great video. i was sceptical of youtube vids for physics because i thought they would be over complicated because many of the creators have degree level knowledge and ability but these are perfect so far also my exact exam board. Thanks a lot please keep this style
Thanks a lot for the comment! Much appreciated, while I love complicated stuff, the most fun is breaking it down to an accessible level : ) Glad you found it useful and drop me a comment if something doesn't make sense!
Thank you sooooooo much, that was helpful, I'm forever grateful and ecstatic beyond measure that I found your channel, You're not making physics seems intricate as my physics teacher
Good summary. But you made a major mistake at 21:30. The electron will actually hit the lower plate after 9.5 nanoseconds so it will never leave. In the time you calculated it would actually travel a vertical distance of 2.6 m!
I used the formula s = ut + 1/2at² to find the time taken to hit the lower plate , with the values of s being 0.1, a = 4.4 x 10^15, and u being 0 ( as when it enters, there is no vertical velocity). I'm getting the value of 6.7 nanoseconds is it correct or have I made a mistake?
Something I was taught is that instead of using '1/4x pi x epsilon naught' you can use a constant k which is equal to 9x10^9. I find it easy to remember and gives you less numbers to imput. Thank you for the video!
The dimensions of this question is faulty, no? 21:20 because the vertical distance is 2.68 m, but the question gives it as 0.2 m. The answer wouldn’t exactly be giving the speed right after it exists the field then. So, anyway with the given dimensions, resultant velocity should be around 1.54 x 10^8 ms^1 right? Anyway, thank you so much for this. I love your explanations. You connect the concepts very fluidly, cover all most all the parts in the lesson in a breeze. I actually grasped a few of the lessons I considered super confusing, in less than a day before my test, thanks to you. Thank you so much
Thank you for the comment! I agree, I remember when I filmed this video I made up the problem as I was writing. Glad the video is useful overall and thank you!
hi at 19:48 is u=0? if s=0.2m why cant we use v^2=u^2+2as, I tried this way and got a different value for the vertical velocity thanks for the videos 😄
studying for my exams for foundation year electrical engineering and i just want to thank you so much for this video, such clear and useful information
Hi sir, in 17:20 , since F in Newton's 2nd law is the resultant force would you not need to take into account the gravitational force acting on the electron and substitute F = ma - mg ? Or do we just assume the gravitational force to be negligible? help pls I don't understand :)
This doesn’t really have anything to do with electric fields but when an electron is moving through two parallel plates, if the length was increased, could it travel faster than speed of light, similar to minute 19:58
excellent question. In that particular example btw, the acceleration is above the number of the speed of light, but the acceleration acts for a very small fraction of a second, so it does not accelerate it to that value. To answer your question though - we are always bound by the speed of light. In particle accelerators like CERN though they are using similar principles to accelerate particles as close as possible to the speed of light at higher and higher energies, getting to the speed of light though is impossible for particles with mass.
@@zhelyo_physics again this is very theoretical😂 but if an electron is acting like a wave, does that mean it is massless and so can travel at speed of light or higher?
no worries, I need to make a video about this. Nope, all particles seem to have wave like properties (including the photon), but the massless particles go at the speed of light, the particles with mass are bound below it.
Excellent question, so this actually can be derived and it stems from the fact that E=V/d for uniform field and this is the field for a parallel capacitor. Since Q=CV, Q=CEd i.e. we can see that C=QE/d, to derive it for a parallel plate capacitor you need to find the electric field which is a little beyond the scope of the course but can be found using what is known as Gauss's law. Intuitively, if the plates are closer, d is smaller, you would be able to store more charge per unit volt as the field is stronger. Hope this helps!
I love your videos but I wanted to clear my confusion up if you don't mind all I have is just 1 and a half months now from my a2 physics exam and is the concept covered in your playlist enough to tackle the past paper question?
