Michel van Biezen or should I start calling you professor? I'd just like to thank you and your assistants for providing this educational content on youtube. You have multiple videos with in depth examples and it just about covers everything i'm learning in my first year engineering class at Mcmaster University. Sir, I just got my results back, and I scored an 87% on my midterm. Although I worked my ass off, you made it incredibly easy for me to learn what I needed to know for the test. So thank you once again. I'm going to be asking a lot more questions as we progress through this semester aha. Next up! Circuits!
Tomorrow I am writing my first physics semester test in 2nd semester and I am feeling pretty confident since I have been making use of your assistance. Thank you so very much. We are all grateful.
i tried five different explanations before yours: my professor, my TA, chatgpt, my textbook and a book full of solved problems for reference. this is the only one that gave me steps to follow. thanks to you i can do a single question on my homework (after the last four hours alone of trying to apply the previous five explanations to it). you are saving my life.
Professor, many thanks for this excellent explanation. I totally didn't understand this from the textbook. I guess you already know this, but I will say it again, you can make anyone learn physics.
you are covering exactly what I am seeing in class tghank you sooo much! I shoud have found you before !!! my exam is tomorow and I hope for a miracle!!!!
Why am i like this?? Why do i even start reading the comments of a educational video like why in the hell my fking brain just orders me to read random comments when i'm sitting here tryna learn some electric field Is anyone else like me or i'm the only soul left to posses such immense power!!!
Regarding toward the end, when you discuss the situation where the test charge is NOT equidistant from the ends of the rod: Wouldn't the calculations get more complex, because the Y-components of the field vectors in THIS CASE do NOT cancel each other out. ie. the y-component in one direction will be more than the Y-component in the opposite direction??
At the very end of the video you discuss what would change if the electric field wasn't at the center of the rod. You said we would simply need to change the bounds. However I think there would be some Y components that would otherwise have canceled out. Since the angle from the top of the rod to the electric field would be a different angle from the bottom of the rod to the electric field. Therefore I think that the Y components would not cancel and it would be more complicated. Am I correct in this assumption? Please resond
+Ely Fialkoff Indeed, you would have a vertical and horizontal component. The technique would be the same with one integration for the vertical component and one for the horizontal component.
It is not so much a question of an equation, but of logic. In this case due to symmetry the component in the y direction caused by the upper half of the charged line, will be cancelled out by the component in the y-direction caused by the lower half of the charged line, which will be directed in the opposite direction.
Michel van Biezen, I noticed that you chose to integrate from 0 to L/2 and then multiplying the integral by 2 to get the total charge on the rod. I, however, chose to integrate from 0 to L and believing that I would get the same final answer as you, but I didn't. My answer was: E= [K*(lambda)*L] / [a*(L^2 + a^2)^(1/2)]. The only difference between our answers is that in your denominator, the L is divided by 2, while mine is not. Why am I wrong? Shouldn't we get the same answer? Thank you.
Muhammad, I find it more practical to look at the problem and decide what makes the most sense. Notice that the point of interest is on the perpendicular bisector, which means that the magnitude of the electric field caused by the upper portion is exactly the same as the electric field caused by the lower portion, due to its symmetry. Therefore is is easier to integrate from 0 to L/2 and then double the result. The way you did it places the point of interest at the bottom of the charged rod, which is not what the problem was asking for.
Hey professor, in the end of the video you said if we changed the place of the charge instead of being in the middle to be any where else we would change the integral boundaries and you didn't mention the electric field in the y-axis Is the y-axis field not affected really?
Only if you consider a point directly across the middle of the line charge distribution will the y-component of the electric field cancel from both halves and therefore will be zero. That is because of the symmetric distribution of the charge. At any other point you don't have that symmetry and there will then be a y-component of the electric field.
Why do we have to multiply the integral by a factor of 2? for a infinite line charge we directly integrate from negative infinity to positive infinity so wehy cant we integrate this integral directly from 0 to L?
Integrating from zero to L gives you a different geometric situation then placing the point of interest directly across from the middle of the line charge as shown in the video.
Hi Michel, if the origin and the centre of the rod do not coincide, how does the integral produced in the video account for the newly introduced net y component? You said dE_x = E_total, which makes sense at the centre of the rod due to symmetry arguments, however if it is assymetric, won't the problem of calculating net electric field be more complicated than simply changing the limits of integration?
After some further thought, won't it make sense to produce two integrals - one for x-components (as produced in your video), and one for y-components (similar to one produced in your video, except using sin(theta) instead of cos(theta)), then taking the vector sum both?
you said that when you move the point that your testing far from the middle of the rod you have to change only the limits of the integral but as i know the Ey component will taking into consideration since the R will not be equal (not symmetrical ) so can you clarify more
Yes, when the point of interest is taken across from the middle of the rod, then the y-components will cancel and you will only have an x-component of the electric field. If the point of interest is not in the middle, then the y-components will not cancel out. (We probably should do such an example)
Like it, but I have a doubt here: would anyone explain me why don't we double the final solution since we have two x axis components of E field? Thanks in advance.
The integral is from 0 to L/2, which gives you half the electric field (in the x-direction). So you need to double the answer to get the whole electric field. In the y-direction, the field cancels out.
