(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Would it be possible for someone to check my answer to this past examination question, as I am not sure I'm doing them correctly!

Thanks in advance.

A negatively charged oil drop, mass 2.4 x 10^–12 kg, is stationary between two horizontal metal plates spaced 1 cm apart. Find the number of electrons (to the nearest whole number!) on the oil drop if the potential difference between the plates is 35000 V. Calculate the work done on the oil drop moving it from one plate to the other.

2. Relevant equations

F=mg, F=QE, E=V/d and for the second part W=EQd

3. The attempt at a solution

It says the drop is stationary, so for it to not fall under gravity, the downward force must equal the upward force. Since F=mg and F=EQ and E=V/d,

(V/d)Q=mg so Q=mgd/V

This gives a charge of 6.73x10^-18

Dividing this by the charge on an electron gives the number of electrons to be 42!!

I can't see why this answer would be wrong, but surely there should be more electrons!

For the second part of the question, I said that W=EQd

So this gives 2.36X10^-13J

Thanks again for your help.

Mike

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# Homework Help: Parallel Plates and Electric Fields

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