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

🤣

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

Gonna need it

@Matthew Briggs it did the one about a about parallel plates, wtf

12 hours*

thank you brian chan

How about 2hours?

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.

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

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

+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

+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

same bro (2023 ver.)

Seamless ADR

so, we meet again

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

Because it is discharged through a resistor, different from charging.

If the capacitor is disconnected, where could the charge go?

I always make that mistake! 🙄

@@ScienceShorts you make the mistake of replying after 3 months😂😂😂😂😂😂 JK!

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

intergrate the function

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 😂

😂😂😂😂

You're right.

can someone explain this to me? what does d mean?

Because where would the charge go?

+Kaumudie Athukorale saying "log" is correct for both natural log (ln or "lun") as well as log (base 10)

Ahh, thanks

No, it's not in Edexcel.

@@ScienceShorts It is. It is in the electric and magnetic fields section of Edexcel.

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

same. did you find any other videos on it?

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.

😂

;-;

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

god dammit

Cuz it gets full...

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

I said less accidentally 🙄

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. 👍🏼

👀

edexcel IAL?

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

Splitting hairs. That's the same as saying it's wrong to say that a hydroelectric dam stores water.

isnt that what he said?

:'(

damn that's harsh checking

where are u now ? (just curious)