# Confused by the Question (oil drop charge)

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. Im 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.

Last edited:

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?

Last edited:
andrevdh
Homework Helper
Well, we don't have the given values, so we can't tell you where you went wrong.

Last edited:
the negative value of the charge would indicated an excess of electrons, so i thought dividing by the charge of an electron would give me the number of excess electrons, not sure if thats right however.

still need help one this one...

An electron is negatively charged so I don't think it's a problem getting a negative value for the charge. As for the answer, 2.93 is as close to 3 as makes no difference (considering how small a mass you're talking about) so I'd say thats all ok...You're supposed to get integer multiples of the electron charge so it's fine.

I don't know about the excess electrons sorry.