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

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To calculate the energy stored in a capacitor, we calculate the work done in separating the charges. As we separate more charges, it takes more work to separate even more, due to increased repulsion. Hence, to find the total work done, one needs to integrate. Let's see how we can set up this integral and find the total work done.

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Created by Mahesh Shenoy
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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!

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

tamer
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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.

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

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

jennychong
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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

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

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

gowtham
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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.

jacobvandijk
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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.... : /

dazaiosamu
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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.

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

aestherits
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호이호이-nk
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Thanks a lot for this awesome explanation

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

asgovindarajan
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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

vanitaghogare
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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

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

straightforward
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can you please do more on capacitor and ac circuits

moratasa
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please explain method of images topic too.

rutusodha-csmt