Potential Energy Stored in a Capacitor | Doc Physics

[@BenSilber]

you've taught me more in 45 minutes on youtube than I've learned from my shit professor in a whole semester. And in a more fun way as well. Good stuff!

[@DocSchuster]

Dang, man. I always hate hearing things like this, but I'm happy to be a part of helping out. Keep learning!

[@BenSilber]

exam tomorrow, I'll be watching all night :D

[@yoramstein]

After watching the previous video (Dialectrics =Polarization ability) this video I give you thumbs up before it starts.

[@finalresistance]

This makes so much more sense than the hand-waving nonsense in my text book. Thank you.

[@TuningFreak23]

"u in a cap. me in a cap? what?" lmao

[@j0mezzy]

Squrrrrrrrrr

[@jacobvandijk6525]

@ 8:58 "FIELDS STORE ENERGY!!!" BUT NOT THIS FIELD! The potential energy of a capacitor is IN THE ELECTRONS of the metal wire (not in the field between the plates), because in loading the capacitor the battery is doing work on the electrons! Just connect a capacitor to a circuit and the electrons start moving: potential energy being converted into kinetic energy!

[@avdeshkumar8500]

why DiD you took average charge

[@curiousbit9228]

Well, It's not really the average. The net electric field in between the plates is the sum of the electric field because of the left plate and the electric field because of the right plate. when calculating work in moving the plate on the right side, you should not include the electric field generated by the plate on the right. I other words "Thou Shall not push thyself"

[@Sora-ce1zx]

The strength of electric field created by a single, infinite-large metal plate is exactly half the strength of electric field created by this capacitor. Therefore, the equations should be u=∫Q(E/2)dx*, instead of u=∫QEdx Then, E=Q/ε0A ⇔ Q=Eε0A (note that this E is the strength of electric field created “by the capacitor”, not “by a single, infinite-large metal plate”.) Put this in * and you’ll get u=∫E^2ε0A/2dx

[@Tsarmakli2]

Thank you for your videos they have helped so much!! I have a question though. Why did you divide Q= Eε0A by 2. You say it is to get the Q average but I'm not really understanding the concept of that and why you need to use it instead of just plain Q.

[@DocSchuster]

Ntiana Sakiotis Q represents the charge on the cap when I'm finished charging. However, it started at a charge of zero and ramped up to Q. So on average, the charge during the charging is Q/2

[@Tsarmakli2]

Doc Schuster Thank you!!

[@packrat541]

An electrostatic generator such as the Vandegraf are just capacitors that constantly change their capacitance and they all require work or energy to convert force and distance into an electrostatic field and the reverse is true also and you can make an electrostatic field do work (motor).

[@NotLegato]

is there a way to get the 1/2 that comes from an actual calculation instead of just... putting it in there? it doesn't quite convince me. in fact, why would the charge change over time, anyway? if we store a set amount of charge in the capacitor and then pull them apart, the charge is constant, so there should be no 1/2. a little confused. edit: i found a much easier way to find the energy stored in a capacitor: the work done by a battery to charge it up, so it's just *W = integral U dq from 0 to Q* which is to say, energy to charge from 0 to Q. no need for field strengths, dielectrics or anything.

[@TheGameJammer]

How come the electric field isn't dependent on x? We'd been taught the formula E=V/d where d is the separation of the plates

[@DocSchuster]

TheGameJammer ...An easily misunderstood equation. In a uniform electric field (where your equation is true), V is what depends on x. E is constant (uniform).

[@UpayanM]

How come the energy isn't just QV? In my textbook, voltage is defined as potential energy per unit charge, so wouldn't the energy be Total QV?

[@DocSchuster]

+1steinstein There's an integral. The same reason that the energy stored in a spring isn't kx*x even though the force is kx and the distance stretched is x. The force wasn't always kx!

[@UpayanM]

+Doc Schuster But when the plates reach the maximum charge capacity and there is no current flow, each plate contains a total charge, Q,and there is a voltage, V, between the plates. At this point wouldn't it be QV?

[@DocSchuster]

+1steinstein Do you agree with the perfect analogy to the spring situation? Surely the energy stored must be equal to (or less than) the work that was done to get it there, right?

[@UpayanM]

+Doc Schuster Yes...but then I have had similar problems in which I have had to find the energy required to move a electron across two parallel plates and the formula has been eV. The two seem to contradict. Once the capacitor reaches its full capacity, it contains a charge Q and the voltage between the two plates is V. Then, by definition PE = QV.

[@DocSchuster]

+1steinstein Ooh, you're good. Here's the thing. Discharge the capacitor "halfway." If the fully charged voltage and charge are V_0 and Q_0, what are the voltage and charge when it is "halfway" discharged?

[@antonywagner8079]

ok I feel a little dim here. 2 questions here. first what is the highest electric WATS that can be stored and for how long? this very important for me.

[@DocSchuster]

Antony Wagner Watt is the unit of power.

[@antonywagner8079]

YEAH I know that. sorry I put the question in a wrong format... it has been a hot day. try again. What is the longest that a cap can hold its power for? What is the highest watts that a cap can hold?

[@rohitsanjay1]

Hey Doc, help me out. I've written this in my notes, but I can't seem to make sense of it: 'When a positively charged plate is kept near a negatively charged plate, the voltage of the positive plate decreases'

[@DocSchuster]

Well, if you change "kept" to "brought," it works. In bringing them closer, you are reducing the electrostatic energy but keeping the charge constant. We're assuming the capacitor is isolated, of course. Since voltage is energy per charge, it decreases.

[@rohitsanjay1]

How is the electrostatic energy decreasing?

[@rohitsanjay1]

Can you please explain the part in bold? Say we have a large plate and we give a positive charge to it. There is a limit to the amount of charge that can be given to the plate because as charge is given its potential rises and beyond a certain limit the charges start leaking. If we get another plate and place it next to this positively charged plate then negative charge will pulled towards the side of this plate which is closer to the positively charged plate and positive charge on the further side. *This negative charge on plate 2 will reduce the potential on plate 1. At the same time the positive charge on plate 2 will try to increaser the potential of plate 1. However the effect of the closer side of plate 2 holding negative charge will be more. This leads to reduction in the potential of plate 1. So now more charge can be given to plate 1* Now if we earth the outer side of the second plate. Then the positive charge on this side will go to Earth.With this plate 1 will be able to hold even more positive charge. (source: tutor4physics.com/tutorialcapacitors.htm)

[@DocSchuster]

If you bring a positive and a negative charge nearer, the system loses potential energy because it is doing work.

[@abhishekcherath2323]

I didn't like your taking average charge as final/2 there, why not just do the integral?

[@alexandgarciacalle]

AWESOME!

[@ananyapathak8701]

2:48. 😂🤣

[@CleverHavok]

Can you be my teacher?

[@avdeshkumar8500]

please reply

[@DocSchuster]

Hand-waving nonsense? Yikes. Whose text are you using?!? I'll have to avoid it!

[@abhishekcherath2323]

Doc Schuster my state board books are like that. A bunch of crap, followed by a number of formulas, followed by templates for problems