Hi The Watcher. I don't know if I am a genius, but what I try to do is to unlock those that could be, by clearing the path a little. I am glad you enjoyed my work!
Hello Xeku. I am happy I put you on the right track! it is one of the goals of the channel, provide students with the wings they need to fly by themselves!
Thanku for this amazing explanation.....💜 I am very confused with some topics of physics because i am not able to feel what is going on ..but ur videos help me a lot to understand each and every thing 💕💕 thank u again Make more content we love it... 👍
At about timestamp 5:45 there is a, somewhat unknown, use for this circuit (since the cap is first in line). Suppose the switch is in the MIDDLE position and the capacitor is totally discharged. IOW the potential on both cap terminals is 0V. Now, keeping in mind that the VOLTAGE across a cap can not change instantly (just as the CURRENT in an inductor can't change instantly), if you flip the switch to the battery (say it's 12v), then the BOTTOM of the cap, as well as the top is instantly 12v, since there is no potential DIFFERENCE between the plates. Now the top plate of the cap STAYS at 12v, while the bottom plate 'charges' the cap through the resistor by going toward 0v via the RC rate. This is a GREAT way to apply a fast 12v pulse to switch on an SCR, etc. without any timing circuitry...here a hi-value resistor can be placed across the cap for auto-reset (the cap used in this type of circuit is usually very small...~0.01uF. to 0.1uF). I have used this method, and have also seen complicated circuitry where this would have greatly simplified it. A great example of combined knowledge in the academic and typical circuit app of a capacitor from the FIELD point of view! THANKS AGAIN, --dALE
Please correct me if I am wrong. Can the Capacitor have more charge's density in its plate, compared to that of the of the battery's side?. And if so, how the electric field would allow such un-equal distribution of charges on the same wire ( from higher potential of battery to the higher potential of the capacitor's plate. Also, is it the reason why the capacitor can produce a lot of power in a very short time?
Hi ggg. Higher the charge density, higher the energy each coulomb of charge will carry. So to answer your question, no. The charge density of a capacitor's plate, once it has reached its steady state (= once fully charged) should be the same than that on the positive plate of the battery. In other words, the electric potential should be the same. The reason why a capacitor can deliver a lot of energy in a short amount of time (high power), in the right conditions dependent on capacitance and resistance of the circuit, is because the charges are already on the capacitor's plate, ready to go. In a battery these are generated by a chemical process that has its own kinetics.
Thank you for your kind comment. I checked your channel. 8 years old, coding an interesting game from scratch like Mathonoids. Wow! I am stunned! You beat me, my first game coded, I was 11 (in 1982!). But I was made to stop coding at around 14: My school grades were dropping, and my parents seriously limited access to the home computer (a TRS80). When I was older, around 17-18 yrs, windows 3.1 and PCs had arrived (beurk, I didn't like this), and in the meantime, I had developed other hobbies, so never went back to game coding. Conclusion: continue, but make sure your grades at school remain OK to preserve your momentum :-)!
This was a little more complicated video on circuits and rather a rapid introduction to different aspects of electricity while the first 2 videos in this serie started slow and beginner friendly
teacher at 8:21 we considered that Vc=0 because no charges has arrived yet on the capacitor .but why Vr has a value? , why we didnt consider that no charges have arrived on the resistor also ? i hope that you clarify this for me .
Hello Anass, This is a good question. We are in a transition mode. Charges have not moved yet, but they want to. Before it charges, the capacitor can be considered like an equipotential wire (it will let pass current with no resistance)… so the electric potential will be the same on both plates: that of the positive side of the battery. V (before C) is epsilon, V (After C) is epsilon, so Vc = 0, thus Vr = Epsilon. Then as current passes, the electric potential of the lower plate will decrease, because some of the charges on the lower plate are pushed by those that get stuck in the higher plate : that is when Vc starts increasing. It is very important that you understand clearly what is an electric potential to understand my answer (If you cannot explain what an electric potential in a few words in your mother tongue, then watch the video about that on this channel). I hope this clarifies things !
