This is a short question, from my understanding, to determine the charge of an object we can use the equation [tex]q=\frac{mg}{E}[/tex] , 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?)(adsbygoogle = window.adsbygoogle || []).push({});

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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# Homework Help: Confused by the Question (oil drop charge)

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