THANK YOU. you are seriously a life saver to so,so many students out here. Most of the time, physics doesn't make sense to me until I watch your videos
Watching this a second time and it is more beautiful on the second viewing. Thanks so much for making these lectures available. Certainly is helping with E&M.
Sir I'm from India and had searched for many lectures already where I could find the best one but then founds you here.Your lectures are very nice and the way you explains fits in my mind permanently......... Stay tuned like this always 😇
I needed this for my physics homework. Actually I just need the equation at 10:53. Now I need to go back and learn everything up to that timestamp... Thank you professor.
Sir, thanks for all you do and more grease to your elbow. I have learned a lot from your videos, especially whenever I need more class time on these topics, your videos are always there for me. Thank you and God bless. With my background in math, I have worked hard to where I am now and few more semesters to graduate, if I can do this, I know you folks out there can too, just put your mind into it and it will come.
comment to the infinity / infinity: On your calculator, let a number like 1,000,000 represent infinity and a much smaller number for a: like a=2. Plug them in at the integral evaluations and you will find you get 1- (-1)=2...this PARTICULAR one is defined as stated!
Thank you very much, i was searching exactly for this, everywere i looked it was explained with Gauss Law and the gaussian surface, i had to search in english to find this video whith the Coulomb's law solution. Greetings from Argentina.
If you are having problem convincing yourself with the last expression, use the fact that the integral is even (integrate from 0 to inf instead) and then draw a diagram of the last expression with the interpretation sin of an angle
The way you would integrate that would be to trig sub x=atan(theta), dx=asec^2(theta) and use Pythagorean identity. tan^2 theta = sec^2 theta - 1 pull the 1/a^2 out You'd end up with sec^2 d theta / sec^3 theta = cos theta. Integrate that to get sin theta, plug back into triangle sin theta = x/sqrt(x^2+a^2) multiply that by 1/a^2 you pulled out. Should get the same answer.
The interpretation of the result is that the vector field are parallels field lines, but perpendicular to the charged line. And the magnitude decays to the power of one with the distance (instead of two). And more important, it is independent of the length of the line.
Pro Tip: limit of x approaches infinity/ infinity is equal to 1. If you have a TI-89, you'll see infinity/ infinity is undef, but if you hit F3, limit(x/x, x, infinty), you'll see one or just do 10^100/10^100. :)
Hi, Thank you so much for your instructional videos! I just had a particular question in your process. At around 7:27, when you defined dEx, why is it that you didn't multiply the equation by 2 given that while the vertical (y) components of the charges above and below the horizontal lines cancel out, their horizontal components match and double? Again thanks for your videos! I always come to watch them after having read the book to reinforce my understanding!
Note that the limits of integration are from negative infinity to positive infinity, therefore all the contributions from the top portion and bottom portion are accounted for.
While inf/inf I don't casually agree with, what I find amazing about this is what was NOT calculated. My thought experiments are making me question the following: (1) if the line charge is symmetrically spherical/circular, the result, I assume, would be the same. (2) (more importantly) if the line charge is NOT perfectly symmetrically (i.e., contains jagged edges), then the Electric field would contain Electric Field components in the Y-Direction, which, effectively, takes away x-component electric field values, REDUCING the overall result. I find the (2) thought experiment more important because, while we want an overall "Electric Field Measurement" that is easy to calculate, we should account for how any impurity would affect our initial measurements. A "kink" on the top of the line charge would cause an E-Y component to exist and that Y-Component would take an electric field value from E_X.
Doing L'opitals will not work or at least I just kept going in a big circle. I divided through by x before taking the limit using infinity and negative infinity. For the life of me, I get 1 for the positive infinity and 1 for negative infinity which subtracting equals zero. Maybe it is me but treating infinity like a real or any kind of number and doing operations with it probably saves time if you know the outcome but confuses the hell out of everybody. Limits in indeterminate forms do take time but I would really like to know how you determined the second fraction as negative. Thanks again
Hi Michael..... once again.. thanks for the excellent presentation. As I said in previous posts there should be an award for your Excellence in Teaching..... I have a question regarding this video. What if the distance "a" was set to ZERO.... since "a" appears in the denominator of the final answer for the E field... that means that the E field strength in the X-direction at "a" = to 0 would be INFINITY.... and yet at a=0 there really isn't an x direction vector at that point. However, if there was an X-directed E field at a=0 and the E field strength was Infinity.. then I guess that means every point up and down the Infinite Line Charge would be at infinity too..... Mind Boggling, YES?..... any opinion on that?... thanks again for fantastic lectures and demonstrations. I no longer refer to my text books for Refresher courses, I simply go to your videos!!!!! THANK YOU...