Thank you for the video, when is the next capacitor video coming? It would be nice if you could also explain intuitively how capacitors work in series and parallel
Hi GodFred, Good suggestion. I'll put in near the top of the wishlist. Unfortunately other priorities showed up, so I had to slow down posting (one 15 minute video is a full time job for at least a week!). Yet, I would like to produce more soon, and guess that Capa in parallel and series could be one of the next videos...
12:24 initally Vc is epsilom but why is Vr epsilom initially if when we charged the capacitor Vr went down to 0. I understand you said kirchoffs 2nd law but i dont get how Vr suddenly changed from 0 to epsilom Thanks
Hi N, When the capacitor is charged (and still connected to the battery: no current flows through the circuit (there is no incentive for charges to move around since the electric potential is the same at the positive terminal of the battery and the positively charged plate of the capacitor. So, no current, hence, no potential drop at the resistor. Once you flip the switch on position B. The capacitor discharges: charges move from the positive plate to the negative plate via the resistor (note that I use here the conventional model, in reality, electrons move from the negative plate to the positive plate). So as soon as you flip the switch, you have a maximum current flowing through the resistor, hence the potential drop Vr. This potential drop will decrease in time at the same rate at the voltage of the capacitor Vc to end up at 0, when the capacitor is fully discharged (no more current flows in the circuit). I hope this helps!
Sir thank you for an excellent explanation . Can you please elaborate at time stamp 2.42, how I= omega C V, what is the frequency of the current calculated?
Hi Wu, The current in a capacitor can be calculated by I = CdV/dt. So for current to pass, the voltage needs to change with time. If the voltage is a sine curve (V = Vmax * Sin(omega*t), you see that omega will come out when you differentiate the voltage. so here you go I(t)=omega * C *V(t). I hope this helps!
New sub here. Thanks for your channel. Can you link me to a clarification of the use of/for 'adjacent capacitors' Some years ago(2013). I inadvertently constructed an LC circuit. At that time I had no idea what I was doing. But now seeing what I've learned in this vid, I may be able to take advantage of my earlier findings. My intent is to charge and discharge two capacitors to do work via 2-coils, each in succession, charging and discharging at a rate of 8.63hz. Thanks in advance for your reply.
Hi, well, I didn't know. It's a great circuit to truly understand how a capacitor behaves, so it doesn't surprise me that it can show up in various high school programs, and their related exams.
Sir I have a small but for me mind boggling confusion about work and energy. For this if we consider an isolated part of the universe where no force is acting just there are one object and me. From a reference point, the box is moving with a uniform velocity 'u', and then I applied a force 'f' on the object when it was moving with uniform velocity at time zero. After sometimes when the displacement 's' occurs at time 't' I stop applying force. So by the formula the work done by me on that object should be f times s, but sir here the object had some initial velocity u, so in the displacement there should be some part of it that is not done by me, by the formula (s=ut+1/2at²). So by conjectures it seems that the formula says that I have done more work than I have actually done. Is it so? Where I am wrong sir?
Consider case A, initial velocity 0, and case B, initial velocity u. In both cases You apply a constant force f on the object for the same amount of time, t. In case B, you will have travelled a longer distance than in A in the time t, because you already had an initial velocity u. Therefore, the distance travelled s has to be larger. Conclusion, In case B, you will have provided more work. If you need to be convinced, just calculate the change in KE in A and compare it to B. for B the change in KE will have extra positive terms than for A. Or if you prefer avoiding too much algebra, you can also do this with numbers that you set yourself…
@Physics Made Easy sir but the quantum mechanics says that gravity is one of the four fundamental forces and its particle is graviton, so if it is true then we should not feel any type of gravitational force from black hole because when ,from inside the event horizon, a black hole emits a particle of graviton it will never exceed that line. So does it resolve that gravity is not a force and if so then why we are still looking for particle of gravity? I have heard on many videos that gravity is made up of particles so that is why this question came in my mind.