Hi sir, at the end of the video when you say "this was pretty much the entire OCR A specification" does that mean there are still things I need to revise in this module?
so I follow all the points on the OCR spec when I make these videos however as always it is a great idea to have the spec at hand and to be ticking off and checking you have covered everything. I also recommend lots of practice questions such as these: th-cam.com/video/Sigq6SJwJgo/w-d-xo.html good luck! : )
nope, the electric field is defined as the force per unit charge the particle will experience at any point. A negative particle will be attracted to the positive plate at each point, whereas a a positive particle will be repelled and attracted towards the negative. Hope this helps! : )
You forgot to mention potential difference from an electric radial field, exact same as electric field strength in a radial field, just replace E with v =, and r^2 is now r.
For the insulator of relative permitivity equation, is that actually in our spec, Bcz it’s not in our formula booklet ( ocr A ), and I have a feeling they will make a question with capacitors and electric fields.
Hi sorry to bother you again, on the ocr 2020 paper 2 q 23 a I can't seem to understand why you have to multiply by sin60, surely the electric force is in all directions of each proton. Why does it only have a vertical magnitude in the question?
excellent question! So in this charge triangle, some the electric force from one of the charges below will be towards P and a little to the right, the other one towards P and a little to the left. The horizontal components will be equal and opposite and they will cancel out leaving only a repulsive vertical component. Hope this makes sense! Let me know if it doesn't.
@@zhelyo_physics Sorry i don't really understand that, Is there a different way you could explain or? If I'm using up too much of your time don't worry btw!
No worries! I would draw out the two vectors along the triangle. Then resolve them into vertical and horizontal components (the horizontal ones will be equal and opposite and should cancel.)
@@zhelyo_physics Hi sorry I just saw this comment, So basically I just have to imagine theres only a horizontal and vertical component of electric force, (but the horizontal cancels)? This does make sense to solve the question however I just dont get why the force directing towards the top point in the diagonal direction is not the same as the vertical component as I thought they were always the same in every direction.
1) yep! The whole field of electrostatics is based around it and the force is given by coloumb's law. 2) a stationary charge will always have an electric potential, as another charge around it will have the potential to move. The force is transmitted by the electric field, so it doesn't matter if it is vacuum of or not. Hope this helps! : )
So if you are referring to the potential, V=Energy/Charge and Energy is proportional to 1/r rather than 1/r^2 as introduced at 24:21 . Hope this helps!
Calculate the speed of an electron accelerated from rest through a distance of 40 mm by a uniform electric field of 3.0 x 10 3 NC^-1. I tried doing the question but the answer I got isn’t matching with the given answer.
so find it's acceleration using F=ma, EQ=ma, giving us a=EQ/m , after this, use the suvat equations: v^2=u^2+2as, initial speed=0, so v=sqrt(2aS)=sqrt(2EQ/m * S) plug in all the numbers converting the mm to m and you will get the right answer : )
@@binodtharu4910 did you get around 6.5*10^6 m/s? That should be the right answer: sqrt(2*3.0*10^3*1.6*10^(-19)*40*10^(-3)/9.11*10^(-31)) probably a calculator mistake
Argh, the question is wrong then. The speed cannot exceed the speed of light (well questions that showcase this shouldn't be designed). Must be a typo in the mark scheme especially, is it from a past paper?
So this is from the A Level syllabus not AS. Depending on the school though, as it's a two year course many institutions might teach it at a different time.
Hi guys! I have also filmed some past paper questions for practice: th-cam.com/video/Sigq6SJwJgo/w-d-xo.html
Yeah,am a surgical physicst 😎.
Finishing so many chapters the night before the exam wouldn't have been possible without you :3
Good luck tomorrow! Glad this was useful!
LMFAOOOOOO were all out here😭😭😭
Good luck guys! You've got this!
It is 2am right now and I am doing the same thing and I hav exam in the morning
@@mhasultanmohammed4552 Good luck in the exam!
"Welcome back physicists " oh my God this phrase is another whole motivation , thank youuu
heh thanks for the comment! Glad to hear so!