Philip That is an excellent question and from a mathematical point of view it does present a strange situation. But in real life there isn't such a situation of zero distance to a charge distribution. Imagine that you get closer and closer. Eventually you get down to the molecular level and to the actual individual charges that produce the electric field. These are the free electrons that are constantly going to be moving (like waves) and providing some actual distance between them and the "probe" or observation point. Yes the electric field will be very large as it represents the change in potential divided by distance and that can be very large being very close to charges.
That is the difference between pure mathematics and how mathematics is applied in the real world. There is no such thing as "infinity". Nothing is infinite in the universe. So what we do is make the numbers bigger and bigger and see what the ratio becomes as the numbers become very big.
Oh yeah, something I get confused often: The symbol for charge, is it Q or C. Since C is usually "Capacitance", I would guess "Q". Which is a symbol representing "Farads" which is symbolized by "F" ie C=10F. ?? But you used "C" in the expression for "charge per unit length: C/m.
In physics many symbols are used several times, so you have to figure out the meaning through the context. The units of charge is Coulombs, so I indicated the number of coulombs per unit length.
Sir, my doubt is that why u took components to find electric field?? Basically u first found out it to be k (lambda) dx / (a^2 + x^2). Why not u directly integrate this term from minus infinity to plus infinity? Plz clear my doubt..
Michel van Biezen , Sir, that integration comes out to be [k (lambda) tan inverse (x/a) ] / a. If I put the limits from minus infinity to plus infinity, answer is coming different. It is coming (pi) k ( lamda) / a. ??? Am confused.
What if we wanted to figure out the electric field strength at a point, not right in the middle of the rod? Would we need to integrate electric field in a y direction too?
I was thinking since your line charge is always positive and thus can only produce a positive field, would that negate using negative infinity or negative values to evaluate with? Even so, I can find anything that would equate the second fraction to a negative one. Thanks
in an ideal physical problem like this we can assume that the line is soooooo long that it might as well be infinite, but physically speaking that isn't possible, so as long as you assume equal distances from top and bottom of the 'ends' of this infinite line charge they would cancel out exactly and leave 1
oh kid you know nothing... This is a very very basic calculation that we first encountered in the first week of the second semester physics bsc at bme hungary
@@gubazoltan99 Whatever, in my program "Engineering Physics & Electrical Engineering" in Sweden we were busy with math and physics courses during our first year and didn't have room for electromagnetics that year. I took this course on my second year, and only because my first year was already fully booked, and line charges were some of the very first topics there.
infinity / infinity is indeed indeterminate. However if you allow the variable to reach infinity in the limit, what does the fraction converge to? That is a better way to look at it in the "real" world.
Good day Mr.Binzen, can you please help me this question please. a. Using Coulomb’s law, find the electric field a distance z above one-third of one end of a straight line segment of length L, which carries a uniform line charge λ. b. What would the electric field due the straight line charge in (1a) at z be if we consider the line charge to be infinitely long?
If the point of interest is not directly in the middle of the line segment then the vertical component does not cancel out and you'll have to integrate twice. Also the limits of integration will change. (Part b is straight forward, then it acts as if it is in the middle, see the video for that case).
hello Prof may you please tell me how to approach this question A uniform line charge of 2 μC/m is located on the z axis. Find the electric field E in Cartesian coordinates at P(1, 2, 3) if the charge extends from - infinity < z < infinity ? ? ?
Since the charged like goes from infinity to infinity on the z-axis, the z-location doesn't matter and you can work the problem with only the x and y coordinate point. The distance from the charged line = sqrt (1^2 + 2^2). The rest is just the same as on the video.
thank you very much Prof i think i get it, i do since the uniform line charger extends to infinity the location of where the the force is felt is the same always and i only need to know the distance from the uniform line charge which by pythagoras rule is sqrt(1^2+2^2) beautiful thank you very much Prof.
In a purely mathematical sense, yes, but in a real world application sense, no. In the real world, infinity does not exist, and we use the approach: "as the variable approaches infinity", which allow us to do that.
That is the mathematical way of looking at it. In the real world infinity doesn't exist, so we see what happens when the distance becomes very large. In the limit the ratio becomes one.
Usually you try to memorize a handful of integrals. If you are interested in how to integrate we have a playlist with many integrals: CALCULUS 2 CH 15 HOW DO YOU INTEGRATE?
