This is a godsend. Our professor skipped justification of every step, and without this I would have to just memorize his proof rather than understand it
If you let u = (x^2 + a^2) then the differential of u (=du) = 2x dx. Since we only have x dx in the integral we need to multiply by 2 to get 2x dx, but then we also have to divide by 2 otherwise we have changed the integral.
you have helped me a lot. i started watching your videos last semester and you're the best. I actually understand the lessons after I watch you explain them. thank you
Can you please explain why you're using the area of a circle for the ringlet for A in dA? Shouldn't it be the outer radius - the inner radius since the shape is not a circle?
+Hadi Doudar Cause first you have to visualize that the way the problem will be solved is by imagining concentric rings. in order for the equation to "understand" what the integration will be you need to leave the X in since it will not be constant it will vary with the integral from 0 to R...
@@MichelvanBiezen Yes, I subscribe your channel and I am also very glad that you are still active here to answer me and my friends Questions regarding the Topic...☺️🙃
Michel van Biezen Michel van Biezen I don’t get this either. The area of a ring is the area of the bigger circle minus the area of the smaller circle. The radius of the larger circle is r+dr and the radius of the smaller circle is r. The area of the ring should be pi(r+dr)^2-pi(r)^2=pi(r^2+2rdr+dr^2)-pi(r^2)= pi(2rdr+dr^2) NOT 2pi(rdr)
Thank you Mr. Biezen, your lectures have helped me so much in my physics courses. I noticed how similar these videos are compared with calculating the moment of inertia of certain objects. Using the formula of the disc of charge you calculated in this video would it be possible to integrate the discs to derive the electric field from a sphere?
When solving for the electric field of a Disk of charge problem where the point of interest is the midpoint between two disk (one being positively charged and the other negatively charged) separated a distance apart. so I do 2 calculations, one for the positive charged disk and another calculation for the negative charged disk then add up the two calculated electric fields, Do you take the absolute value of Q when plugging in to the disk formula?
The problem you describe sounds a lot like 2 capacitor plates, and that would be the case if the two disks are close enough together. Ignoring the end effects, the electric field magnitude would be constant. The farther you place the plates the more it would be like you suggested.
Also, depending on the charges on the plates. If both plates carry positive charge then the electric field in the middle would be zero. If they are oppositely charged, then the direction of the electric field cause by both plates would be in the same direction.
This might sound like a stupid question, but I'll ask anyway, for dQ = sigma2pixdx @ 2:23 , how come we just don't right dx^2? Doesn't x change depending on how big our ring is? I think I'm overthinking it, but any clarification would be greatly appreciated. :)
Charge is equal to charge per unit area x area = sigma x 2 pi x dx where 2 pi x is the circumference and dx is the width of the circular strip. This is the typical way in which we set up integrations which is the summing up of (in the limit) an infinite number of strips adding up to the whole area and thus the total charge on the disk.
In this case if you take the strip cut it and stretch it into a linear shape and lay it flat you would have a rectangle where the length would be 2*pi*x and dx would be the width
How can we find the electric field and hence electric potential at any point on the circular disc of surface charge density sigma?Should we go for Guass’s Law ?
It could be done in two integrals, but the small area of charge was taken as a strip in the shape of a circle and thus we didn't need the first integral.
I'm on board with the solution - though I'm frustrated (as are my students) as I expect to see a 1/a^2 dependence for E for values of a>>R. The expression does yield that E --> 0 as a gets very large, but for lots of other charge distribution situations I'm able to see that fields approximate point-charge fields when we get really far away. Any thoughts?
It is actually "just" algebra. If you factor out an a^2 from the radical in the denominator you will end up with 1 - 1/((R^2/a^2) - 1)^(1/2) and then you can see the relationship to the usual electric field equation.
I appreciate the "quotes" as I like to have a laugh w/ my kids about how the word "just" has little place in the work they do! I'm embarrassed to say that
Hy sir, i would like to ask, if we are asked to get the magnetic field instead of the electric field and we have a current instead of charges, this method still applicable? or i must just integrate the result of 1 ring? thank you in advance
The magnetic field is typically calculated differently, since it tends to be circular. We have a number of examples in the magnetic field playlist. PHYSICS 44 MAGNETIC FIELD SOURCES ruclips.net/video/D8TyClG8d5k/видео.html
Sir this is an awesome video! Thanks a lot! I think you've multiplied with an extra 2, after the integration I mean it was supposed to be sigma*k*pi*a*[-1/(x^2+a^2)^(1/2)] and not sigma*k*pi*a*2*[-1/(x^2+a^2)^(1/2)]
hi sir, firstly i wanna thank you so much. i have a question: what if a>>R0? right side of the last formulation becomes zero, so does that mean that the disk doesn't act like a point charge?
electric field is a vector but ELECTRIC POTENTIAL is a scalar right? So if we are asked to find Electric Potential on the Axis of a Uniformly Charged Disc what would be the formula?