A black hole is a sphere from which nothing can leave because of the limitations of motion in our Universe’s space time. Causality from the inside to the outside is broken. This is because the inside of a black hole is NOT in our space time (our Universe… ). So the singularity itself does not interact gravitationally. What does, is the surface. You have to consider the part of the back hole in our universe like a big black sphere (or more precisely, a sphere of radius just a little larger). What exerts gravity in our Universe, is not the singularity itself, it is the surface of that sphere … So if gravitons are real, they will be exchanged between an object in our universe and the surface of the black hole. But again, as I told you in a previous comment, that object (the surface of a black hole) is macroscopic… so trying to understand things with quantum mechanics here is not a good idea (until we have something better). Here you should use general relativity… Oh, and when you watch a science video, always pay attention to the context of the video and be careful when extrapolating. what you learn elsewhere (it might not apply). Also keep a critical eye... More the title looks appealing, more you should be cautious... I see things sometimes that make me jump out of my chair lol!...
@Physics Made Easy sir and also I have one doubt about angular displacement. Sir when our vehicle makes a sharp turn we feel that we are pulled on the side opposite to the direction of turning and generally the reason for this is given as due to inertia, our body tends to move in one direction so we feel that pseudo force also called centrifugal force. In this case if we just look at our vehicle so then it makes an Arc of a circle, and if you want to find the angular velocity of the car , which is v/r or theta/time, we will require its radius because we can easily measures the velocity on speedometer said so in this case how can we calculate the radius? And if sir we can just measure the angular displacement then we can easily measure all the angular quantities so how can we calculate that how much of degree or radian that car has covered because we don't know the radius?
If you know the coefficient of friction (MU) and the linear velocity, you can figure out the angular velocity (w). N = mg f = MUxN = MU x mg f is centripetal, so MU x mg = m v^2/R so MU x g = v^2 /R From that you can deduce R, and then w.
Hi , i i want know one thing is it compulsory to make notes of physics ? Is essential if i want to know physics in deep clear all my doubt i study because love studying physics but i am serious about physics so can you just answer my question , by the way your videos are just awesome
Hi Haider, Your question is more about How to Study in general than about a Physics. Since I have been tutoring for years many students between 15 and 20 years old, I have learned that… it depends on the student! Most students take really bad notes during a lesson, and taking these takes away their attention from what the teacher is saying and leaves them confused when they try to understand their notes later. Some others, more rarely, will take excellent notes and will be able to use them to recollect what they have seen in class. What I would advise when a student is a bad note taker: minimal notes - just note down the main points (like titles of sub-topic, main definitions and main equations), so you can focus your attention to what the teacher is saying. Then, the same day, at home, take your notes, and a text book next to you. The notes are just for remembering what you need to study and know.. With the help of the book, write your own lesson like if you were trying to teach it. That is for the lesson, when it comes to exercises and worked examples, I would try to take deeper notes that at least guide you towards the solution when you try and redo the exercise. For my videos, don’t take notes the first time you watch them. Just focus on the video. You can always take some if you want when watching a second time. By the way, thanks for the kind words :-)
@@courage936 @Haider Abbas Kazmi It's funny, sometimes I have to scold my students...not to take notes while I am teaching! What I do, is that when I finish a demonstration or the presentation of a concept, i give them a few minutes to take notes. If the demonstration is too long, I will cut it in little bits... to give them multiple note taking periods. It allows them to make a synthesis of what was said, And I let them ask questions (because it's often then that they realize that they have some to ask...). Then sometimes, I ask them to read their notes to me. So I can check if they really got it! Full focus of the student during the lesson itself + full focus of the student when he takes note (the student can digest the info more effectively). Magic technique maybe you should ask your teachers to apply this?
