there a difference in the forces it sees comparing the two shapes?
     
  2. jcsd
  3. Feb 24, 2010 #2

    rl.bhat

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    Homework Helper

    By changing the area, the charge density will increase but the total charge on the irregularly shaped body remains the same. If you find the electric field at the point where the electron is situated, by using Gauss's theorem, you can see that, it is the same before and after changing the area.
     
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