Prof may you please do this for me it tried doing it but i am missing something i just cant get to the answer. . . its not as easy as it seems i suppose A uniform line charge of 2 μC/m is located on the z axis. Find the electric field E in Cartesian coordinates at P(1, 2, 3) if the charge extends from - infinity < z < infinity ? ? ?
the problem i encounter is in which direction will the resulting E be, cos " was used as a unit vector in the example above i cant figure out what to use
after getting dE = K (~dx/(x^2 + a^2)^2) because sum of E in y-axis cancel / add to 0, cosx was used to find the resulting E in one direction . this problem has resulting E in the y axis and x-axis, how can i tackle that ? thank you very much for all your help Prof.
Can you explain how do you change the sign of the -infinity part to + and get 1+1@10:35? I get that 1-1 doesn't make sense, so 1+1 is right. But i'm confused about the sign changing process.
Actually the expression is evaluated as the number approaches infinity. (infinity doesn't exist) As far as the sign is concerned. The numerator will be negative and the denominator positive since it is squared. Subtracting a negative number makes it positive.
In mathematics that is correct. However in the real world there is no such thing as "infinity" and it is better to replace the limit by a large number and you can make it as large as you want.
i am a bit confused. Although the explanation is great. I had read in some book that electric field is equal to i.e E= ⍴L/2π⍴∈∘ ⍴L is uniform charge density of line charge. ⍴ is charge density Please reply soon
Excuse me, if we assume that there is another dE that cancel the effect of dE in Y direction, why in X direction shows only dE not 2dE? Don't they add to each other?
What if the infinite line charge were to be drawn horizontally, would you multiply the magnitude of dE by sine theta since the components of dE change if you do so? If so, would sine theta= x/(x^2 + a^2)3/2?
I have doubt about that integral and I am stuck on it cuz I have used reverse power rule and I am getting -2/√(x^2+a^2) and keeping aside the constant that is lambda and k•a
What sometimes helps is reversing the process. Take the result of the integral and take the derivative and then see why you get the original integrand back.
Hi sir. I found the sample above electric field problem is somehow similar and applicable also to radiometry ray tracing e.g. light with d²P = L(x, x1→x2)(dA₁cosθ₁)(dA₂cosθ₂/r²) where L(x, x1→x2) is the power transfer from x1 to x2 surfaces area per dA₁cosθ₁ of x1 per dAcosθ₂/r² solid angle that it subtends on x2. A key difference here is dA₁cosθ₁ of x1 the infinite conductor part. It requires foreshortening from x1 where dA₁cosθ₁ is the foreshortened area viewing angle with the normal of x1 surface while the test charge side x2 is similar to solving for dP/dA₂ instead of electric field. My question sir is why do we need to foreshorten the x1 part when it comes to energies such as light and cannot it behave just like electric field lines? Was it due to wave cancelling out interferences at these narrowed foreshortened angles and if so does it mean losing energies because they are not reflecting back equally from all angles? Thank you!
biot savarts law gives the direction of magnetic field as dl×ar . similarly, is there any way to get electric field direction through vector products.??
There is no such thing as "infinity" in the real world, it is just a mathematical concept. In the real world when we work with physics concepts think of infinity as a really big number.
In this example the charge is distributed along a thin line. The lambda symbol represents the charge per unit length, which is appropriate for this type of geometrical arrangement.
That particular integral appears a lot in E&M type of physics problems. I remember a course I took (a very long time ago) where this integral appeared dozens of times.
kreweta00 Infinity / infinity is off course zero. But in this problem, you have to take the limit. As you keep increasing x, as x reaches infinity, the fraction converges to 1. (try it)
Mathematically that is correct. But in the real world there is no such thing as "infinity". Therefore when we calculate such quotient, we must take the limit and see what happens when the numbers become very big and APPROACH infinity, but they can never be infinity. (That same principle is used in calculus)
In mathematics yes. But in the real world there is no such thing as infinity. Replace infinity by a really large number like a million or a billion and see what happens.
for this integral we have to substotute x=a*tan(u) and dx will be dx=asec^2(u)du Then (a^2 + x^2)^3/2 = (a^2*tan(u)^2 + a^2)^3/2 = a^3 * sec(u)^3 and u=tan(x/a)^-1........ No integral tables can expain that ********, and you call that "Not that hard"? pls O.o i will pass Eletromagnetic field(i think xD), but not thank to you...
when one has infinity/infinity aren't you supposed to take limits and use L'hopital rule? because i don't know if one can just cut out infinity and infinity as they are "undefined" quantities i.e no real value such as say 7/7 or x/x. maybe i am missing something however
It depends on the equation. From a mathematics point of view, when you strictly look at infinite / infinity, you can correctly claim that it is undefined. But in the real world, there are situations like this example, where you need to converge to the limit and see what happens. For example, what is sin(theta) / theta when theta equals zero, and the answer is undefined. But what is sin(theta) / theta as theta approaches zero, and the answer is "one".
