# Confused by the Question (oil drop charge)

• linuxux

#### linuxux

This is a short question, from my understanding, to determine the charge of an object we can use the equation $$q=\frac{mg}{E}$$ , where m is mass of object and E is voltage of electrical field. I am asked to find the charge on a drop of oil suspended in the field. I was given both the mass and the applied voltage of the field, but i was also given distance between the field and also told that the lower plate is "at a lower potential". How do the other pieces of information contribute to the charge since they are not factors in the equation? i also have to find number of excess of deficit electrons, how do i do that? (perhaps by divide by charge of electron?)

the givens were oil drop mass: 2.6 x 10^15 kg, plate separation = .005 m ,potential difference between plate = 270 V , and it also says the lower plate is at a lower potential.

to determine E, i used E = v / d , where v is voltage, and d is plate separation, then i used the above equation to determine q.

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This is Millikan's Oil Drop Experiment - where the oil drop is 'suspended' between parallel plates. Normally the electric field is: E=N/C, ie force per coulomb of charge. In parallel plates the field is basically constant, and can instead be calculated by E=V/d. So if you have the distance and the voltage that gives you the electric field E. Stick that into your first equation and hopefully you'll get an integer multiple of electron charge :D

i calculated the E using the given voltage distance between plates, but the q i got was $$-4.7\cdot10^{-19}$$ . i used -9.8 for my g value.

When i divide this number by the charge of an electron , $$e=1.60\cdot10^{-19}$$ , i get 2.93 !

What went wrong?

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Well, we don't have the given values, so we can't tell you where you went wrong.