Yes, it is the negative charges that moves, but in the conventional model used by all, a current is defined by the motion of positive charges. (in other words, negative charges moving to the left is mathematically identical to positive charges moving to the right)
Hey there, thank you for noticing and asking. It's appreciated. I was actually moving country, and at the same time, moving from a large city to a countryside village, recreating my teaching company in the new country, buying a first house + land, working on it (there was a lot to do!), taking care of my real life students (that consequently became online students), socializing myself with the locals. I basically changed life! With all this happening, I had to slow down with the Physics Made Easy channel for a while , and I stopped also all music production activities. Note, that I kept keeping the channel alive by responding to all comments and questions of you guys! This month was the 1st month where I regained a sense of stability: Things are settling down a little so I could go back to the joys of producing new videos (there is another one cooking right now!), and producing the next intergalactic pop hit ;-)! Be well raste-ka
Thanks for video on capacitor I was eagerly waiting at you channel. It is a great video but I was looking for displacement Current. Kindly explain in your next video. Also why Maxwell's 3rd and 4th are not similar. In 3rd equation time varying magnetic field produces circulating Electric field but in 4th equation for circulating magnetic field, in addition to time varying Electric field, Current is also included. Current is moving electrons, electrons have electric field and moving electrons produce time varying Electric field. Can Current be replaced with electric field of electrons. This will make 3rd and 4th equation similar.
Hi Zakir, your question is above high school level so I am far from being an expert in this. What I see is that in the 4th equation, there is the property of the medium, free space, that is involved (permeability and permittivity), 3rd looks similar but is not the same 4th. Another thing, In high school terms, think about the interaction of a charge with an electric field F = qE, but in a magnetic field, it needs to move to interact with it (F = qvB). Each field has its distinct properties relatively to each other…
@@PhysicsMadeEasyThank you for replying. My guess is that Electric and Magnetic fields are produced by entangled microscopic electric and magnetic particles which reside at 90 degrees in 2 of 3 spatial dimensions. Motion of one moves the other. Same microscopic particles produce displacement current b/w capacitor plates and magnetic field around the plates.
@@PhysicsMadeEasy that's the video ID. You enter it in the search box, or replace the ID in the url of any video you currently have open in your browser.
Plz listen my voice make videos on structure of atoms how electron protons and neutrons discover and how they had given their charges how the mass discover what is spectrum
![[DOKKAN BATTLE] Worldwide Campaign Announcement Video Part 2!](http://i.ytimg.com/vi/JpT5Voak6WA/mqdefault.jpg)
I never seen a teacher like you sir ,you are a genius sir thankyou sir for clearing my concept
Hi The Watcher. I don't know if I am a genius, but what I try to do is to unlock those that could be, by clearing the path a little. I am glad you enjoyed my work!
This is the best physics channel one could ever get on RUclips....everything is perfect abt these videos
Thank you, I am glad you enjoy my work!
@@PhysicsMadeEasy please make videos on physics jee problems too...sir....It will be really helpful to students across India
I am a jee aspirant and this video helped me to improve my conceptual understanding of capacitors . Thank you
Hello Xeku. I am happy I put you on the right track! it is one of the goals of the channel, provide students with the wings they need to fly by themselves!
me to bro. Btw were you able to crack jee if yes which college are you currently in ?
Please make more videos on electricity.Your video helps to understand the concept better.
It's on my list Kuldeep, thx for the suggestion.
Thankyou for this. Much appreciation for ur hardwork and help in understanding these concepts clearly
Hello Avadh, You are welcome. I am happy my work reaches its goal!
Sir you explain all the basic phenomenat very clearly ..thank u
Thank you Jabbar!
your videos are very explicit in explanation sir
Thank you for your kind words. I am happy that my work helps!
Thanku for this amazing explanation.....💜 I am very confused with some topics of physics because i am not able to feel what is going on ..but ur videos help me a lot to understand each and every thing 💕💕 thank u again
Make more content we love it... 👍
Hi Anjali, thank you for your kind words! I am happy that my work helps you gain a better understanding of physics :-)
big thanks and respect for the easy explanation
You are welcome. I am glad my video helped your understanding of capacitors.
