Polarization Dielectrics The Van de Graaff More on Capacitors Romanian translation courtesy of Mihai Olteanu. View the complete course at: ocw.mit.edu License: Creative Commons BY-NC-SA More information at ocw.mit.edu More courses at ocw.mit.edu
Video Rating: 4 / 5
Is that a banana stuck to his pocket?
SCREAM! SCREAM FOR THE PHYSICS PEOPLE!
I am starting to hate my prof.
PLEASE watch my video on the Electrinium battery.
Well, great job! But missed some demonstrations which are really interesting.
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Here’s a guy who knows he stuff and enjoys teaching. Seems to be very well planned in his approach and uses a good mix of lecture, real examples, and some excitement to keep students brains working.
That guy has a funny accent, some chalkboards and he knows how to use em.
@henf5671 what u said was“avatar was not that entertaining“ and i said that’s only ur point of view
(because i completly disagree , avatar was a real success !!!)
@yoyaya007 i never said avatar was entertaining. read previous comment one more time. i said walter lewin’s lectures are a lot more entertaining than anything else that would be regarded as entertainment.
@henf5671 avatar was not that entertaining …. well that’s only ur point of view!!!
@luzzie9 well avatar wasn’t that great. but professor lewin’s lectures are actually more entertaining than anything imaginable that would be classified as entertainment on this planet.
@Compact3 what i am saying is, that by increasing the distance, its like disassembling the two plates, and because they are not infinitly big, disassembling means: having two charged objects which are indepentend from each other.
@Compact3…that means,that the ONE sphere has arrived at his maximum voltag that the charges on it create in this spherical object, its like having ONE sphere alone (since the other is infitly far away). and what is the potential in ONE sphere alone with respect to the charge?not billion! its the potential of a charged sphere (lecture 4 i think)!
ok, one problem solved,but you have another problem
a full explaination would take too long,i hate this limit,if you want to know the rest, write me
@Compact3
you wrote: ” Does that mean that if you make d a few meters that you have several millions/billions of volts? Isn’t that weird?”
no, look what i am saying: suppose you have no plate capacitor, but a sphere-capacitor (because its easier to unterstand). so you have 2 charged spheres, just like you had 2 capacitorplates… soppose you increase the distance to infinite, that doesnt mean that you have tremendously increased the voltage to billion or something, (continuation…)
@luzzie9
Do you even understand the mathematics of this class?
@cyorks90 what I cannot understand is that how did he touch 30000 v without being hurt? I think he did that at the end of the lecture but the camera did not show that.
@rahulilrplac
That’s what I’m thinking too. Glass is an insulator, so some of the induced charge on it will remain in place. This phenomenon is known as hysteresis. When the capacitor is reassembled, the dielectric will induce a charge on the metal and hence a potential difference will be present. The electric field on the plates will be equal to the electric field on the glass.
@Compact3
Once the plates are far enough from each other the plates can be considered point charges. Once this happens, you can find the potential with the equation V=q/(4*pi*epsilon*r).
@Compact3
The capacitance of a parallel-plate capacitor decreases as you increase the distance between the two plates, but that doesn’t matter if you keep Qfree stable, as he said. Think of it as two finite nonconducting sheets with a fixed charge sigma. The electric field due to an infinite nonconducting sheet is sigma/2*epsilon. The sheets don’t have infinite area, but if they did, you would increase the potential difference V as the sheets are pulled further and further apart.
@Compact3
Remember that V=E*d. The units of an electric field are N/C and V/m (Volts/meter is what you should think about in this case).
As you increase the distance between the two plates, you are doing positive work and you are increasing the potential difference between the two plates. He is talking about the case where the plates are close enough that the electric field remains constant. However, when the distance between the two plates increases significantly the electric field is weaker.
I don’t understand something:
if you keep Qfree stable, but you increase d then V will go up he said (and showed).
Does that mean that if you make d a few meters that you have several millions/billions of volts? Isn’t that weird? Does that mean that if you connect the two plates with a resistor when d is little then nothing special will happen, but if you connect the two plates with the same resistor when d is big then the resistor will melt? Wtf?
whts up with the bananas and donuts he puts on his shirt??
@xKorax ops cameras*