Energy stored in capacitor derivation (why it's not QV) | Electrostatic potential | Khan Academy

[@fraekfyr1000]

I randomly came across one of your videos on electricity and now I can't stop watching! And I am not even studying for an exam. What a rare talent you have for explaining things clearly and enthusiastic!

[@Sanskar_Raii]

Physics doubts don't trouble me anymore... Thanks to Khan Academy and especially Mahesh sir🙏

[@not_enough_data_]

For hopefully a bit more intuition on why it's 1/2: In the naive case, let the x-axis be the charge moved so far, and then draw a graph of y = voltage (ie a constant here). Then to compute the work done, you can take an integral of voltage dq, or in other words the area under this graph from x=0 to x=Q. For this naive case, that's just the area of a rectangle = base x height = charge x voltage In the more sophisticated case, you know that the voltage isnt constant, and instead it's proportional to the charge moved so far. That means the graph will slope upwards away from the origin, cutting the rectangle you had before in half. Your total work done is still the same integral, so it's still the area under the graph, which in this case is the area of a triangle = 1/2 base x height = 1/2 charge x final-voltage If you see a factor of 1/2 anywhere else, for example in kinetic energy, you can often interpret it in a similar way

[@klevisimeri607]

Or sometimes I like to tell people that if one parameter is changing with respect to another parameter in the equation it should make that triangle and add up to 1/2 . W=Σ V Δq (=∫ V dq more precisely). Because V changes with respect to charge q and it is not constant when you take the sum than you and up with that 1/2. Conclusion: If you find an equation when something is changing with respect to something inside the equation, you will usually see this 1/2.

[@NestedLump]

It happens to be just because the potential across the capacitor is 'Linearly' proportional to Charge on it assuming constant capacitance. So u get a straight line with 'some slope' on plotting Voltage-Charge graph and work is nothing but area under that curve. Because of the linear proportionality the area happens to be of a triangle which is why there is a 1/2 factor.

[@tamer1474]

Crystal clear explanation with an excellent flow of reasoning. Great job!

[@EA-tc6kb]

No it's not, Power = Current X Voltage. We know a capacitor charges fast at the beginning and slows down, that gives you time. Now you can convert power into energy/work done. He just made it complicated.

[@furiousarpan7473]

@@EA-tc6kb you're explaining that using formula and he is explaining that from core level, don't talk shit here

[@EA-tc6kb]

@@furiousarpan7473 You're the only one talking shit here.

[@Mundane_task]

​@@EA-tc6kbthat doesn't even make sense bruh

[@YohansSeife]

you are making my life as a student so fascinating so i would like to offer you my humble gratitude mister mahish and of course kehan academy.

[@호이호이-n8k]

This is crazy. There are things we should wonder as students but most of us does not recognise and just accept it as given oracle. I love this channel. There are so many who explains wrong. I used to find them unsatisfying. But This channel gives the real knowledge.

[@jacobvandijk6525]

THE USUAL ERROR: There is no energy in this field. The potential energy is only with the displaced electrons!!! The battery did work on THEM, not on some field that wasn't even there in the first place. Think of this: If you roll a ball upon a hill, then the ball acquires potential energy, not the gravitational field.

[@bhavyanayak4342]

That's a nice point

[@jacobvandijk6525]

@@bhavyanayak4342 Thanks.

[@shamirajsurvase8032]

Thx

[@rickgrimes47]

You are wrong. The battery does the work which is equal to charge_flown*V = CV². But the energy stored in the capacitor is CV²/2, it is half. This energy is stored in between the plates of the capacitor i.e the electric field. Now you may ask where'd the other half go? The other half of energy (i.e half of the work done by the battery) is used or wasted in moving the electrons from one plate of the capacitor to the other plate.

[@goldy_on_pc930]

Tooth fairy ❌ Charge Fairy ❌ mahesh ✅

[@gowtham5168]

That video helped me figure out what was my doubt in the first place , and cleared it ... excellent

[@ourtamilnadu9011]

I just completed my chemistry studies now, I busted nothing understood,now I'm gonna study physics,I am so cool now because I gonna see mahesh sir's teaching.

[@dazaiosamu3660]

I wonder... why are there so less subscribers here compared to other youtube channels? It deserves much more views and subs compared to some other channels.... : /

[@vanitaghogare3954]

Why this video has not got any likes.... it's exceptionally excellent and all my marks in exam is because of you Thank you so so much sir Are you from Mangalore side...if it is I want to meet you sir

[@jennychong7972]

You are simply the best! More people should subscribe & follow you. Thank you so much for helping all of us understand better!

