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## Homework Statement

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.

## Homework Equations

|E|=V/d

F

_{net}=ma

F

_{g}=mg

F

_{e}=q|E|

## 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?