First of all, you are correct. Electrical potential is a scalar. But when you are asked to find the electrical potential, it is always in reference to another location. We always find the DIFFERENCE in electrical potential. There is no such thing as the electrical potential at a particular location that is not referenced to another electrical potential. See the videos here: PHYSICS 38 ELECTRICAL POTENTIAL ruclips.net/p/PLX2gX-ftPVXXFqBJixIbQcyXZD6kQnAQH
It depends on the shape of the dA. In this case we used rings, so we integrate from the inner ring to the outer ring in a radial direction. If you have a wedge for dA, then you must integrate around the circle from 0 to 2 pi
The small dA is a small ring with radius x and small width dx. If you cut it and stretch it, it would be a thin strip which is a thin rectangle. The area of that rectangle would be L x W where the L = 2 * pi * x and the width would be dx.
point charge, we consider a sphear shape. and its electric field KQ/ r2. for line and surface charge, however small the 'dq' element is, the shape of the distribution is different from point charge but we use same relation dq/r2. does it mean shape of the distribution is immaterial???
That depends on how far away you are. When you are very close to an irregular shape that is charged, it will make a difference. But when you move far enough away from the object, the electric field will act as if it comes from a point charge.
thank you sir. so, in our case of line charge and surface charge the 'dq' element distribution is assumed to be point charge?? should we assume E field point is faraway from the charge distribution??
man, you saved my academic life, thanks a lot. i m recommending your channel to lotta engineering students friends of mine. but i have 2 questions. as you lecture, 2 point particles can cancel each others electric field out. ok but why not that doesnt happen in electric Force(or E times q). 2nd question can you make a short explanation of the potential difference in between point A and B, with using the partial integration formula (final to initial) as it goes integrate(- E field times dl)
+Ahmet Omer Ozgen Answer to first question. If you place two opposite charges together there will not be any electric field around them, thus the electric fields will cancel out and another nearby charge will not experience a force. Answer to 2nd question. I believe there are a number of examples on that topic in the playlist on potential difference PHYSICS 38 ELECTRICAL POTENTIAL
+Michel van Biezen thanks. i will watch it. so on the electric field they cancel each other out. but for the F, electric charge force on the second particle(signs are opposite) is not zero as i have been told by my peers. shortly, the same idea, does it work in the Force(column) too,(when the latter particle is charged too). thanks again
+Hamza Ali Since we are integrating small (thin) rings, the limits are: r = 0 which is the center of the disk to r = Ro which is the radius of the disk.
I'm not quite convinced with the statement that the area of an infinitesimally thin ring is 2*pi*R*dR. Wikipedia explains this as "onion proof". But the area of a ring is pi*(R^2 - r^2). And from that the area of an infinitesimally small ring should be 2*pi*R*dR + pi* dR^2. (A difference of pi*dR^2). Integrating this over the radius of a circle does not give pi*radius^2 though. Can you explain?
+Tuuskis That is the difference between a mathematician and a physicist. Mathematicians work out the details and physicists are content with the equations that work. Although in the end it becomes more philosophy than science. To me it is perfectly fine to call the dA = 2*pi*R*dR and when you integrate it we get the correct answer. That is good enough for me.
yes. there are lots of questions in the gaussian shepers and so on. if you go deep, when we integrate a simple linear rod as dx, to find the e filed at a specific point, well i thought about it and, at the center of the rod e field is max and at the ends its minimum and when you think that this rod is X axis you see a Sine function out of E fields on the Y axis that are producted with Cosine Alpha(angle between), well then i ask the question, is there anything wrong on the simple intergration. well i dont know, 1 out of 10000 can be neglected easily
1) if a > > x then the charge will act like a point charge 2) if the charge density is negative then the direction of the electric field will be in the opposite direction
Sir i have a doubt that if we find the Ef at the centre of disc it doesn't came zero but logically if we thinks it should add upp to zero as we make the disc out of infinte rings and ring have zero ef at centre Sir pls explain Great explanation though
When "a" goes to zero, the electric field strength becomes 2 sigma / epsilon sub knot just like we would expect, since that is the electric field above an infinite sheet of charge.
If you let the disk become infinite in size, you will get the solution of the electric field of an infinite plane. (and that is constant, regardless of the distance away from the plane)
@@MichelvanBiezen For a finite disk, as the distance from the disk approach negative infinity, the final equation does not give the correct answer. Instead it gives sigma/epsilon0 which is incorrect as the true answer by intuition is supposed to be 0.