Sir, please make a video on Inductor. It's the most confusing concept for many. Please.
It's on my list (you'll need to be patient though)
My physics teacher himself cannot explain like this... thank u
Hi Noh, thank you for your kind comment :-)
@@PhysicsMadeEasy welcome😊... but i said the truth my physics teacher is such a weirdo
At about timestamp 5:45 there is a, somewhat unknown, use for this circuit (since the cap is first in line). Suppose the switch is in the MIDDLE position and the capacitor is totally discharged. IOW the potential on both cap terminals is 0V. Now, keeping in mind that the VOLTAGE across a cap can not change instantly (just as the CURRENT in an inductor can't change instantly), if you flip the switch to the battery (say it's 12v), then the BOTTOM of the cap, as well as the top is instantly 12v, since there is no potential DIFFERENCE between the plates. Now the top plate of the cap STAYS at 12v, while the bottom plate 'charges' the cap through the resistor by going toward 0v via the RC rate.
This is a GREAT way to apply a fast 12v pulse to switch on an SCR, etc. without any timing circuitry...here a hi-value resistor can be placed across the cap for auto-reset (the cap used in this type of circuit is usually very small...~0.01uF. to 0.1uF).
I have used this method, and have also seen complicated circuitry where this would have greatly simplified it.
A great example of combined knowledge in the academic and typical circuit app of a capacitor from the FIELD point of view!
THANKS AGAIN,
--dALE
Hi sir ,can u please make a video on wave mechanical model and schroginder wave equation ( quantum physics)??!!
I am more in an "electricity mood" these days, but I'll keep your suggestion. Thank you.
Please correct me if I am wrong. Can the Capacitor have more charge's density in its plate, compared to that of the of the battery's side?. And if so, how the electric field would allow such un-equal distribution of charges on the same wire ( from higher potential of battery to the higher potential of the capacitor's plate. Also, is it the reason why the capacitor can produce a lot of power in a very short time?
Hi ggg.
Higher the charge density, higher the energy each coulomb of charge will carry. So to answer your question, no. The charge density of a capacitor's plate, once it has reached its steady state (= once fully charged) should be the same than that on the positive plate of the battery. In other words, the electric potential should be the same.
The reason why a capacitor can deliver a lot of energy in a short amount of time (high power), in the right conditions dependent on capacitance and resistance of the circuit, is because the charges are already on the capacitor's plate, ready to go. In a battery these are generated by a chemical process that has its own kinetics.
these comments need more likes this video needs to be known
Thank you for your kind comment. I checked your channel. 8 years old, coding an interesting game from scratch like Mathonoids. Wow! I am stunned! You beat me, my first game coded, I was 11 (in 1982!). But I was made to stop coding at around 14: My school grades were dropping, and my parents seriously limited access to the home computer (a TRS80). When I was older, around 17-18 yrs, windows 3.1 and PCs had arrived (beurk, I didn't like this), and in the meantime, I had developed other hobbies, so never went back to game coding.
Conclusion: continue, but make sure your grades at school remain OK to preserve your momentum :-)!
@@PhysicsMadeEasy thx for the compliment! By the way,, I keep forgetting to change my description to mention I'm 9 years old now oh nooooooo
@@user-qu1jl8nk2z 9 years old!!! I'm so disappointed now, you are sooo old! 😉
This was a little more complicated video on circuits and rather a rapid introduction to different aspects of electricity while the first 2 videos in this serie started slow and beginner friendly
teacher at 8:21 we considered that Vc=0 because no charges has arrived yet on the capacitor .but why Vr has a value? , why we didnt consider that no charges have arrived on the resistor also ?
i hope that you clarify this for me .
Hello Anass,
This is a good question.
We are in a transition mode. Charges have not moved yet, but they want to.
Before it charges, the capacitor can be considered like an equipotential wire (it will let pass current with no resistance)… so the electric potential will be the same on both plates: that of the positive side of the battery. V (before C) is epsilon, V (After C) is epsilon, so Vc = 0, thus Vr = Epsilon.
