# Parallel Plates and Electric Fields

1. Jan 8, 2010

### physicsfun_12

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!

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

Mike

Last edited: Jan 8, 2010
2. Jan 8, 2010

### Stonebridge

The answer gives the number of excess electrons on the drop - not the total number.
(It's the basis of Millikan's experiment to measure the charge on the electron)

Last edited: Jan 8, 2010
3. Jan 8, 2010

### physicsfun_12

Ah, of course! The drop has an overall negative charge meaning more electrons than protons!

Thanks alot, take care

Mike

4. Jan 8, 2010

### RoyalCat

In neutral matter, the number of positively charged particles is equal to the number of negatively charged particles. The electrical force only acts on the net electrical charge, and that is what you found.

As for the work done, remember that gravity's happy to assist the drop on the way down, but doesn't like being resisted. Take that into account to get a more accurate expression for the work done in moving the oil drop.