THANK YOU. you are seriously a life saver to so,so many students out here. Most of the time, physics doesn't make sense to me until I watch your videos
Watching this a second time and it is more beautiful on the second viewing. Thanks so much for making these lectures available. Certainly is helping with E&M.
You are welcome. Glad you are enjoying our videos. 🙂
Loved how casually the guy said infinity over infinity is one xD xD. If you want to succeed in life, you need this much of confidence
Actually,he has given the explanation in one of the comments
we know ..but hey dont discourage someone ....
literally youre the only reason I am not failing physics, thank you!!!
Sir I'm from India and had searched for many lectures already where I could find the best one but then founds you here.Your lectures are very nice and the way you explains fits in my mind permanently......... Stay tuned like this always 😇
pro status! Note to self - always trust physics advice from a man in a bowtie.
I needed this for my physics homework. Actually I just need the equation at 10:53. Now I need to go back and learn everything up to that timestamp... Thank you professor.
You came to the right place if you want to learn physics. (There are over 100 playlists covering every topic in physics).
Sir, thanks for all you do and more grease to your elbow. I have learned a lot from your videos, especially whenever I need more class time on these topics, your videos are always there for me. Thank you and God bless. With my background in math, I have worked hard to where I am now and few more semesters to graduate, if I can do this, I know you folks out there can too, just put your mind into it and it will come.
This will always be one of the best references for future generations of learners. Kudos to you, mr. Biezen.
Extremely clear and intuitive derivation. If only you were my professor.
10:42 all mathematicians: WTFFFFFFFFFf
:)
I was like how is that possible ?
Fr
comment to the infinity / infinity:
On your calculator, let a number like 1,000,000 represent infinity and a much smaller number for a: like a=2. Plug them in at the integral evaluations and you will find you get 1- (-1)=2...this PARTICULAR one is defined as stated!
Best teacher on TH-cam.
Thanks again it looks like i will be with you until i graduate from Mechatronics Engineering.Greetings from Türkiye.
Glad you found our videos and they are helping.
Who else watched video in hope of explanation of solving the integral.
you can do inverse angle subtituion
@Connor Rylan lmao
Me
Put X=a tan theta from figure, limits will then vary from - pai/2 to + pai/2.
How do I go about it?
Thank you very much, i was searching exactly for this, everywere i looked it was explained with Gauss Law and the gaussian surface, i had to search in english to find this video whith the Coulomb's law solution. Greetings from Argentina.
I can't not even find a right word to thank you sir.
You just did. Much appreciated.
genius and with a best simplification
If you are having problem convincing yourself with the last expression, use the fact that the integral is even (integrate from 0 to inf instead) and then draw a diagram of the last expression with the interpretation sin of an angle
missed a couple classes and am so confused during class. this is very helpful, thank you
The way you would integrate that would be to trig sub x=atan(theta), dx=asec^2(theta) and use Pythagorean identity. tan^2 theta = sec^2 theta - 1
pull the 1/a^2 out
You'd end up with sec^2 d theta / sec^3 theta = cos theta.
Integrate that to get sin theta, plug back into triangle
sin theta = x/sqrt(x^2+a^2) multiply that by 1/a^2 you pulled out.
Should get the same answer.
Hi. I know it's a very long time ago since you made this comment but can you explain more simply where sec^3 came from in the denominator. Thank you.
@@joshuaomar2992 in the denominator we have (a^2*sec^2)^(3/2) so when we do the math we will have a^3*sec^3 in the denominator
The interpretation of the result is that the vector field are parallels field lines, but perpendicular to the charged line. And the magnitude decays to the power of one with the distance (instead of two). And more important, it is independent of the length of the line.
Pro Tip: limit of x approaches infinity/ infinity is equal to 1. If you have a TI-89, you'll see infinity/ infinity is undef, but if you hit F3, limit(x/x, x, infinty), you'll see one or just do 10^100/10^100. :)
These are FABULOUS! Thank you! I am sharing with my students!!