A negative distance does not make any sense. For example if you stand on the street and you measure the distance to the nearest house, can that ever be negative? If you let a approach infinity, the equation does indeed converge at zero.
I tried using Gauss law to derive the formula for a disc.I got electric field as that of a plane sheet of charge.I do not understand why Gauss law is giving me the wrong formula for a disc.Can anyone help me with this?
when we can use the Guass' law versus this method to calculate E? what about the electric field on the other side(left) of the disc unless you are considering the charges are only on one side. thank you
Dan, take a look at the playlist on Gauss' law, it shows the typical 6 charge distributions that lend themselves to being able to be solved by Gauss' law method. This isn't one of them. Note that this is the electric field calculated for that location only. If you want it for a different location, you need to do the process all over again.
i still have a question: why do you have to name the thickness of the ring "dx" I see that it is the only way that the integral can be solved but suppose that i did not get that information and i name it "dl" the integral gets confusing and i dont know to what variabel i have to integrate. Is it just dx because it is an extension of x or how do i know? (sorry for my bad english!!) plz answer
Why do you not multiply by cos(theta) to get the electric field in the x-direction? You did that when it was only a ring of charge, so what makes this case any different?
when you integrated you put 2 next to the x in the numerator in the integral and a 2 in the denominator outside of the integral, i don't understand why. Great videos by the way!
In order to integrate that integral, you need the correct differential which is 2 x dx. That means we needed to multiply by 2 and divide by 2. (I could have moved the 2 in the numerator into the integral, but I wanted to illustrate why we need to do that).
Why is the area of the ringlet 2*pi*r. I'm really having difficulty understanding (conceptually) why you're substituting perimeter for area when the formula calls for an area.
There are often multiple ways to do a problem, but as it is presented in the video is probably the best. Why don't you try it and see if you get the same answer?
Sir in previous video we have taken the ds as element and its equal to x. dfie. Here y u have taken the dx as element we have integrated from x=0to r if we take ds instead of dx then whole integration changes
How you set up the integration depends on how you can best set up the dq so that the electric field of that dq can be most easily represented. (There are multiple ways to do this). In this case it appeared easiest to take the whole ring as a dq. That way only the horizontal component of the electric field survives. (the perpendicular component will cancel out).
You need to find a small segment dA with charge dQ on the disk that you can then integrate over the entire disk. There are different integration techniques you can use. This is one of them.
This is an integration technique and based on the definition of integration . When you integrate you let the thickness approach zero and integrate over an (in the limit) infinite number of small strips of thickness dx. Take a look at this video for a reference: Calculus - Integration: Finding the Area Between Curves (1 of 7) Ex. 1: y=e^x, y=x^2, x=0, x=2 ruclips.net/video/mzdWhFh3050/видео.html
It is not in the integral. (It is in the result of the integral). In this case, looking ahead, it makes sense because you end up with 1 - (small number) in the answer which makes it easier to interpret the answer.
I didn't quite get your question, but I think you are asking why need a differential to integrate? You always need a differential to integrate according to the rules of integration. (for example you can't integrate x, but you can integrate x dx, where dx is the differential of x.
@@MichelvanBiezen sorry I'm a bit confused, in the last video where you calculated the ring of charge, you didn't include the variable x and a into the integration. Thus I am confused why you included in the charge of disk example. Thank you for taking the time to reply to me 🙏
Don't try to compare the 2 videos. Look at each video and determine if the setup makes sense. The key in each case is to find the electric field at a particular point P, due to the presence of a small charge element, and how to define that charge element and the distance from the element to the point P
my professor doesn't have your level fluidity and confidence with the material, so you can bet this video is helping! thanks from Montreal, Canada
Welcome to the channel! Thanks for your comment.
I love this! Especially cause I'm also from Montreal, Canada
This is a godsend. Our professor skipped justification of every step, and without this I would have to just memorize his proof rather than understand it
thank you for all that you do. You make physics easier to understand and i appreciate the hard work you've put into this channel. Its appreciated.
amazing! I cannot thank you enough. I was so lost and confused with this until I saw this video.
You are the best! Probably got tired by hearing that from me and others who love your teaching.
We appreciate the comments.
This guy is saving my future :D Thank you!!!!!!!!!
big ups! First time I'm integrating multiple variables so its getting tangled in my head but you're making it clear!
Excellent! I understand this so much better on second viewing. Thanks for for all your excellent lectures we appreciate them.
