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A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 7840 N/C. The mass of the water drop is 3.84 10-9 kg.
(a) Is the excess charge on the water drop positive or negative?

  • physics -

    Directed upward? Then that is the direction a + would experience a force, so since the force is upward countering gravity, it must be positive? Correct?

  • physics -


  • physics -

    The downward force from the weight of the drop is mg, where m is the mass of the drop and g is the acceleration due to gravity.

    This must be balanced by the upward force from the field.

    Let's say that the total charge on the drop is Q Coulombs,

    then 11000Q is the force directed upwards. Since this must be equal to mg, we have

    Q = mg/11000
    = (4.79 x 10^-9)*(9.81)/11000

    =4.27 x 10^-12 C

    which is very small!

    Note that an electron (or proton) has a charge of about 1.60 × 10^-19 C, so there are (4.27 x 10^-12)/(1.60 x 10^-19) = 26700000 or about 2.67 x 10 ^ 7 excess protons or electrons

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