Thank you, I struggled with this problem in the Freedman and Young book.
Glad we can help.
Hi, Thank you so much for your instructional videos! I just had a particular question in your process. At around 7:27, when you defined dEx, why is it that you didn't multiply the equation by 2 given that while the vertical (y) components of the charges above and below the horizontal lines cancel out, their horizontal components match and double?
Again thanks for your videos! I always come to watch them after having read the book to reinforce my understanding!
Note that the limits of integration are from negative infinity to positive infinity, therefore all the contributions from the top portion and bottom portion are accounted for.
Wonderful! Thanks for making this easy to understand I appreciate your lectures.
You are welcome. You will be ready when you start your summer course.
These videos are incredible and you're a great educator. Thank you!
While inf/inf I don't casually agree with, what I find amazing about this is what was NOT calculated.
My thought experiments are making me question the following:
(1) if the line charge is symmetrically spherical/circular, the result, I assume, would be the same.
(2) (more importantly) if the line charge is NOT perfectly symmetrically (i.e., contains jagged edges), then the Electric field would contain Electric Field components in the Y-Direction, which, effectively, takes away x-component electric field values, REDUCING the overall result.
I find the (2) thought experiment more important because, while we want an overall "Electric Field Measurement" that is easy to calculate, we should account for how any impurity would affect our initial measurements. A "kink" on the top of the line charge would cause an E-Y component to exist and that Y-Component would take an electric field value from E_X.
In the real world infinity does not exist. Therefor in physics infinity is replaced by a very large number, then it works.
Holy Cow, you, you, you are a master !!! Many thanks
i officially love you!! I will check your video during all my semester
Sir thanks for sharing this lecture for free with us
It's our pleasure. We remember as students it was difficult to find good references for help. (Time to give something back).
This was a very helpful explanation. I am definitely less confused now. Thank you :)
I have a test in 45 mins and this hlped me a lot🤗🤗
Thanks for making it super easy and fun :)
Doing L'opitals will not work or at least I just kept going in a big circle. I divided through by x before taking the limit using infinity and negative infinity. For the life of me, I get 1 for the positive infinity and 1 for negative infinity which subtracting equals zero. Maybe it is me but treating infinity like a real or any kind of number and doing operations with it probably saves time if you know the outcome but confuses the hell out of everybody. Limits in indeterminate forms do take time but I would really like to know how you determined the second fraction as negative. Thanks again
Hi Michael..... once again.. thanks for the excellent presentation. As I said in previous posts there should be an award for your Excellence in Teaching..... I have a question regarding this video. What if the distance "a" was set to ZERO.... since "a" appears in the denominator of the final answer for the E field... that means that the E field strength in the X-direction at "a" = to 0 would be INFINITY.... and yet at a=0 there really isn't an x direction vector at that point. However, if there was an X-directed E field at a=0 and the E field strength was Infinity.. then I guess that means every point up and down the Infinite Line Charge would be at infinity too..... Mind Boggling, YES?..... any opinion on that?... thanks again for fantastic lectures and demonstrations. I no longer refer to my text books for Refresher courses, I simply go to your videos!!!!! THANK YOU...
Philip
That is an excellent question and from a mathematical point of view it does present a strange situation.
But in real life there isn't such a situation of zero distance to a charge distribution.
Imagine that you get closer and closer. Eventually you get down to the molecular level and to the actual individual charges that produce the electric field. These are the free electrons that are constantly going to be moving (like waves) and providing some actual distance between them and the "probe" or observation point.
Yes the electric field will be very large as it represents the change in potential divided by distance and that can be very large being very close to charges.
Exceptional teacher. Thanks a ton!
Glad you liked it!
Thanks you very much master I finally gotta the right person you're the best I swear.
Youre a genius professor, this really helped me
I have not yet formally learned improper integrals, but from my understanding, saying infinity/infinity = 1 is slightly misleading?
That is the difference between pure mathematics and how mathematics is applied in the real world. There is no such thing as "infinity". Nothing is infinite in the universe. So what we do is make the numbers bigger and bigger and see what the ratio becomes as the numbers become very big.
Oh yeah, something I get confused often: The symbol for charge, is it Q or C. Since C is usually "Capacitance", I would guess "Q". Which is a symbol representing "Farads" which is symbolized by "F" ie C=10F. ?? But you used "C" in the expression for "charge per unit length: C/m.