Great that you went back and reviewed it again. 👍
Thanks from India a Jee advanced aspirant this is very helpful in solving Advanced questions.....
You are welcome. You'll find this a good channel to prepare for the physics portion. 🙂
I like this teacher because he learns effectively! Thanks!
I feel obligor to support your channel, at least by typing comments and liking videos because you share this legends with us for free
We appreciate that!
i watched your videos when i was a junior in high school. Now i am a junior in college and you're still teaching me!
what do you mean the differential denominator is a 2x? (@time 4:15)
If you let u = (x^2 + a^2) then the differential of u (=du) = 2x dx. Since we only have x dx in the integral we need to multiply by 2 to get 2x dx, but then we also have to divide by 2 otherwise we have changed the integral.
Michel van Biezen thank you
You're really good at explaining the concepts thanks so much for posting!
you have helped me a lot. i started watching your videos last semester and you're the best. I actually understand the lessons after I watch you explain them.
thank you
I don"t know how can i thank you. You are an awesome teacher.
excellent explanation, you sir are amazing
THANK YOU!!
Can you please explain why you're using the area of a circle for the ringlet for A in dA? Shouldn't it be the outer radius - the inner radius since the shape is not a circle?
The area of the small ring = (circumference) (thickness) =( 2 pi x) ( dx)
Oh gosh thank you so much ! i cant count how many times you saved my life
At 3:32 why isn't "x" a constant as well?!
+Hadi Doudar
x represents the distance from the center of the disk. (It is not a fixed value)
+Hadi Doudar Cause first you have to visualize that the way the problem will be solved is by imagining concentric rings. in order for the equation to "understand" what the integration will be you need to leave the X in since it will not be constant it will vary with the integral from 0 to R...
Thanks so much for showing all the steps, especially in this one which would have been very difficult without your help.
Glad it helped. 🙂
Thank you for these videos, they have been very helpful!
You're welcome!
Thanks A Looooooooooot❤❤❤ Teacher.... I got the knowledge from you in a simplest way.
You are welcome. Glad you found our videos. 🙂
@@MichelvanBiezen Yes, I subscribe your channel and I am also very glad that you are still active here to answer me and my friends Questions regarding the Topic...☺️🙃
really the best physics teacher
Thank you.
Thank you sooooooo much you are the best, but I have to admit that I watched this video two times to understand the integration :) :)
1:37
How do you get the area of the little ring? I don’t understand that...
The area = circumference x width
Michel van Biezen Michel van Biezen I don’t get this either. The area of a ring is the area of the bigger circle minus the area of the smaller circle.
The radius of the larger circle is r+dr and the radius of the smaller circle is r.
The area of the ring should be pi(r+dr)^2-pi(r)^2=pi(r^2+2rdr+dr^2)-pi(r^2)= pi(2rdr+dr^2)
NOT 2pi(rdr)
@@studywithjosh5109 if you go using that don't forget to use binomial at end as dr is infinitesimal
Or u can use it as
A=pi r^2
dA=2pi rdr
Thank you Mr. Biezen, your lectures have helped me so much in my physics courses. I noticed how similar these videos are compared with calculating the moment of inertia of certain objects. Using the formula of the disc of charge you calculated in this video would it be possible to integrate the discs to derive the electric field from a sphere?
When solving for the electric field of a Disk of charge problem where the point of interest is the midpoint between two disk (one being positively charged and the other negatively charged) separated a distance apart. so I do 2 calculations, one for the positive charged disk and another calculation for the negative charged disk then add up the two calculated electric fields, Do you take the absolute value of Q when plugging in to the disk formula?
The problem you describe sounds a lot like 2 capacitor plates, and that would be the case if the two disks are close enough together. Ignoring the end effects, the electric field magnitude would be constant. The farther you place the plates the more it would be like you suggested.
Also, depending on the charges on the plates. If both plates carry positive charge then the electric field in the middle would be zero. If they are oppositely charged, then the direction of the electric field cause by both plates would be in the same direction.
Shouldn't we multiply the dE with cos(theta) to find dEx as the y-axis cancel?
That is already included in the expression used for dE. (See the development of that expression in the previous video 8 of 16)
Michel van Biezen Thanks a lot!
This might sound like a stupid question, but I'll ask anyway, for dQ = sigma2pixdx @ 2:23 , how come we just don't right dx^2? Doesn't x change depending on how big our ring is? I think I'm overthinking it, but any clarification would be greatly appreciated. :)
Charge is equal to charge per unit area x area = sigma x 2 pi x dx where 2 pi x is the circumference and dx is the width of the circular strip. This is the typical way in which we set up integrations which is the summing up of (in the limit) an infinite number of strips adding up to the whole area and thus the total charge on the disk.
