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Another Electric Field Question

  1. Jan 7, 2005 #1
    So the question reads:

    "A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 8480 N/C. The mass of the water drop is 3.50 x 10^-9 kg. (a) Is the excess charge on the water drop positive or negative? Why? (b) How many excess electrons or protons reside on the drop?"

    The correct answers are:
    (a) Positive, so that the electrostatic force points upward.
    (b) 2.53 x 10^7 protons

    I really wasn't sure where to begin, so I wrote some equations down.
    I tried using F=Eq, or E = F/q where the E = 8480, but I wasn't sure about F or q. There was an equation that said something = q/A but I don't have A or q.

    I think the main part i'm missing is converting the 3.50 x 10^-9kg of H2O into how much charge it has. Can anyone give me some direction as to where to go for this problem? Any help is greatly appreciated, Thanks. :smile:
  2. jcsd
  3. Jan 7, 2005 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    You are correct that F=qE gives the upward force. What is the downward force (hint: think mechanics not electricity)?

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