In physics many symbols are used several times, so you have to figure out the meaning through the context. The units of charge is Coulombs, so I indicated the number of coulombs per unit length.
Thanks this a life saver. BTW great teaching going on there
Sir, my doubt is that why u took components to find electric field?? Basically u first found out it to be
k (lambda) dx / (a^2 + x^2). Why not u directly integrate this term from minus infinity to plus infinity? Plz clear my doubt..
The limits of integration can be chosen as you indicated if you prefer to integrate it that way. You will get the same answer.
Michel van Biezen , Sir, that integration comes out to be
[k (lambda) tan inverse (x/a) ] / a. If I put the limits from minus infinity to plus infinity, answer is coming different. It is coming
(pi) k ( lamda) / a. ??? Am confused.
Could we simply say that infinity/infinity = 1 or we must take the limit of x/x^2(x^2+a^2)^0.5 when x approaches infinity
Only in the limit as a value approaches infinity. infinity / infinity is undefined (since infinity does not exist)
Very clear... easy to understand
Thank you so much, sir. I am forever indebted to you.
I guess it will be easy if we take
x= a Tan (theta) for integration.
What if we wanted to figure out the electric field strength at a point, not right in the middle of the rod? Would we need to integrate electric field in a y direction too?
That is correct. For a point that is off-center, you would have to integrate for both the x and y directions.
I was thinking since your line charge is always positive and thus can only produce a positive field, would that negate using negative infinity or negative values to evaluate with? Even so, I can find anything that would equate the second fraction to a negative one. Thanks
Thnk you so much sir ! U are an angel In distress ! Thank you sir!
Very nice explanation
Thanks and welcome
Could someone explain why the direction of E is x hat?
Because of the symmetry, the components in the y-direction cancel and only the components in the x-direction are left.
@@MichelvanBiezen Thank you so much!!
who is the best you
who is number you
who is the best teacher in the world
Saving my grade again!!! Thank you
Isn't the theta just 45 degrees? 7:06
The angle is not constant.
@MichelvanBiezen I mean Sir, the righttriangle has two thetas in it and one 90 degree. So my question is wrong?
Infinity over Infinity is undetermined.
How did you assume that it will equal 1?
in an ideal physical problem like this we can assume that the line is soooooo long that it might as well be infinite, but physically speaking that isn't possible, so as long as you assume equal distances from top and bottom of the 'ends' of this infinite line charge they would cancel out exactly and leave 1
This is more or less second-or-third-year college physics.
Pretty cool.
oh kid you know nothing... This is a very very basic calculation that we first encountered in the first week of the second semester physics bsc at bme hungary
@@gubazoltan99
Whatever, in my program "Engineering Physics & Electrical Engineering" in Sweden we were busy with math and physics courses during our first year and didn't have room for electromagnetics that year.
I took this course on my second year, and only because my first year was already fully booked, and line charges were some of the very first topics there.
if you'd solved the integral it would've been perfect.
sit infinite by infinite is not equal to one but it is an indeterminate form?
infinity / infinity is indeed indeterminate. However if you allow the variable to reach infinity in the limit, what does the fraction converge to? That is a better way to look at it in the "real" world.
Good day Mr.Binzen,
can you please help me this question please.
a. Using Coulomb’s law, find the electric field a distance z above one-third of one end
of a straight line segment of length L, which carries a uniform line charge λ.
b. What would the electric field due the straight line charge in (1a) at z be if we
consider the line charge to be infinitely long?
If the point of interest is not directly in the middle of the line segment then the vertical component does not cancel out and you'll have to integrate twice. Also the limits of integration will change. (Part b is straight forward, then it acts as if it is in the middle, see the video for that case).
Beautifully done.
hello Prof may you please tell me how to approach this question
A uniform line charge of 2 μC/m is located on the z axis. Find the electric
field E in Cartesian coordinates at P(1, 2, 3) if the charge extends from
- infinity < z < infinity ? ? ?
i watched the video that follows this 1 and noticed its not in 3D but this question is . . .
Since the charged like goes from infinity to infinity on the z-axis, the z-location doesn't matter and you can work the problem with only the x and y coordinate point. The distance from the charged line = sqrt (1^2 + 2^2). The rest is just the same as on the video.
thank you very much Prof
i think i get it, i do
since the uniform line charger extends to infinity the location of where the the force is felt is the same always and i only need to know the distance from the uniform line charge which by pythagoras rule is sqrt(1^2+2^2) beautiful thank you very much Prof.