So, basically we are treating x like a "height" in the way we find areas under curves for graphs?
In this case if you take the strip cut it and stretch it into a linear shape and lay it flat you would have a rectangle where the length would be 2*pi*x and dx would be the width
Oh, ok I think I understand now. Thank you for the clarification!
Hello, I was wondering why we don’t use use a double integral, instead integrating dE over 0 to R then 0 to 2pi.
That is essentially the same thing.
nice way of teaching...
How can we find the electric field and hence electric potential at any point on the circular disc of surface charge density sigma?Should we go for Guass’s Law ?
We wouldn't find the electric field ON the disc, but directly above it, and yes, for that you can use Gauss's law which makes it a lot easier.
how is area of small ring area with charge dq equal to 2 pi x dx?
If you take a thin ring, cut it and straighten it, it will look like a rectangle. The area of a rectangle = length x width
Thanks for answering! I was confused on that part as well.
@@tziw9912 I am also confused in this part.. very very thanks Michael sir.. I am from India
Thank you sir. This video help me a lot on studying Electromagnetic Field.
you are a great teacher michael
This is very nice explanation
Hello from Turkey!It ıs very helpful.Thank youu :)
Welcome to the channel.
Why isn't it 2 integrals, 0 < theta < 2pi and 0 < x < R0, but only 1?
It could be done in two integrals, but the small area of charge was taken as a strip in the shape of a circle and thus we didn't need the first integral.
Thank you, Michel
I can't understand how you have time to respond :D
Excellent job
How can we prove in that case using math, and not symmetry on the ringlet, that the vertical components are add up to zero?
Integrate it in four sections from 0 to 90 degrees around the circle. The 4 integrals will add up to zero.
I'm on board with the solution - though I'm frustrated (as are my students) as I expect to see a 1/a^2 dependence for E for values of a>>R. The expression does yield that E --> 0 as a gets very large, but for lots of other charge distribution situations I'm able to see that fields approximate point-charge fields when we get really far away. Any thoughts?
It is actually "just" algebra. If you factor out an a^2 from the radical in the denominator you will end up with 1 - 1/((R^2/a^2) - 1)^(1/2) and then you can see the relationship to the usual electric field equation.
I appreciate the "quotes" as I like to have a laugh w/ my kids about how the word "just" has little place in the work they do! I'm embarrassed to say that
Thank you so much for this video! it is incredibly useful and clear!
Mr. Professor can we use the electric field Instensity of semicircle and multiply it by 2 ?
Yes you could.
Hy sir, i would like to ask, if we are asked to get the magnetic field instead of the electric field and we have a current instead of charges, this method still applicable? or i must just integrate the result of 1 ring? thank you in advance
The magnetic field is typically calculated differently, since it tends to be circular. We have a number of examples in the magnetic field playlist. PHYSICS 44 MAGNETIC FIELD SOURCES ruclips.net/video/D8TyClG8d5k/видео.html
@@MichelvanBiezen Ah okey, I saw all of them it's only about the disk point for me, thank you sir.
Wouldn't the field be in the z direction since the perp components are the only ones that survive? assuming the disk is on the xy plane.
That depends on how the x,y, and z axis are defined. In this problem the positive x direction is to the right.
@@MichelvanBiezen No it's not. You already defined x to be in plane of the disc.
4:30 how did you integrate it sir?
Substitute either using x=r tan⊙
Or u=(r^2+x^2)^3/2
Then calculate du and apply limits
Awesome explanation !!
How does this reduce to Culombs law at large distances?
At large distance it will look like a point charge
@@MichelvanBiezen but from the formula, when a>>R, E-->0 (the fraction reduces to 1 )
You would expect the electric field to go to zero, when the distance from a point source becomes very large.
You would expect the electric field to go to zero, when the distance from a point source becomes very large.
Very helpful, professor!
Sir this is an awesome video! Thanks a lot! I think you've multiplied with an extra 2, after the integration
I mean it was supposed to be sigma*k*pi*a*[-1/(x^2+a^2)^(1/2)] and not sigma*k*pi*a*2*[-1/(x^2+a^2)^(1/2)]
It is correct in the video. That 2 is needed for the correct differential in the integral.
Yeah, i think i get it now. Thank you for your reply.
hi sir, firstly i wanna thank you so much. i have a question: what if a>>R0? right side of the last formulation becomes zero, so does that mean that the disk doesn't act like a point charge?
Yes indeed. That is one way to check if the result is reasonable, to go to the limits and see what happens.
I did not get that integration part. I don't understand help
Hey what are you doing in life today?
