Robert Millikan Eperiment!

My Robert Millikan video explaining how he did his experiment to find the charge on an electron!

Category:

Science & Technology

Tags:

• robert
• millikan
• omg
• electricity
• physics
• charge
• electron
• science
• awesome

[layfs]

It's great that people go to so much trouble to make stuff like this. I'm teaching about Millikan and things like this always help. Why do some people make negative comments about stuff like this? If you have to criticise make it positive (unlike the charge on an electron).

Thanks dukeuke.

[dukeuke]

@layfs You're a teacher?

[ultimatemusicfreak]

My teacher showed our whole Physics class after showing that boring one, and I think the whole class got it after that. Oh, the guy dancing made us all laugh. Thanks dude, you saved me from my last test. 100!!

[dukeuke]

hahaha Thanks! It's great to know that stupid little video is being put to good use lol

[Rockas360]

Thanks dude this helped soooo much I got it, I mean even my college chem teacher couldn't help me understand it but you did thx sooo much XD!

[19perox05]

I wish my classes were as fun as this video... why it's always hv to be so formal n boring~

[vixvapourub]

@kessinglandude i meant no net forces sorry. but thanks, its clearer now

[kessinglandude]

@vixvapourub There ARE forces acting on it. One downward force (which is the weight) and two upward forces (viscous drag and upthrust). When the drop BEGINS to fall, these are unbalanced - there is much more downward force than upward - but eventually, as it speeds up, these become equal in size. So no UNBALANCED forces at this stage. Weight = drag + upthrust. As upthrust is negligible, weight= drag.

[KenzieLien24]

This video was shown to me and my physics class by our teacher... I LOVE this video!!!!

[kessinglandude]

@vixvapourub The terminal velocity bit is when the droplet is falling. You don't equate Eq to mg in this part. This part is used to obtain a value for the radius of the drop as it cannot be obtained any other way. The radius is needed to calculate some of the other values needed to get the final value for the charge on the electron. This falling bit is only one part of the experiment. The other part is the 'holding still' bit. It's here that the weight and electrostatic forces are equal.

[vixvapourub]

@kessinglandude that's exactly my point. If there are no unbalanced forces acting, why would you equate q.E, the force required to suspend the droplet, to mg?

[kessinglandude]

@vixvapourub At terminal velocity there is no unbalanced force. There are the forces of weight; upthrust and viscous drag acting on the drop as it falls, but these all balance out to give no overall force.

[vixvapourub]

great vid ! :D but if the oil droplet is falling at terminal velocity already there are no forces acting on it?

[CallumBerryman]

Haha made me laugh :D good vid mate :P