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

A 1.50x10^{-14}kg oil drop accelerates downwards at a rate of 1.80 m/s^{2}when placed between two horizontal plates that are 9.40 cm apart. The potential difference between the two plates is 980 V. Determine the magnitude of the charge on the oil drop.

2. Relevant equations

|E|=V/d

F_{net}=ma

F_{g}=mg

F_{e}=q|E|

3. The attempt at a solution

So first I find the strength of the electric field.

|E|=V/d

=(980 V)/(0.094 m)

|E|=1.04x10^{4}V/m

Then I find the net force.

F_{net}=ma

=(1.50x10^{-14}kg)(-1.80 m/s^{2})

=-2.70x10^{-14}N

Then I find the force of gravity on the drop.

F_{g}=mg

=(1.50x10^{-14}kg)(-9.81m/s^{2})

=-1.47x10^{-13}N

Here is where I think I am going wrong. I try to find the electric force.

F_{net}=F_{g}+F_{e}

F_{e}=f_{net}-F_{g}

=(-2.70x10^{-14}N)-(-1.4715x10^{-13}N)

=1.20x10^{-13}N

Then I find the charge.

F_{e}=q|E|

q=F_{e}/|E|

=(1.2015x10^{-13}N)/(1.04x10^{4}N/C

=1.16x10^{-17}C

Now this answer is obviously wrong. Where am I messing up?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Millikan oil drop problem

**Physics Forums | Science Articles, Homework Help, Discussion**