A spherical water drop, 1.79 ?m in diameter, is suspended in calm air due to a d

Question

A spherical water drop, 1.79 ?m in diameter, is suspended in calm air due to a downward-directed atmospheric electric field of magnitude E = 422 N/C.
What is the magnitude of the gravitational force on the drop?

How many excess electrons does it have?

NOTE*** A lot of people are having difficulties with this one, although I think they are mostly numerical. Hopefully everyone sees the important physics here: this is a statics problem, and there are only two forces (Fe going up and Fg going down). Now Fe =q *E, and Fg = mg so we should be all set. The only detail is to find the mass of our droplet. To do this we only need to recall that we know a volume of the drop and can lookup its density (1000 kg/m^3). Note that we are given a DIAMETER and not radius, and we need to convert some units. For arguements sake a sphere has a volume of (4/3)pi*r^3 and density times volume gives us mass.

If you still are not generating the correct number, send me a feedback showint a few of your steps (ie what is the mass you calculate) and I will check the specifics of your problem.

Explanation / Answer

for the drop to be suspended, the total force acting on it must be zero;

this means that the weight down must be countered exactly by the electrical force up

to calculate the weight down, calculate the mass of the drop knowing that that m= density x volume, and the volume of a sphere is 4/3 pi r^3, where r is the radius of the sphere.

weight = mg where g is the acceleration due to gravity

then, this must be equal to the upward directed electric force, which is:

F=qE

knowing that qE=mg (and knowing E, m and g) you can solve for q, the net charge of the sphere

knowing the charge of 1 electron (which if memory serves me correctly is 1.79x10^(-19) C),

you can divide your value of q by this factor and get the total number of charges

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