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!

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.

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

2 жыл бұрын

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

Жыл бұрын

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.

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

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

@EA-tc6kb

2 жыл бұрын

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

Жыл бұрын

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

@EA-tc6kb

Жыл бұрын

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

@Mundane_task

4 ай бұрын

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

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

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

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

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

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.... : /

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

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 the potential energy, not the gravitational field. Keep on thinking for yourselves!!!

@bhavyanayak4342

2 жыл бұрын

That's a nice point

@jacobvandijk6525

2 жыл бұрын

@@bhavyanayak4342 Thanks.

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.

This is a gem !

veryyy clear explanation

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.

Thanks a Lot!.

THANK YOU SO MUCH!!!

Thanks a lot for this awesome explanation

that was sooo elegant

Awesome Big fan....

Thanku sir

Thankyou sir ..

Great

can you please do more on capacitor and ac circuits

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

Thank you Khan academy

please explain method of images topic too.

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

Thank you sir

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

Please explain how to use the formula in different cases.

wow man

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.

Sounds like Koothrapalli is teaching!

man... I never understood physics so well

But there is no charge flow across a parallel plat capacitors

Is there any video on capacitor?

hat's off....

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

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

2 жыл бұрын

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

2 жыл бұрын

@@petar7867 thank you so much!

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.

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

2 жыл бұрын

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

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

@amruthaailukunje163

Жыл бұрын

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)

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 😅😢

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

Жыл бұрын

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