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! ;)
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!
@@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
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
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?
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".
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 :)
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.
@@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
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
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!
Sharpie should really sponsor this man
Idk... it might be a bit of an anti-advert - the pen never seems that good in these videos.
@@namename3130 he uses yhem so much the ink dries out so looks faded which shows his intense undying love for sharpies
🤣
@@namename3130 probably cause he uses them so much lmao
Please ignore my comment about energy being 1/2QV due to resistance - nonsense!
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?
Sorry if the answer is really obvious, we haven't been taught capacitance yet.
@@Annie-xb8xx It's probably derived using differentiation.
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.
Brain : No , that's the only thing I am gonna retain
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
It's deffo going to be in multiple choice.
we get advance info before exams?? damn I didnt know that. is this for all exam boards? (I do AQA)
@@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.
Good luck for everybody sitting A level Physics in 9 days time :)))
Gonna need it
@Matthew Briggs it did the one about a about parallel plates, wtf
12 hours*
thank you brian chan
How about 2hours?
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!!!
+themexican sob Thanks man! No, I don't have a website, but feel free to watch the adverts all the way through to give me an extra few pennies! :)
@@ScienceShorts I won't skip ad in your videos hereafter!!! =D
@@ScienceShorts better yet, click on the ads, cause its the clicks that give the money
@@tychophotiou6962 yes
Thank you so much for these videos. Prepping for my exam Friday!
same bro (2023 ver.)
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
One of the best videos I’ve seen on capacitance
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! ;)
@@ScienceShorts how can i donate to you
@@attedau6235 link in description
Thank you so much for the video! I struggled with it a lot but now it is as clear as it can be!
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!
Boys we got just over 1 day, best of luck
I am dead 2morrow
@@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
Zeyn which exam board are you
@@zeyn4792 hehe i will try. Best of luck man. I will come here tomorrow afternoon to complain about my faliure lol
@@danielb1500 AQA, bit shitty they pushed the exams into may rather than june as it always has been
I always appreciate that you explain things differently so I can have 2 sets of notes on the same topic :)))
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
your video has being extremly helpful to me,thx man appreciate the hard word and the effort you put in your video
Hey man thanks for that vid! You just explained how to derive those formulas so much more simply than my lectures do
if I get an A in physics I owe you a pint
How did you do?
?
Relationship questions you missed the word “if” at the beginning of the comment.
Relationship questions yikes
He is not replying
he probably failed with flying colours
here for last minute revision
In 15 min this guys literally explained what my teacher couldn't explain in 2 weeks.
Lesgoooo, in'sha'Allah gonna get an a*
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?
Would you be able to do one for lenses, particularly ray diagrams? Thank you
you are better than my teacher jeez
Thank you for the video. By the way, I think the most common way to denote the time constant is using tau.
Thank you so much, I always struggle with this type of topic and this video was a big help!
13:30 nice voice over :0
Seamless ADR
at 2:00 you make the switch which is open it doesn't mean the electrons cant flow back to the +side of the capacitor?
7:20 I honestly don't know what I'm talking about but what about a superconducting circuit. I wonder whether energy stored would increase to near QV
That’s what I was thinking
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
that was good revision, thanks!
Why does the current decrease exponentially as the capacitor gets charged and discharged could you please elaborate that? Would be a great help TIA.
Brilliantly explained, thank you!
THANK YOU FOR THE AMAZING CONTENT!
That’s a hell of a video
man this is really helpful
Capacitors are the most interesting things in the world to me idk why
Really nice work!
Hi Sir, please skip to 13:33 I think you had some editing issues. But dont take video down as I got my exam in 2 weeks. after that then feel free.
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. 👍🏼
FINALLY UNDERSTOOD IT! THANK YOU!
Lads good luck for the final exam
love your videos but ive got a question :why does the capacitor discharge slowly in the rectified ac current
Because it is discharged through a resistor, different from charging.
