Wait so what is the correct reason then? Why is it 1/2QV not QV? And why is it the area under the graph not just the charge at that point multiplied by the potential difference at that point?
It's because of the fact that the original E = QV assumes charge and voltage are constant, but they are not. Obviously to charge up a capacitor, you would need to have started at 0 and then increase as you supply it with voltage. It's the average.
8 days till my physics paper 2 and even though this topic isn't in the advance info watching this video made me feel extra prepared for the curveball I know is gonna come. thanks
@@puddleduck1405 This was for last year's students. We won't get any this year. Apparently the advance information for last year didn't help too much anyways.
Hey I stumbled upon your videos and though that they were actually better than my physics teacher. And I was wondering if you were thinking about making a website with all your videos where you could for example accept donations like the examsolutions guy does so that at least you can get a little extra for your incredible effort. Also just wanna thank you for your videos P.S. I'm on my last year pleaseee go ham on the uploads I need them!!!
noticed most of these top comments are years old, just wanted to say there is new blood coming here too! These videos will never stop helping years of under taught kids
You, sir, are excellent. I think making your videos accessible through payments will only make the channel less favourable. I think your videos so far are of brilliant quality and the fact that they don't require us to pay, makes it so much better.
+Burry21 Thanks! Yes, I wouldn't have wanted to pay when I was doing my A-levels, so I don't make anybody pay. Please feel free to donate enough to buy me a cup of tea to keep making videos though! ;)
@@hi44098 na don't worry man, get the easy marks with thermal and nuclear. Get like 13/25 on the multiple choice and answer the wordy questions, bound to get a few points right, decent mark right there. All the best
bro where would we all be without you? I started to get so demotivated because my teacher teaches solely through equations with no derivations or real explanations. It's been bugging me so much why half of the energy was just disregarded but now I get it. Thanks so much!
After seeing a lot of other videos about the same subject i got simply truly totally astonished by the different point of views used to explain it ! As i always say, there's never complete understanding if you can't describe things in different ways
When you increase distance between the plates whilst a battery is connected, you're still putting work so would the energy increase slightly but resultantly decrease?
I assume that because superconductors repel magnetic feilds, the capacitor would just act as a break in the circuit. Because the magnetic repulsion from the negative plate won’t be able to repel the electrons from the positively charged plate meaning there would be no current in that half of the circuit and the other half would be a dead end
@@ScienceShorts Thor's voice, is the same as yours but a little thicker. I mean the accent, the speaking style sounds almost the same but his voice is a little lower pitched than yours you could say... anyways, I'm just glad I got a reply from you... unreal moment for me
Aditya Misra Obviously he has made a mistake on his initial audio and has corrected it with a voice over instead of redoing that section. It does sound a bit weird but it gets the point across fine. 👍🏼
Hi I was just wondering at the end why is charge constant when the battery is disconnected? Is that assuming the the charge of the plates has got maximum charge /capacitance?
The AQA spec says you need to be able to describe the action of a simple polar molecule that rotates in the presence of an electric field. I'm not sure what this means and I don't think this video covered it?
I just revised this, so I can give you a brief description. Basically, the particles in a dialectric are called polar molecules, which means they have a positive and a negative end. Normally, these particles are jumbled up and facing in random directions, but when charge is created across the capacitor around them, an electric field is created. As you know, each side of the capacitor has an opposing charge, so each of the particles in the dialectric orientates themself, so that the positive end faces the negative end and vice versa. When this happens, each one of these particles has an electric field as they themselves have a difference in charge by virtue of being orientated, but because they're facing in the opposite direction to the capacitors electric field, it's an opposing electric field, which causes the overall electric field to decrease - this means that a lower voltage is needed to charge the capacitor, which in turn increases the capacitance through C = Q/V (i.e. as V decreases, C increases). Hope I was of help :)
One side has electrons (negative charge) the other side has a lack of electrons (positive charge). The electrons are attracted to the positive plate as it's oppositely charged and therefore flow towards it.
Using the equation Q=Q0e^-t/RC, if the capacitor is initially uncharged i.e Q0 = 0, then surely the equation doesn't work, as for all values of t, Q will be 0. Can somebody please explain this?
Muhammad. Husnain the charge moves to the positive plate. The charge moving to the positive plate decreases as time increases and as I=q/t, the current will decrease as a result. I hope that helps
Please post explanation of CIE syllabus Chapter 25 - Electronics Seems like you've missed it. I'm really struggling in it and need help in amplifiers and sensing devices.
6:56 to 7:27 Nope. The slower a capacitor is charged, the less energy is wasted. The one half factor is in there because each additional charge increases the amount of work needed to add subsequent charge, for an average work per charge of V_f/2 (assuming linearity). Meanwhile, energy loss can be modeled as a function of ESR or Equivalent Series Resistance, which is not at all directly related to the Capacitance. By charging the capacitor slowly, you keep the voltage drop IR due to resistance R low relative to the voltage of the capacitor, allowing most of the energy to be stored in the capacitor rather than wasted in the resistance.
If you apply *constant* voltage V to a capacitor C with equivalent series resistance R, then half of the energy gets dissipated in the resistance and the other half of the energy gets stored in the capacitance. Why: For a *constant* voltage, you can represent the total input energy as simply the product of voltage V and charge Q, which on an x-y chart looks like a rectangle, where the height of the rectangle is the applied voltage (V) and the width of the rectangle is maximum charge stored on the capacitor (Q=C*V). The voltage on a capacitor is typically proportional to Q (i.e. C is typically constant). So the energy stored in the capacitor increases simply with the square of the voltage, and this energy stored can be represented by the lower-right diagonal half of the rectangle with "area" (1/2)V*Q (where Q = C*V). The upper-left diagonal half of the rectangle represents the Joule heat loss incurred as the capacitor was being charged. So: If you apply a *non-constant* voltage in such a way that it is maintained only slightly above the voltage of the capacitor, up until the desired voltage is reached, then you can minimize the amount of energy you lose when charging the capacitor. In such a case, the area of representing the input energy would no longer be as simple as a rectangle with uniform height because the initial applied voltage would need to be lower than the final applied voltage. This cuts away from the "upper-left diagonal half of the rectangle" which "represents the Joule heat loss incurred as the capacitor was being charged".
bro where would we all be without you? I started to get so demotivated because my teacher teaches solely through equations with no derivations or real explanations. It's been bugging me so much why half of the energy was just disregarded but now I get it. Thanks so much!