I've been hearing it wrong this entire time..I thought he said "welcome back to physics"
great video. i was sceptical of youtube vids for physics because i thought they would be over complicated because many of the creators have degree level knowledge and ability but these are perfect so far also my exact exam board. Thanks a lot please keep this style
Thanks a lot for the comment! Much appreciated, while I love complicated stuff, the most fun is breaking it down to an accessible level : ) Glad you found it useful and drop me a comment if something doesn't make sense!
@@zhelyo_physics will do thanks again
watching this before my paper 2 exam so i can compensate my horrible performance in paper 1
REAL
Good luck
is electric fields not a part of the paper 4 syllabus?
@@anujbhattarai21paper 4? Which exam board are you?
What did you get at the end?
What are the hardest topics?
Thank you sooooooo much, that was helpful, I'm forever grateful and ecstatic beyond measure that I found your channel, You're not making physics seems intricate as my physics teacher
I am glad you are finding them useful! : )physics is fun!
@@zhelyo_physics You’re right,physics fun while your teaching thank you again
lol
YOUR VIDEOS HELPED ME GETTING A* in PHYSICS！！！！！
Amazing result! Well done and thanks a lot for your comment!
@IslaBetPoker what tips would you give to ppl who are doing the exam?
Good summary. But you made a major mistake at 21:30.
The electron will actually hit the lower plate after 9.5 nanoseconds so it will never leave. In the time you calculated it would actually travel a vertical distance of 2.6 m!
Of course, I was testing if any of my viewers will spot this : ) Thanks a lot for mentioning this!
@Tycho Photiou How did u get those numbers
i got 7.5 ns as it only travels 0.5 m, how did you work out the distance
I used the formula s = ut + 1/2at² to find the time taken to hit the lower plate , with the values of s being 0.1, a = 4.4 x 10^15, and u being 0 ( as when it enters, there is no vertical velocity). I'm getting the value of 6.7 nanoseconds is it correct or have I made a mistake?
Honestly you deserve so many more followers, can't wait till your account blows up remember me when it does haha :)
Haha of course! Thanks a lot for the kind words and support to the channel!
Something I was taught is that instead of using '1/4x pi x epsilon naught' you can use a constant k which is equal to 9x10^9. I find it easy to remember and gives you less numbers to imput. Thank you for the video!
Yet another EPIC ZPhysics W
thank you!
The dimensions of this question is faulty, no? 21:20 because the vertical distance is 2.68 m, but the question gives it as 0.2 m. The answer wouldn’t exactly be giving the speed right after it exists the field then.
So, anyway with the given dimensions, resultant velocity should be around 1.54 x 10^8 ms^1 right?
Anyway, thank you so much for this. I love your explanations. You connect the concepts very fluidly, cover all most all the parts in the lesson in a breeze. I actually grasped a few of the lessons I considered super confusing, in less than a day before my test, thanks to you. Thank you so much
Thank you for the comment! I agree, I remember when I filmed this video I made up the problem as I was writing. Glad the video is useful overall and thank you!
@@zhelyo_physics Absolutely, thank you for the great content.
hi at 19:48 is u=0? if s=0.2m why cant we use v^2=u^2+2as, I tried this way and got a different value for the vertical velocity
thanks for the videos 😄
hmm, same here. Let me investigate this. Very interesting.
its because you dont have S it wont necessarily have travelled 2cm down because it has entered the field at some unknown height
Hey! Just wanted to say thanks. Your videos are very helpful! 😊
Thanks a lot for your comment! Much appreciated!
thank you so much , you explained it very beautifully
Glad to hear this is useful!
studying for my exams for foundation year electrical engineering and i just want to thank you so much for this video, such clear and useful information
thank you so much for the comment and best of luck on your exams!! : )
Hi sir, in 17:20 , since F in Newton's 2nd law is the resultant force would you not need to take into account the gravitational force acting on the electron and substitute F = ma - mg ? Or do we just assume the gravitational force to be negligible? help pls I don't understand :)
Yes that is indeed the case. The force of gravity is 10^36 times smaller so it is practically negligible
You are the best explainer. I am from India
Thank you for the comment! Much appreciated!