Can you please make a video on graphs related to magnetic flux in a transformer and hall probe?!
Your videos are very good!
Can you make a video on forces exerted by pistons please?(By the way love your videos)
Why does disconnecting the battery create a constant charge? I don't know if you still reply on these older videos
Because where would the charge go?
Very helpful. thank you.
so, we meet again
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?
If the capacitor is disconnected, where could the charge go?
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.
Lol.
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".
@@ScienceShorts absolute legend
l watched your video, also I made some electronic using proteus 8, thanks.
Where did the equation E=0.5Q/C² COME FROM
Does this topic come in edexcel A level physics? My edexcel cgp revision guide doesnt cover capacitors.
No, it's not in Edexcel.
@@ScienceShorts It is. It is in the electric and magnetic fields section of Edexcel.
Love your videos but i hate the sound of sharpie! Can you find different pens?
😂
BEST VIDEOS EVER
Thank you, very helpful!
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 :)
Thanks! It's something that wasn't covered by my teacher so I was a bit confused when reviewing the specification
At 10:27 did you PUT ln accidentally or SAY log accidentally because I'm a bit confused ..
+Kaumudie Athukorale saying "log" is correct for both natural log (ln or "lun") as well as log (base 10)
Ahh, thanks
In a certain question, it required to find area under the exponential graph of current/time .. how can I calculate this?
intergrate the function
These videos are great!!! - can you do some on OCR A-level chemistry if thats possible :)
Thanks
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.
same. did you find any other videos on it?
Professor I thought the energy is half because it doesn't start with q and v, but dq and dv? Like the calculus?
You're right.
can someone explain this to me? what does d mean?
life saver !!
He sounds like Thor (Chris Hemsworth) but with a thinner voice
Thinner?!
@@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
I can put it on like he does if that's what you want 😂
😂😂😂😂
*Dialectric* i love this!
I always make that mistake! 🙄
@@ScienceShorts you make the mistake of replying after 3 months😂😂😂😂😂😂
JK!
8:38
god dammit
AMAZING
What happened at 13:33? 🤔
I said less accidentally 🙄
0:23 why does the battery take electrons from one plate to the other?
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.
Can you explain safe pd?
LOVE IT
i lov3 you so much it hurts
rip 2020 seniors
;-;
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?
Charging and pd follow the equation Q = Q0 (1-e^-t/RC)
can someone explain to me what e is
Q0 is max charge- Q is the charge at a given time. When t=0, Q=Q0(1-e^-0/RC), Q=Q0(1-1), Q=Q0x0=0
where does the equation at 14:22 come from?
rearrange the earlier Q=VC for V=Q/C, put it in E=0.5xQV to get E=0.5xQx(Q/C), which is the same as E=0.5xQ^2/C
why does current reduce when a capacitor charges?
Cuz it gets full...
We don't have capacitance in GCSE
But we do have it in A2
edexcel IAL?
Can you please do a video on kinematics graph questions for as levels
Without u I’d fail my a levels
At 1:56 the circuit is not complete and glow the bulb
why does current decrease when capacitor discharges
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
Im so fucked
13:33 Greater than XD
👀
my man sound like the rightful king of asgard..
Teach slowly , informative video
ty dud
I don't think you can state "Charge stored". Only energy is stored .The net charge in a capacitor is zero.
Splitting hairs. That's the same as saying it's wrong to say that a hydroelectric dam stores water.
U legend
Re upload ?
13:33
Decent
Fundamental error in this video.....time constant is "RC" NOT 1/RC!!!!!
isnt that what he said?
Mr j day?
@5.00 I think you mean CHARGE stored per unit potential difference, not 'charged' XD
:'(
damn that's harsh checking
if i draw this 3d
proceed to draw a 3d image
physics 4 in 17 days...
Sir a little bit confusing
Edexcel peeps sitting for unit 4 tmr, where you @
Helpful but youre going to fast
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!
where are u now ? (just curious)