Walmart James Murray really helped me with my physics problem
Did you integrate an improper integral at 9:06
In a purely mathematical sense, yes, but in a real world application sense, no. In the real world, infinity does not exist, and we use the approach: "as the variable approaches infinity", which allow us to do that.
10:25 isn't infinity divided by intifinity undefined ? I am confused
That is the mathematical way of looking at it. In the real world infinity doesn't exist, so we see what happens when the distance becomes very large. In the limit the ratio becomes one.
should the answer be (4kh)/a because it is the sum of two vectors not just one?
thanks
+mohamed shata
No, the answer is correct. That was taken care of by taking the limits form + infinity to - infinity
+Michel van Biezen yeah you are right. Sorry. i just got confused. thanks for replying 😊
+Michel van Biezen i am really relying on your explanations. because it is really awesome
Thank you sir so much !
I have a question though, how did you find the antiderivative when doing the integral ?
Thank you in advance for your answer!
Usually you try to memorize a handful of integrals. If you are interested in how to integrate we have a playlist with many integrals: CALCULUS 2 CH 15 HOW DO YOU INTEGRATE?
I couldn't find that exact integral on his channel but here is the solution th-cam.com/video/iwL60KHp0Po/w-d-xo.html&ab_channel=IntegralsForYou
In x axis ,there r 2dE right? Then why we r writing only one dE ?please give me an easy answer sir 😔
We only utilize one of the two dE components to find the solution. (Two were drawn to show the symmetry and to show why the y-component cancels).
i love your videos
Prof may you please do this for me it tried doing it but i am missing something i just cant get to the answer. . . its not as easy as it seems i suppose
A uniform line charge of 2 μC/m is located on the z axis. Find the electric
field E in Cartesian coordinates at P(1, 2, 3) if the charge extends from
- infinity < z < infinity ? ? ?
can i use gauss's law for this ? ? ?
the problem i encounter is
in which direction will the resulting E be, cos " was used as a unit vector in the example above i cant figure out what to use
The direction of the electric field is found by theta = arc tan (2/1) = 63.4 degrees from the x-axis.
after getting dE = K (~dx/(x^2 + a^2)^2) because sum of E in y-axis cancel / add to 0, cosx was used to find the resulting E in one direction . this problem has resulting E in the y axis and x-axis, how can i tackle that ? thank you very much for all your help Prof.
Can you explain how do you change the sign of the -infinity part to + and get 1+1@10:35? I get that 1-1 doesn't make sense, so 1+1 is right. But i'm confused about the sign changing process.
Actually the expression is evaluated as the number approaches infinity. (infinity doesn't exist) As far as the sign is concerned. The numerator will be negative and the denominator positive since it is squared. Subtracting a negative number makes it positive.
Oh. Now i see. Thank you so much.
Thank you Sir. It help a lot!!!
When you have infinite limits you have to treat them as an improper integral. At least that’s what my prof taught me.
In mathematics that is correct. However in the real world there is no such thing as "infinity" and it is better to replace the limit by a large number and you can make it as large as you want.
What an actual legend
EXCELLENT explanation professor! thank you!
i am a bit confused. Although the explanation is great. I had read in some book that electric field is equal to i.e
E= ⍴L/2π⍴∈∘
⍴L is uniform charge density of line charge.
⍴ is charge density
Please reply soon
is there a video on how to integrate this equation? thank you
We may have covered that in the calculus videos, but I can't remember.
th-cam.com/video/2sCmuwTrgDU/w-d-xo.html
Excuse me, if we assume that there is another dE that cancel the effect of dE in Y direction, why in X direction shows only dE not 2dE? Don't they add to each other?
That is taken care of by choosing the limits at + infinity and - infinity
Sir....is there any lecture on potential due to dipole....plz share its link.....plz
From the home page of the channel type: ILECTUREONLINE DIPOLE and they will all show up.
Thank you so much sir😊
Thank you so much for this!
What if the infinite line charge were to be drawn horizontally, would you multiply the magnitude of dE by sine theta since the components of dE change if you do so? If so, would sine theta= x/(x^2 + a^2)3/2?
+jandon603
If you draw the infinite line charge horizontally, the problem would be exactly the same.
I know it is a late reply but,
Just rotate the page and you get a vertical infinite rod :D
@@tushargahlaut5812 replying for the other 2 year mark
@@ismail.a4015 😳
Thank you very much Sir.
You are welcome.
