If a drop of mass 1.49061e-12 kg remains stationary in an electric field of 1.1e

Question

If a drop of mass 1.49061e-12 kg remains stationary in an electric field of 1.1e6 N/C, what is the charge on this drop? The acceleration due to gravity is 9.8 m/s^2. Answer in units of C.
PART B
How many extra electrons are on this particular oil drop (given the presently known charge of the electron?

Explanation / Answer

F = mg = Eq ==> q = mg/E You're given m and E and you know g. When you solve for q, divide by the charge of one electron, also given, to get the number of electrons. This problem is not very well set up (or you made a typo), because the answer is not very close to an integer as it should be. 1.4906 *10^-12 *9.8/1.1 * 10^6 q=1.327 * 10^-17 extra electrons = 1.327*10^-17/1.602* 10^-19 =82.83 electrons

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