Millikan oil drop experiment charge determination

[Jacob Pilawa]

Howdy y'all!

If you could help with the following question, my physics class and I would be extremely grateful.

A charged oil droplet is suspended motionless between two parallel plates (d=0.01m) that are held at a potential difference V. Periodically, the charge on the droplet changes as in the original oil drop experiment. Each time the charge changes, V is adjusted so that the droplet remains motionless. Here is a table of recorded values of the voltage:

i. 350 V

ii. 408.3 V

iii. 490 V

iv. 612.5 V

From the data above, determine the charge on the dorplet for case (i) above. What assumptions do you need to make? (Hint: the ratio of voltages = ?)

Thanks a ton, we've been stumped.

I'm going to be honest here, me and 2 friends have been working on this for about 4 hours, and we don't really have any substantial work to show. Any help would be great. Thanks.

 

[Charles Link]

I can give you a hint, but I haven't solved it myself yet: $350* \, Q_1=408.3* \, Q_2=490* \, Q_3=612.5* \, Q_4$. $Q_4<Q_3<Q_2<Q_1$. Find some $Q_o$ so that $Q_4=n_4 \, Q_o$, $Q_3=n_3 \, Q_o$, etc., $n_4, n_3,...$ integers (hopefully small ones). Sorry, I edited a couple of times because I read it incorrectly.

 

[Jacob Pilawa]

I can give you a hint, but I haven't solved it myself yet: $350 \, Q_1=408.3 \, Q2=490 \, Q_3=612.5 Q_4$. $Q_4<Q_3<Q_2<Q_1$. Find some $Q_o$ so that $Q_4=n_4 \, Q_o$, $Q_3=n_3 \, Q_o$, etc., $n_4, n_3,...$ integers (hopefully small ones).

Okay, this makes sense. However, where can we go from here? Is there anyway to solve for the integers?

 

[Charles Link]

I have it, but I'm not allowed to give the solution. I can give you a hint though. The smallest number, $Q_4$ is greater than 3. Another hint is the numbers are exact enough, that I think the data is probably simply constructed by the professor as a good learning exercise. One additional hint=let $Q_4=n_4$ (Ignore the $Q_o$ part mentioned previously.) Please let us know if you figured out the answer.

 

[Jacob Pilawa]

I have it, but I'm not allowed to give the solution. I can give you a hint though. The smallest number, $Q_4$ is greater than 3.

Okay, so we talked it out a little bit. So does this mean that the answer is 7e=1.12x10^-18 coulumbs?

 

[Charles Link]

Okay, so we talked it out a little bit. So does this mean that the answer is 7e=1.12x10^-18 coulumbs?

Yes. One additional question for you=what did you get for the other 3 integers? And were the calculations almost exact?

 

[Jacob Pilawa]

Yes. One additional question for you=what did you get for the other 3 integers?

We got all the integers as 7,6,5, and 4. Thank you so much! We just screamed in excitement and relief.