More Coulombs Law and Electric Fields!

Hello there, as in the previous post I would just like to check my answer to this past examination question.

Any help much appreciated.

A negatively charged oil drop of mass 1.5x10^-14Kg is observed to remain stationary in the space between two horizontal charged metal plates. Calculate the number of electrons on the drop if the potential difference between the plates is 50V and their distance apart is 15mm.

I calculated the elctric field strength using E=V/d giving 3.33x10^3.

Then I said that since it was stationary, F=mg and F=QE give QE=mg which rearranges to give Q=mg/E giving a charge on the oil drop of 4.42x10^-17C

Dividing this by the charge on an electron I get the number of electrons to be 276!

I am pretty sure there should be alot more electrons than this!!

Thanks again for any assitance.

Mike

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Originally Posted by mikewhant

Hello there, as in the previous post I would just like to check my answer to this past examination question.

Any help much appreciated.

A negatively charged oil drop of mass 1.5x10^-14Kg is observed to remain stationary in the space between two horizontal charged metal plates. Calculate the number of electrons on the drop if the potential difference between the plates is 50V and their distance apart is 15mm.

I calculated the elctric field strength using E=V/d giving 3.33x10^3.

Then I said that since it was stationary, F=mg and F=QE give QE=mg which rearranges to give Q=mg/E giving a charge on the oil drop of 4.42x10^-17C

Dividing this by the charge on an electron I get the number of electrons to be 276!

I am pretty sure there should be alot more electrons than this!!

Thanks again for any assitance.

Mike

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I get the same calculation ... remember that the 276 electrons is in excess of the number of protons, not the actual number of electrons.

Oh yes, because it is a net negative charge meaning there are more electrons than protons!

Thanks very much

Mike