I have doubt about that integral and I am stuck on it cuz I have used reverse power rule and I am getting -2/√(x^2+a^2) and keeping aside the constant that is lambda and k•a
What sometimes helps is reversing the process. Take the result of the integral and take the derivative and then see why you get the original integrand back.
@@MichelvanBiezen ohhh I have to practice more on integration and thanks for the advice.
Ohh yeah now I understood that you have used trig substitution to find out the integral of 1/(a^2+x^2)^3/2.
technically our universe is infinite, its still expanding and will continue to do so according to our current knowledge of physics.
Hi sir. I found the sample above electric field problem is somehow similar and applicable also to radiometry ray tracing e.g. light with d²P = L(x, x1→x2)(dA₁cosθ₁)(dA₂cosθ₂/r²) where L(x, x1→x2) is the power transfer from x1 to x2 surfaces area per dA₁cosθ₁ of x1 per dAcosθ₂/r² solid angle that it subtends on x2. A key difference here is dA₁cosθ₁ of x1 the infinite conductor part. It requires foreshortening from x1 where dA₁cosθ₁ is the foreshortened area viewing angle with the normal of x1 surface while the test charge side x2 is similar to solving for dP/dA₂ instead of electric field. My question sir is why do we need to foreshorten the x1 part when it comes to energies such as light and cannot it behave just like electric field lines? Was it due to wave cancelling out interferences at these narrowed foreshortened angles and if so does it mean losing energies because they are not reflecting back equally from all angles? Thank you!
biot savarts law gives the direction of magnetic field as dl×ar . similarly, is there any way to get electric field direction through vector products.??
Yes you can.
tell us Sir for infinite line charge case. we have dx=dz in +ve direction and unit vector ar ; in cylindrical coordinate system
Can we divide infinity to infinity?
Why we don't use limit to solve that improper integral Sir?
There is no such thing as "infinity" in the real world, it is just a mathematical concept. In the real world when we work with physics concepts think of infinity as a really big number.
Sir when we are using the λ symbol because we are using cylindrical gaussian surface too. I couldn't understand the idea could you help me, please?
In this example the charge is distributed along a thin line. The lambda symbol represents the charge per unit length, which is appropriate for this type of geometrical arrangement.
Make theoritical video about String theory
At 09.00 I don't understand how you deduced the integral like that. can you help me?
That particular integral appears a lot in E&M type of physics problems. I remember a course I took (a very long time ago) where this integral appeared dozens of times.
Why there is n't repulsion force between charges of the rod
There are and thus the charges will distribute themselves as far away from each other as possible.
so the field must be perpendicular to the rod
All nice and great, videos are cool. But I wonder how many people covered ears when you said "inf / inf is one" xD (Indeterminate form)
kreweta00
Infinity / infinity is off course zero.
But in this problem, you have to take the limit. As you keep increasing x, as x reaches infinity, the fraction converges to 1. (try it)
Sir, infinity divided by infinity is undefined
Mathematically that is correct. But in the real world there is no such thing as "infinity". Therefore when we calculate such quotient, we must take the limit and see what happens when the numbers become very big and APPROACH infinity, but they can never be infinity. (That same principle is used in calculus)
Can anyone tell me what would happen if there were an infinite number of these exact wires, placed along the x axis, with point P on on the y-axis?
same
But infinity divide by infinite is undefined isn't it ?
In mathematics yes. But in the real world there is no such thing as infinity. Replace infinity by a really large number like a million or a billion and see what happens.
professor, pls explain to me how you did that integral :/
for this integral we have to substotute x=a*tan(u) and dx will be dx=asec^2(u)du
Then (a^2 + x^2)^3/2 = (a^2*tan(u)^2 + a^2)^3/2 = a^3 * sec(u)^3 and u=tan(x/a)^-1........
No integral tables can expain that ********, and you call that "Not that hard"? pls O.o
i will pass Eletromagnetic field(i think xD), but not thank to you...
when one has infinity/infinity aren't you supposed to take limits and use L'hopital rule? because i don't know if one can just cut out infinity and infinity as they are "undefined" quantities i.e no real value such as say 7/7 or x/x. maybe i am missing something however
It depends on the equation. From a mathematics point of view, when you strictly look at infinite / infinity, you can correctly claim that it is undefined. But in the real world, there are situations like this example, where you need to converge to the limit and see what happens. For example, what is sin(theta) / theta when theta equals zero, and the answer is undefined. But what is sin(theta) / theta as theta approaches zero, and the answer is "one".
@@MichelvanBiezen ah that explains it, thanks