What is the charge on the oil drop in Millikan's oil drop experiment?

[BobRoss]

Homework Statement

A 1.50x10^-14 kg oil drop accelerates downwards at a rate of 1.80 m/s² 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|=1.04x10⁴V/m

Then I find the net force.
F_net=ma
=(1.50x10^-14kg)(-1.80 m/s²)
=-2.70x10^-14N

Then I find the force of gravity on the drop.
F_g=mg
=(1.50x10^-14kg)(-9.81m/s²)
=-1.47x10^-13N

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^-14N)-(-1.4715x10^-13N)
=1.20x10^-13N

Then I find the charge.
F_e=q|E|
q=F_e/|E|
=(1.2015x10^-13N)/(1.04x10⁴N/C
=1.16x10^-17 C

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

 

[gneill]

What makes you think that your answer is "obviously wrong"?

 

[BobRoss]

Doesn't the charge have to be an integer multiple of the elementary charge (1.60x10^-19C) ?

 

[gneill]

Doesn't the charge have to be an integer multiple of the elementary charge (1.60x10^-19C) ?

Sure. But what experiment has perfect accuracy?

Divide by the fundamental charge and see how close you come to a round number.

 

[BobRoss]

Okay, so then my original answer is correct?

 

[Liquidxlax]

Doesn't the charge have to be an integer multiple of the elementary charge (1.60x10^-19C) ?

well if you will ever do the actual Millikan experiment you will find that you will not exactly get 1.6x10^-19, but around 1.6x10^-19 ± some error x10^-17

 

[gneill]

Okay, so then my original answer is correct?

Your method and numbers look fine to me.

 

[BobRoss]

Great, thanks for the help!

 

[Saad Ahmed]

What happen If we place negative charge between parallel opposite charge plate ?
answer is negative charge attract toward positive plate...

Then why Oil drop is suspended btw opposite plates in Millikan's oil drop experiment??

 

[gneill]

What happen If we place negative charge between parallel opposite charge plate ?
answer is negative charge attract toward positive plate...

Then why Oil drop is suspended btw opposite plates in Millikan's oil drop experiment??

What forces are operating on the oil drop? Draw a Free Body Diagram.