Then as current passes, the electric potential of the lower plate will decrease, because some of the charges on the lower plate are pushed by those that get stuck in the higher plate : that is when Vc starts increasing.
It is very important that you understand clearly what is an electric potential to understand my answer (If you cannot explain what an electric potential in a few words in your mother tongue, then watch the video about that on this channel).
I hope this clarifies things !
Thank you for the video, when is the next capacitor video coming? It would be nice if you could also explain intuitively how capacitors work in series and parallel
Hi GodFred, Good suggestion. I'll put in near the top of the wishlist. Unfortunately other priorities showed up, so I had to slow down posting (one 15 minute video is a full time job for at least a week!). Yet, I would like to produce more soon, and guess that Capa in parallel and series could be one of the next videos...
Wonderful video 👍
Thank you Man!
12:24 initally Vc is epsilom but why is Vr epsilom initially if when we charged the capacitor Vr went down to 0. I understand you said kirchoffs 2nd law but i dont get how Vr suddenly changed from 0 to epsilom Thanks
Hi N,
When the capacitor is charged (and still connected to the battery: no current flows through the circuit (there is no incentive for charges to move around since the electric potential is the same at the positive terminal of the battery and the positively charged plate of the capacitor. So, no current, hence, no potential drop at the resistor.
Once you flip the switch on position B. The capacitor discharges: charges move from the positive plate to the negative plate via the resistor (note that I use here the conventional model, in reality, electrons move from the negative plate to the positive plate). So as soon as you flip the switch, you have a maximum current flowing through the resistor, hence the potential drop Vr. This potential drop will decrease in time at the same rate at the voltage of the capacitor Vc to end up at 0, when the capacitor is fully discharged (no more current flows in the circuit).
I hope this helps!
Sir thank you for an excellent explanation . Can you please elaborate at time stamp 2.42, how I= omega C V, what is the frequency of the current calculated?
Hi Wu, The current in a capacitor can be calculated by I = CdV/dt. So for current to pass, the voltage needs to change with time. If the voltage is a sine curve (V = Vmax * Sin(omega*t), you see that omega will come out when you differentiate the voltage. so here you go I(t)=omega * C *V(t). I hope this helps!
Thank you very much sir, you rock.
Thank you, sir.
You are welcome Antonio :-)
Thanks
Thank you very much Jaswanth for you gift. It is very much appreciated :-)
New sub here. Thanks for your channel. Can you link me to a clarification of the use of/for 'adjacent capacitors' Some years ago(2013). I inadvertently constructed an LC circuit. At that time I had no idea what I was doing. But now seeing what I've learned in this vid, I may be able to take advantage of my earlier findings. My intent is to charge and discharge two capacitors to do work via 2-coils, each in succession, charging and discharging at a rate of 8.63hz. Thanks in advance for your reply.
8:13 question ask in neet 2016,2019 exam
Hi, well, I didn't know. It's a great circuit to truly understand how a capacitor behaves, so it doesn't surprise me that it can show up in various high school programs, and their related exams.
Excellent video
Thank you Kadari
Sir I have a small but for me mind boggling confusion about work and energy.
For this if we consider an isolated part of the universe where no force is acting just there are one object and me. From a reference point, the box is moving with a uniform velocity 'u', and then I applied a force 'f' on the object when it was moving with uniform velocity at time zero. After sometimes when the displacement 's' occurs at time 't' I stop applying force.
So by the formula the work done by me on that object should be f times s, but sir here the object had some initial velocity u, so in the displacement there should be some part of it that is not done by me, by the formula (s=ut+1/2at²).
So by conjectures it seems that the formula says that I have done more work than I have actually done.
Is it so? Where I am wrong sir?
Consider case A, initial velocity 0, and case B, initial velocity u. In both cases You apply a constant force f on the object for the same amount of time, t.