This was toooooo helpful
Thank you so much 🥺
Thanks a lot!
more helpful than my teacher fr
For the answer at 21:00 shouldnt it be 0 since it moves upwards by 0.23m during its time in the accelerator and thus hits the plate?
This video helps me a lot. Thanku😊
thanks a lot! much appreciated!
for the equation in 8:11 if we want to find electric field strength for q we need to put big Q in the equation?
In E=Q/4pier^2 Q is the source of the electric field, in F=Eq q is what feels the electric field. Hope this helps!
underrated
thanks a lot! I have a playlist here of similar videos on all the topics: th-cam.com/play/PLSygKZqfTjPC3hJ7nRSnnXTw3tI_o67dR.html
Thanks very much sir this video really help me
Anytime! Glad this was helpful!
thank you very much .
anytime!
This doesn’t really have anything to do with electric fields but when an electron is moving through two parallel plates, if the length was increased, could it travel faster than speed of light, similar to minute 19:58
excellent question. In that particular example btw, the acceleration is above the number of the speed of light, but the acceleration acts for a very small fraction of a second, so it does not accelerate it to that value.
To answer your question though - we are always bound by the speed of light. In particle accelerators like CERN though they are using similar principles to accelerate particles as close as possible to the speed of light at higher and higher energies, getting to the speed of light though is impossible for particles with mass.
@@zhelyo_physics again this is very theoretical😂 but if an electron is acting like a wave, does that mean it is massless and so can travel at speed of light or higher?
no worries, I need to make a video about this. Nope, all particles seem to have wave like properties (including the photon), but the massless particles go at the speed of light, the particles with mass are bound below it.
@@zhelyo_physics okay thank you, looking forward to the video!
great vid !!
thank you for the comment!
Hi, Could you please explain why the capacitance of a capacitor is inversely proportional to the distance between the plates?
Excellent question, so this actually can be derived and it stems from the fact that E=V/d for uniform field and this is the field for a parallel capacitor. Since Q=CV, Q=CEd i.e. we can see that C=QE/d, to derive it for a parallel plate capacitor you need to find the electric field which is a little beyond the scope of the course but can be found using what is known as Gauss's law.
Intuitively, if the plates are closer, d is smaller, you would be able to store more charge per unit volt as the field is stronger. Hope this helps!
@@zhelyo_physics thank you 👍🏾
I love your videos but I wanted to clear my confusion up if you don't mind all I have is just 1 and a half months now from my a2 physics exam and is the concept covered in your playlist enough to tackle the past paper question?
I generally recommend supplementing my videos with a textbook, the syllabus as a checklist and also every possible past paper question out there : )
13:45. Shouldn't the field lines at each end be bent slightly?
only around the edges potentially. We assume the field between plates is perfectly uniform.
Hi
At 26:57 you substituted the electric potential for the potential difference. Are these two the same?
In this context yes.
Hi sir, at the end of the video when you say "this was pretty much the entire OCR A specification" does that mean there are still things I need to revise in this module?
so I follow all the points on the OCR spec when I make these videos however as always it is a great idea to have the spec at hand and to be ticking off and checking you have covered everything.
I also recommend lots of practice questions such as these: th-cam.com/video/Sigq6SJwJgo/w-d-xo.html good luck! : )
watched this in 2x speed to copmlete in 15 mins
Amazing! Glad this is helpful, excellent bit of speedy revision there!
🐐
does anyone know the application he i using to make these videos?
I do! It's just Microsoft Whiteboard, which is a free software that often comes with windows.
@@zhelyo_physics thanks a lot!
Sir, when an electron enters a uniform field, will its downward motion towards +ve plate not be opposed by the direction of the electric field?
nope, the electric field is defined as the force per unit charge the particle will experience at any point. A negative particle will be attracted to the positive plate at each point, whereas a a positive particle will be repelled and attracted towards the negative.
Hope this helps! : )
@@zhelyo_physics thank you sir. So when finding the speed of the particle in both instances, theoretically they should be equal right sir?