@@salonisingh7159 Today we all pray for the death of annoying orange
what would be the results in case of charge density on a ellipse
That would be much more difficult to calculate, but it can be done.
electric field is a vector but ELECTRIC POTENTIAL is a scalar right? So if we are asked to find Electric Potential on the Axis of a Uniformly Charged Disc what would be the formula?
First of all, you are correct. Electrical potential is a scalar. But when you are asked to find the electrical potential, it is always in reference to another location. We always find the DIFFERENCE in electrical potential. There is no such thing as the electrical potential at a particular location that is not referenced to another electrical potential. See the videos here: PHYSICS 38 ELECTRICAL POTENTIAL ruclips.net/p/PLX2gX-ftPVXXFqBJixIbQcyXZD6kQnAQH
the refference point has to be infinity, I think we can define electric potential like that, right?
Sir
I had one small doubt
Why did you integrate dx from 0 to R
But in case of ring you did it from 0 to 2πR
Hoping for a quick reply Thanks..!
It depends on the shape of the dA. In this case we used rings, so we integrate from the inner ring to the outer ring in a radial direction. If you have a wedge for dA, then you must integrate around the circle from 0 to 2 pi
Hi sir, I am unable to figure out how you determined this:
dQ = sigma*dA
giving
dQ = sigma*2*pi*x*dx
How did you know that dA = 2*pi*x*dx?
Thank you!
The small dA is a small ring with radius x and small width dx. If you cut it and stretch it, it would be a thin strip which is a thin rectangle. The area of that rectangle would be L x W where the L = 2 * pi * x and the width would be dx.
Thank you so much sir! Your videos are awesome! I have been watching them since 2014 :)
Michel van Biezen OMG THANKS YOU SO MUCH
Literally u are very weak in studies
@@tarannumkhan7539 this comment shows that you are not about learning, but that you are truly about putting others down. Shameful
point charge, we consider a sphear shape. and its electric field KQ/ r2.
for line and surface charge, however small the 'dq' element is, the shape of the distribution is different from point charge but we use same relation dq/r2.
does it mean shape of the distribution is immaterial???
That depends on how far away you are. When you are very close to an irregular shape that is charged, it will make a difference. But when you move far enough away from the object, the electric field will act as if it comes from a point charge.
thank you sir.
so, in our case of line charge and surface charge the 'dq' element distribution is assumed to be point charge??
should we assume E field point is faraway from the charge distribution??
Really great explanations, thanks!
Sir can we just substitute (x^2+a^2)^3/2 as t and the solve the intergal and find nee upper limit and new lower limit then substitute them and solve.
man, you saved my academic life, thanks a lot. i m recommending your channel to lotta engineering students friends of mine.
but i have 2 questions.
as you lecture, 2 point particles can cancel each others electric field out. ok but why not that doesnt happen in electric Force(or E times q).
2nd question
can you make a short explanation of the potential difference in between point A and B, with using the partial integration formula (final to initial) as it goes integrate(- E field times dl)
+Ahmet Omer Ozgen Answer to first question. If you place two opposite charges together there will not be any electric field around them, thus the electric fields will cancel out and another nearby charge will not experience a force. Answer to 2nd question. I believe there are a number of examples on that topic in the playlist on potential difference PHYSICS 38 ELECTRICAL POTENTIAL
+Michel van Biezen thanks. i will watch it. so on the electric field they cancel each other out. but for the F, electric charge force on the second particle(signs are opposite) is not zero as i have been told by my peers. shortly, the same idea, does it work in the Force(column) too,(when the latter particle is charged too). thanks again
+Ahmet Omer Ozgen Yes, the two particles will exert a force on one another.
What is the big "Q" by itself ? Is that the charge of the entire disc? Also is this typically given in the problem??
I don't understand how you got the limits of that integral. R(o) and 0, how does this affect the problem and the electric field?
+Hamza Ali Since we are integrating small (thin) rings, the limits are: r = 0 which is the center of the disk to r = Ro which is the radius of the disk.
+Michel van Biezen Thanks a bunch!
You sir are my hero !
I'm not quite convinced with the statement that the area of an infinitesimally thin ring is 2*pi*R*dR. Wikipedia explains this as "onion proof". But the area of a ring is pi*(R^2 - r^2). And from that the area of an infinitesimally small ring should be 2*pi*R*dR + pi* dR^2. (A difference of pi*dR^2). Integrating this over the radius of a circle does not give pi*radius^2 though. Can you explain?