In case B, you will have travelled a longer distance than in A in the time t, because you already had an initial velocity u. Therefore, the distance travelled s has to be larger. Conclusion, In case B, you will have provided more work.
If you need to be convinced, just calculate the change in KE in A and compare it to B. for B the change in KE will have extra positive terms than for A. Or if you prefer avoiding too much algebra, you can also do this with numbers that you set yourself…
@Physics Made Easy sir but the quantum mechanics says that gravity is one of the four fundamental forces and its particle is graviton, so if it is true then we should not feel any type of gravitational force from black hole because when ,from inside the event horizon, a black hole emits a particle of graviton it will never exceed that line.
So does it resolve that gravity is not a force and if so then why we are still looking for particle of gravity?
I have heard on many videos that gravity is made up of particles so that is why this question came in my mind.
A black hole is a sphere from which nothing can leave because of the limitations of motion in our Universe’s space time. Causality from the inside to the outside is broken. This is because the inside of a black hole is NOT in our space time (our Universe… ). So the singularity itself does not interact gravitationally. What does, is the surface.
You have to consider the part of the back hole in our universe like a big black sphere (or more precisely, a sphere of radius just a little larger). What exerts gravity in our Universe, is not the singularity itself, it is the surface of that sphere … So if gravitons are real, they will be exchanged between an object in our universe and the surface of the black hole.
But again, as I told you in a previous comment, that object (the surface of a black hole) is macroscopic… so trying to understand things with quantum mechanics here is not a good idea (until we have something better). Here you should use general relativity…
Oh, and when you watch a science video, always pay attention to the context of the video and be careful when extrapolating. what you learn elsewhere (it might not apply). Also keep a critical eye... More the title looks appealing, more you should be cautious... I see things sometimes that make me jump out of my chair lol!...
@Physics Made Easy sir and also I have one doubt about angular displacement.
Sir when our vehicle makes a sharp turn we feel that we are pulled on the side opposite to the direction of turning and generally the reason for this is given as due to inertia, our body tends to move in one direction so we feel that pseudo force also called centrifugal force. In this case if we just look at our vehicle so then it makes an Arc of a circle, and if you want to find the angular velocity of the car , which is v/r or theta/time, we will require its radius because we can easily measures the velocity on speedometer said so in this case how can we calculate the radius?
And if sir we can just measure the angular displacement then we can easily measure all the angular quantities so how can we calculate that how much of degree or radian that car has covered because we don't know the radius?
If you know the coefficient of friction (MU) and the linear velocity, you can figure out the angular velocity (w).
N = mg
f = MUxN = MU x mg
f is centripetal, so MU x mg = m v^2/R
so MU x g = v^2 /R
From that you can deduce R, and then w.
Hi , i i want know one thing is it compulsory to make notes of physics ? Is essential if i want to know physics in deep clear all my doubt i study because love studying physics but i am serious about physics so can you just answer my question , by the way your videos are just awesome
Hi Haider,
Your question is more about How to Study in general than about a Physics. Since I have been tutoring for years many students between 15 and 20 years old, I have learned that… it depends on the student!
Most students take really bad notes during a lesson, and taking these takes away their attention from what the teacher is saying and leaves them confused when they try to understand their notes later.
Some others, more rarely, will take excellent notes and will be able to use them to recollect what they have seen in class.
What I would advise when a student is a bad note taker: minimal notes - just note down the main points (like titles of sub-topic, main definitions and main equations), so you can focus your attention to what the teacher is saying. Then, the same day, at home, take your notes, and a text book next to you. The notes are just for remembering what you need to study and know.. With the help of the book, write your own lesson like if you were trying to teach it.
That is for the lesson, when it comes to exercises and worked examples, I would try to take deeper notes that at least guide you towards the solution when you try and redo the exercise.
For my videos, don’t take notes the first time you watch them. Just focus on the video. You can always take some if you want when watching a second time.