You forgot to mention potential difference from an electric radial field, exact same as electric field strength in a radial field, just replace E with v =, and r^2 is now r.
thanks for mentioning
For the insulator of relative permitivity equation, is that actually in our spec, Bcz it’s not in our formula booklet ( ocr A ), and I have a feeling they will make a question with capacitors and electric fields.
It's not in the formula booklet but I have definitely seen it asked in questions. We definitely need to know it.
when u worked out Vy was the t meant to be squared in the suvat equation
which bit of the video exactly? for the suvat equation I used v=u+at in the y direction and no square there. Hope this makes sense!
Hi sorry to bother you again, on the ocr 2020 paper 2 q 23 a I can't seem to understand why you have to multiply by sin60, surely the electric force is in all directions of each proton. Why does it only have a vertical magnitude in the question?
excellent question! So in this charge triangle, some the electric force from one of the charges below will be towards P and a little to the right, the other one towards P and a little to the left. The horizontal components will be equal and opposite and they will cancel out leaving only a repulsive vertical component. Hope this makes sense! Let me know if it doesn't.
@@zhelyo_physics Sorry i don't really understand that, Is there a different way you could explain or? If I'm using up too much of your time don't worry btw!
No worries! I would draw out the two vectors along the triangle. Then resolve them into vertical and horizontal components (the horizontal ones will be equal and opposite and should cancel.)
@@zhelyo_physics Hi sorry I just saw this comment, So basically I just have to imagine theres only a horizontal and vertical component of electric force, (but the horizontal cancels)? This does make sense to solve the question however I just dont get why the force directing towards the top point in the diagonal direction is not the same as the vertical component as I thought they were always the same in every direction.
@@thevinecompany7733 my man got aired
Hello sir! In Fg, we do put the negative sign. But do we have to answer it in negative or just simply take the magnitude?
excellent question, in this case we were looking at the magnitude only to compare them the forces.
@@zhelyo_physics oh ok ok thank you so much sir! Your videos are really helping me with A2 physics!
Can a stationary charge experience an electric force? and does a stationary charge in a vacuum have zero electric potential?
1) yep! The whole field of electrostatics is based around it and the force is given by coloumb's law.
2) a stationary charge will always have an electric potential, as another charge around it will have the potential to move. The force is transmitted by the electric field, so it doesn't matter if it is vacuum of or not. Hope this helps! : )
Mock in 20 mins lads, 1.5x speed lol
Hope it goes well!
@@zhelyo_physics it didn't, further mechanics hurts my soul
25:32 why isn't "r" squared this time?
So if you are referring to the potential, V=Energy/Charge and Energy is proportional to 1/r rather than 1/r^2 as introduced at 24:21 . Hope this helps!
@@zhelyo_physics how come Energy is proportional to only 1/r, rather than 1/r^2? Thank u so much for this video!!
Calculate the speed of an electron accelerated from rest through a distance of 40 mm by a uniform electric field of 3.0 x 10 3 NC^-1. I tried doing the question but the answer I got isn’t matching with the given answer.
so find it's acceleration using F=ma, EQ=ma, giving us a=EQ/m , after this, use the suvat equations: v^2=u^2+2as, initial speed=0, so v=sqrt(2aS)=sqrt(2EQ/m * S) plug in all the numbers converting the mm to m and you will get the right answer : )
@@zhelyo_physics still I’m not getting the required answer
@@binodtharu4910 did you get around 6.5*10^6 m/s? That should be the right answer: sqrt(2*3.0*10^3*1.6*10^(-19)*40*10^(-3)/9.11*10^(-31)) probably a calculator mistake
@@zhelyo_physics I got that as answer but the answer given is 1.03 x 10 ^ 31
Argh, the question is wrong then. The speed cannot exceed the speed of light (well questions that showcase this shouldn't be designed). Must be a typo in the mark scheme especially, is it from a past paper?
is this for AS Level/ 1st year of A Levels?
So this is from the A Level syllabus not AS. Depending on the school though, as it's a two year course many institutions might teach it at a different time.