+Tuuskis
That is the difference between a mathematician and a physicist. Mathematicians work out the details and physicists are content with the equations that work. Although in the end it becomes more philosophy than science. To me it is perfectly fine to call the dA = 2*pi*R*dR and when you integrate it we get the correct answer. That is good enough for me.
yes. there are lots of questions in the gaussian shepers and so on. if you go deep, when we integrate a simple linear rod as dx, to find the e filed at a specific point, well i thought about it and, at the center of the rod e field is max and at the ends its minimum and when you think that this rod is X axis you see a Sine function out of E fields on the Y axis that are producted with Cosine Alpha(angle between), well then i ask the question, is there anything wrong on the simple intergration. well i dont know, 1 out of 10000 can be neglected easily
sir, is it possible to use the limit from zero(0) to area of the disk (pi*r^2)?
As long as the area element you choose has a uniform electric field element that you can integrate.
Michel van Biezen thanks a bunch
can this can be solved using gauss's law?
what is the typical unit of the described EF??
What is "EF" ?
N/C. = Newton/ coulomb
Hello teacher
1. Can you Discussion if a >> x
2. If the surface charge density is negative?
1) if a > > x then the charge will act like a point charge 2) if the charge density is negative then the direction of the electric field will be in the opposite direction
Thank you.
Why is there a net field when a=0? At the centre of the disk. Thanks
Sir i have a doubt that if we find the Ef at the centre of disc it doesn't came zero but logically if we thinks it should add upp to zero as we make the disc out of infinte rings and ring have zero ef at centre
Sir pls explain
Great explanation though
When "a" goes to zero, the electric field strength becomes 2 sigma / epsilon sub knot just like we would expect, since that is the electric field above an infinite sheet of charge.
Dear Michel
I guess that what you have drawn is an annulus not a disc of charge. I got a little bit confused.But thank you very much for this video.
Professor, if the distance a appraches negative infinity then the electric field approaches a non-zero constant. This doesn't seem to make sense.
If you let the disk become infinite in size, you will get the solution of the electric field of an infinite plane. (and that is constant, regardless of the distance away from the plane)
@@MichelvanBiezen For a finite disk, as the distance from the disk approach negative infinity, the final equation does not give the correct answer. Instead it gives sigma/epsilon0 which is incorrect as the true answer by intuition is supposed to be 0.
A negative distance does not make any sense. For example if you stand on the street and you measure the distance to the nearest house, can that ever be negative? If you let a approach infinity, the equation does indeed converge at zero.
@@MichelvanBiezen I meant in the negative x-hat direction
I tried using Gauss law to derive the formula for a disc.I got electric field as that of a plane sheet of charge.I do not understand why Gauss law is giving me the wrong formula for a disc.Can anyone help me with this?
What was the shape of the disk?
@@MichelvanBiezen The same disc for which you derived the above formula.A finite uniformly charged circular disc.
Gauss' law only works if the disk is infinite.
Or If you are close enough to the disk so that the disk "appears" infinite.
@@MichelvanBiezen Oh,I get it now.Thanks a lot for replying.
when we can use the Guass' law versus this method to calculate E? what about the electric field on the other side(left) of the disc unless you are considering the charges are only on one side. thank you
Dan, take a look at the playlist on Gauss' law, it shows the typical 6 charge distributions that lend themselves to being able to be solved by Gauss' law method. This isn't one of them. Note that this is the electric field calculated for that location only. If you want it for a different location, you need to do the process all over again.
thank you!
i still have a question: why do you have to name the thickness of the ring "dx"
I see that it is the only way that the integral can be solved but suppose that i did not get that information and i name it "dl" the integral gets confusing and i dont know to what variabel i have to integrate.
Is it just dx because it is an extension of x or how do i know? (sorry for my bad english!!) plz answer
Btw this video is so helpful thank you for uploading :))
Why do you not multiply by cos(theta) to get the electric field in the x-direction? You did that when it was only a ring of charge, so what makes this case any different?
+Daniel Shats I did. cos (theta) = adjacent / hypothenuse = a / (x^2 + a^2)^(1/2) (which is part of the dE term)
why did you use here area charge density and in the the other video of the ring you used Area charge density ?
Thanks in advance
You mentioned "area charge density" for both? Is that what you meant to do? THanks,
Sorry i meant linear charge density in the ring question
It makes more sense to talk about linear charge density when the charges are on a thin strip, (even if the strip is in a circular shape).
Thanks for the help :) most Of the students I know use your videos
when you integrated you put 2 next to the x in the numerator in the integral and a 2 in the denominator outside of the integral, i don't understand why. Great videos by the way!