[@straightforward3402]

And generation will understand when you will be ahead in time AGAIN as you are now .... Truly remarkable

[@alirezaamooei6874]

Beside clear explanation , you made the presentation really attractive. wish the best for you :)

[@aestherits]

This video is truly a gem! great explanation from a great tutor

[@marvelcalculus6307]

You can take the battery is doing work to move the charged particle e^- by spending energy which is stored as chemical energy in it and by this you can neglect the attraction since electrolyte is present in battery does not create attraction to e^- and so it is applicable in reality

[@jonathantran429]

Thank you, Mahesh!!

[@asgovindarajan5597]

Wow! The explanation was extremely clear! 🔥🎉 Thank you so much for uploading this video, sir!

[@ikashyap5]

Beautiful explanation

[@ViktorGoossenshasseltedition]

veryyy clear explanation

[@parthkhanayat420]

Thanks a lot for this awesome explanation

[@lawrencewinter]

well explained, thanks. One small point though, when you describe moving the charges, it seems as if you are suggesting that the charge moves directly across the dielectric gap, as opposed to going round the circuit. May seem obvious to you or I, but poeple will carry that misconception which is fundamental and could derail their core understanding of the capacitor. The dielectric is an insulator.

[@ranjitchinnu6371]

i think charges won't go from one plate to another plates, my answer is yes there is an external force (mahesh), acting on only one of the plates due to which that plate gets negatively charged, at this time in the same rate other plate gets positively charged.

[@ahonasarkar1293]

Sounds like Koothrapalli is teaching!

[@moratasa9741]

can you please do more on capacitor and ac circuits

[@someone2007ash]

man... I never understood physics so well

[@sehrishjabeen2447]

Thank you Khan academy

[@parahumour4619]

that was sooo elegant

[@green3487]

THANK YOU SO MUCH!!!

[@shubhamsaha6130]

This is a gem !

[@rutusodha-cs6mt]

please explain method of images topic too.

[@pingakshasahay7080]

Thank you sir

[@vijayanand2783]

But there is no charge flow across a parallel plat capacitors

[@YamanAyhan-e9w]

You don't add additional positive charges. The negative charge build up on one plate repels the negative free moving charges on the other plate leaving positive charges.

[@eshawilson771]

Thanks a Lot!.

[@r.p.blifestile8088]

Hello mister, second plate is induced by charge deposition in 1st plate,

[@piyushkumarsingh0694]

Is there any video on capacitor?

[@anilsurjewala3606]

Awesome Big fan....

[@ashmitgupta8039]

Thankyou sir ..

[@anikaantara4836]

Thanku sir

[@FearlessS21]

Please explain how to use the formula in different cases.

[@sheikhsheheemrizwan]

hat's off....

[@praiseprince_]

OMG!!! I wanted to say "That's what she said" soooo many times.

[@jsathibabu8822]

Sir when charge travels through the battery it gains potential energy so we have to add that potential energy also IS IT RIGHT OR WRONG ❓❓

[@jacobvandijk6525]

The potential energy is with the displaced electrons, NOT in the field!

[@azrielshacham6829]

Great

[@asgovindarajan5597]

I've a doubt (rather a request). Could you please explain / derive the expression of "Electric Potential" (Or Voltage) from coulomb's law of electrostatics? In other words, can you express electric field as a consequence of coulomb's law of electrostatics?

[@petar7867]

take the integral of the E field (E=Q*k/r^2 for point charge) and dr from the distance between the charge and the point of interest to infinity. Q and k are constants so you can move them outside of the integral which leaves you with 1/r^2, when you integrate you'll get -1/r, now just plug in your boundaries

[@asgovindarajan5597]

@@petar7867 thank you so much!

[@ojasvi-gm2ff]

wow man

[@Elitephysicszone]

How charges move in between capacitor plates??. There is free space between plates..

[@amruthaailukunje163]

According to me.... When the electron accumulates on one plate, the electron from the neutral atom of other plate moves away from plate and hence that atom becomes positively charged and hence the plate become positively charged ( the charges do not actually move)

[@Okapi_92]

Sirrrrrrrrrrrrrrr plzzzzzz could u please clarify the realtion of capacitance with potential energy bcz in one equation it is direct while in other it is inverse driving me crzy 😅😢

[@prabhatkumarsingh7563]

Wrong explanation, when initially first plate has Q charge and second plate has zero charge then there must be potential difference although carrying dq charge first would not be repelled. So method of eqplation is wrong however you got correct result. Tell me earth is considered having zero potential and if you have plate with positive charge then what is potential difference obviously not zero.

[@amruthaailukunje163]

Sir, he mentioned at first that both the plates are neutral and hence there is zero potential difference. In the presentation he just represented one side to make us understand.....i think u missed this part 3:46

[@r.p.blifestile8088]

Golu!how can you transfer a+ve charge from a neutral point shown by you? Only charge transfered are electrons only by emf not by your common sense Golu, did you understand Beta. Read carefully before teaching😢