By the way, thanks for the kind words :-)
at school I was forced by my teachers to take notes and it was the worst thing ever because I could never concentrate on the lecture
@@PhysicsMadeEasy thanks for your advice
@@courage936 same brother
@@courage936 @Haider Abbas Kazmi
It's funny, sometimes I have to scold my students...not to take notes while I am teaching! What I do, is that when I finish a demonstration or the presentation of a concept, i give them a few minutes to take notes. If the demonstration is too long, I will cut it in little bits... to give them multiple note taking periods.
It allows them to make a synthesis of what was said, And I let them ask questions (because it's often then that they realize that they have some to ask...).
Then sometimes, I ask them to read their notes to me. So I can check if they really got it!
Full focus of the student during the lesson itself + full focus of the student when he takes note (the student can digest the info more effectively).
Magic technique maybe you should ask your teachers to apply this?
After a long time😄.
So it was haha! but I'm back :-)
what are these positive charges? I understood that it was the negative charges that moved.
Yes, it is the negative charges that moves, but in the conventional model used by all, a current is defined by the motion of positive charges. (in other words, negative charges moving to the left is mathematically identical to positive charges moving to the right)
Sir Where were you all these days it's been almost 11months
Hey there, thank you for noticing and asking. It's appreciated.
I was actually moving country, and at the same time, moving from a large city to a countryside village, recreating my teaching company in the new country, buying a first house + land, working on it (there was a lot to do!), taking care of my real life students (that consequently became online students), socializing myself with the locals. I basically changed life!
With all this happening, I had to slow down with the Physics Made Easy channel for a while , and I stopped also all music production activities. Note, that I kept keeping the channel alive by responding to all comments and questions of you guys!
This month was the 1st month where I regained a sense of stability: Things are settling down a little so I could go back to the joys of producing new videos (there is another one cooking right now!), and producing the next intergalactic pop hit ;-)!
Be well raste-ka
@@PhysicsMadeEasy which country?
And also make videos of the of ipho olympiads only topics with deep not question
That might be a little too specialized for the channel Tara...
Thanks for video on capacitor I was eagerly waiting at you channel.
It is a great video but I was looking for displacement Current. Kindly explain in your next video. Also why Maxwell's 3rd and 4th are not similar. In 3rd equation time varying magnetic field produces circulating Electric field but in 4th equation for circulating magnetic field, in addition to time varying Electric field, Current is also included. Current is moving electrons, electrons have electric field and moving electrons produce time varying Electric field. Can Current be replaced with electric field of electrons. This will make 3rd and 4th equation similar.
Hi Zakir, your question is above high school level so I am far from being an expert in this. What I see is that in the 4th equation, there is the property of the medium, free space, that is involved (permeability and permittivity), 3rd looks similar but is not the same 4th.
Another thing, In high school terms, think about the interaction of a charge with an electric field F = qE, but in a magnetic field, it needs to move to interact with it (F = qvB). Each field has its distinct properties relatively to each other…
@@PhysicsMadeEasyThank you for replying. My guess is that Electric and Magnetic fields are produced by entangled microscopic electric and magnetic particles which reside at 90 degrees in 2 of 3 spatial dimensions. Motion of one moves the other. Same microscopic particles produce displacement current b/w capacitor plates and magnetic field around the plates.
Can you make a reply video/poke holes in this theory uYosC10FoLU ?
who is uYosC10FoLU and what did he say?
@@PhysicsMadeEasy that's the video ID. You enter it in the search box, or replace the ID in the url of any video you currently have open in your browser.
DEAR SIR,
YOU DID NOT ANSWER YOUR OWN QUESTION.
ELECTIC FIELD, MAGNETIC FIELD AND GRAVITATIONAL FIELD.
QUESTION IS,
WHAT IS A FIELD ? ? ?
Did you see the video about fields (what is a field?)?
Plz listen my voice make videos on structure of atoms how electron protons and neutrons discover and how they had given their charges how the mass discover what is spectrum
Thank you for your suggestion. If you like this topic, you should check the playlist on particle physics!
electrical engineering not physics
Electrical engineering is a branch of physics :-)