In order to integrate that integral, you need the correct differential which is 2 x dx. That means we needed to multiply by 2 and divide by 2. (I could have moved the 2 in the numerator into the integral, but I wanted to illustrate why we need to do that).
it's just u sub doesn't matter where you put the constant it in the numerator
Isn't DA the area of the disk and so we use ( Pi*r^2) instead of the circumference formula?
+ACiD Burner No, dA is the area of an infinitesimal thin ring with charge dQ
Why is the area of the ringlet 2*pi*r. I'm really having difficulty understanding (conceptually) why you're substituting perimeter for area when the formula calls for an area.
The are is not 2 * pi * r, but it is 2 * pi * r * dr where dr is the thickness of the ring and thus that represents area.
@Armens cut the ring open. You will see it takes the shape of a rectangle with length 2*pi*r and breadth 'dr'. So the area then becomes (2*pi*r*dr).
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Thank you. Welcome to the channel.
For the integration could you also have used a Trig-Sub by taking x=a*tan(w)?
There are often multiple ways to do a problem, but as it is presented in the video is probably the best. Why don't you try it and see if you get the same answer?
Sir in previous video we have taken the ds as element and its equal to x. dfie. Here y u have taken the dx as element we have integrated from x=0to r if we take ds instead of dx then whole integration changes
How you set up the integration depends on how you can best set up the dq so that the electric field of that dq can be most easily represented. (There are multiple ways to do this). In this case it appeared easiest to take the whole ring as a dq. That way only the horizontal component of the electric field survives. (the perpendicular component will cancel out).
Sir why you take ring width dr into account
You need to find a small segment dA with charge dQ on the disk that you can then integrate over the entire disk. There are different integration techniques you can use. This is one of them.
Why do we take the thickness as dx. Is the thickness related to to the radius x? This has caused much confusion to me.
This is an integration technique and based on the definition of integration . When you integrate you let the thickness approach zero and integrate over an (in the limit) infinite number of small strips of thickness dx. Take a look at this video for a reference: Calculus - Integration: Finding the Area Between Curves (1 of 7) Ex. 1: y=e^x, y=x^2, x=0, x=2 ruclips.net/video/mzdWhFh3050/видео.html
Why did you keep the negative in the integral if -1 is a constant when you divided by -1/2?
It is not in the integral. (It is in the result of the integral). In this case, looking ahead, it makes sense because you end up with 1 - (small number) in the answer which makes it easier to interpret the answer.
Sir, why did we take the differential of the nominator?
I didn't quite get your question, but I think you are asking why need a differential to integrate? You always need a differential to integrate according to the rules of integration. (for example you can't integrate x, but you can integrate x dx, where dx is the differential of x.
Sir why d area of ring is not pir^2 instead 2pirdr
It is basically a rectangle in the shape of a circle. pi x R^2 is the area of a disk (not the ring).
@@MichelvanBiezen thanq sir
Hello sir, I'm not understanding why you used "x" and "a" into the integration if they are constants. Could you please elaborate?
Thank you!
Since they are constants, we could have assigned any character. We chose x and a.
@@MichelvanBiezen sorry I'm a bit confused, in the last video where you calculated the ring of charge, you didn't include the variable x and a into the integration. Thus I am confused why you included in the charge of disk example. Thank you for taking the time to reply to me 🙏
Don't try to compare the 2 videos. Look at each video and determine if the setup makes sense. The key in each case is to find the electric field at a particular point P, due to the presence of a small charge element, and how to define that charge element and the distance from the element to the point P
Thank you for all each letters that you taugh me by your videos Michel Van Biezen :D
how would I do a solid cylinder
Take a look at this playlist for more examples: PHYSICS 37 GAUSS'S LAW
I'm still confused about taking the integration from 0 to R , how ?
You need to account for the contribution of all the charge on the disk, so you add all the rings of charge contributions (integration)
Why we said from0 to R not 2piR ?
The only variable in the integration is r. We don't have to integration around the circle since our area element already is in circular form.
Thank you sir ✨
How can we find out of the electric field of disc due to negative charge ??
Exactly the same way, but the directions will be opposite.
Why it isn't dE cos(the angle) as a component of X?
If you calculate the cos(theta) you will see that it is part of the dE expression.
what if we suppose a ring which radius is equal to Ro?
+Gebran Khan Look at video: Physics - E&M: Electric Field (8 of 13) Ring of Charge
If we write sigma*pi*r^2 instead of dq,we make mistake??
Also thanks for all this helpful videos sir..
+SERDAR KAMİ ÜÇKARDEŞ For the thin ring with chargedA = circumference * thickness = 2*pi*r * dr
+Michel van Biezen no mistakes at all.
Thank you so much!
Notice how it behaves when a tends to infinity