When i take the limit L-->infinity then E = kλ/2a. If the rod was infinite then why the E depend on a? For a infinite rod the E shouldn't be constant? Like the infinite plane?
Unlike the infinite charged plane where the electric field is independent from the distance from the infinite plane, as the mathematics show, that is NOT the case with an infinite line. With an infinite line of charge the magntitude of the electric field does depend on the distance from the infinite line of charge.
What's the point of taking the small segment there in the beginning? I know it has something to do with Integral Calculus but I never got the concept of it.
If the force or the electric field (or the quantity you are interested in) changes with position (as is the case in this example), you find the electric field caused by a very small piece and then do it again and again with the adjacent pieces and sum them all up. That is the same as integrating them.
It is all about calculus. How can you sum infinitely small changes between discrete values? It is valid for any measurement: we use this time for small changes of electric field if we move on the wire, because every little piece of it creates a little electric field dE. When we do this, we can see we have the derivative of what we wanted. Yet, its integral requires tan. subs. i guess.
The vertical components would not cancel, and the limits on the vertical axis would change and would not be symmetrical. The process would be exactly the same.
Two field lines from the same source should not intersect. In this case the field lines come from two different sections (2 different sources) of the line of charge.
I have the same problem except the rod is horizontal and the point is on top of the rob. I looked up how to do is and it says that dEy=dE*cos(theta). Why would is not by dE*sin(theta) since in finding the Y component
Don't automatically associate x with the cos and y with the sin. It depends on the geometry of the problem. Note that if you were to rotate the drawing of this problem by 90 degrees you get exactly the problem you have and it would still be the cos(theta).
I have a doubt..... For every point upwards there is a symmetrical point downwards.. Till a limited length ( eg if upper part is 5cm and lower part is 3cm and we cut a dl at 3 cm in upper portion.... Then won't all the sin components btw -3 & 3 cancel?.... And we will only consider sin components from 3 to 5 cm in upper portion.?
Yes, if the point of consideration is not exactly at the middle (and there is no symmetry) the electric field in the positive direction will not cancel the electric field in the down direction.
A test charge can be both positive or negative. Typically we use a positive test charge such that it experiences a force in the same direction as the electric field.
Thank you for your great videos. Perhaps the question has been asked but I just can't spot it. The direction of the net electric field on the point should be in the + i hat direction? Perhaps your x hat notation meant this also but since you used x as the vertical direction I am a bit confused. The point itself also seems to be labeled x (as opposed to e.g. point P)
I am trying to find the force of a rod like this one on an identical rod standing a distance 'a' away. I know force is F=qE. Would I just do the process you did here and then integrate that over another length of L? I have found lots of resources explaining the force of a distribution on a point charge but not of a distribution on another distribution.
HD MM yea it can it’s the same as integrating from -L/2 to L/2. Your limits of integration should be where on the rod it is charged. Studied for physics midterm that’s today :D
Can any one give some comments according to the below question: Does the the periodic boundary condition interfere with the application of a finite electric field? Thanks in advance
I wouldn't know why a periodic boundary condition is applied to this problem as that is not where a periodic boundary condition technique would be applied. The application is more suited for atoms packed in a repeatable patteren, not the electric field caused by a charge distribution which can be easily calculated with traditional methods.
It would have been better if I had used the y variable here instead of the x variable to indicate the line charge. However that said, it works fine regardless of what variable we use.
The integral of our dEx function should be a^3, rather than the a^2 written on the board. I check with integral calculators and I'm not sure which one to use now. Can anyone verify?
Professor, I have a similar problem I'm trying to solve but the rod is positively charged 0 upwards and negatively charged 0 downwards on the y axis. I realize that in that case, the x components will cancel each other out rather than the y ones, and the total field will be in the y direction, downwards. But I keep getting the answer wrong and I'm not sure what I'm doing wrong. Could you shed some light please? Thank you so much!
That is an interesting problem. I will go ahead and make a video of it. Your assumptions appear correct. What is the answer you get and what should the answer be? That will help me understand where you may have gone wrong.
Thank you for your reply! I unfortunately don't actually know what the correct answer is, it's an online problem portal where we're supposed to type in our answer and submit. I got an "incorrect" but they don't send any feedback on what I might be doing wrong. I got a 55.24x10^4 N/C. The problem states: "A charge per unit length λ = +7.00 μC/m is uniformly distributed along the positive y-axis from y = 0 to y = +a = +0.500 m. A charge per unit length λ = -7.00 μC/m, is uniformly distributed along the negative y-axis from y = 0 to y = -a = -0.500 m. What is the magnitude of the electric field at a point on the x-axis a distance x = 0.311 m from the origin?" I really appreciate the attention!
Thank you, Professor! That seems to be the correct answer. I'm going to go back and review my work to see if I can catch any mistakes I've made. And I'll try to get it to the right answer. Thank you very much!
sir , electric field are parallel to each other because of repulsion , but here u draw electric field in the another direction , i m get u , help me to clear my doubt
The electric field of any particular piece of the total charge, emanates in all directions, but if we have an infinitely long line of charge, the components to the left and right cancel out and the electric field lines will be parallel as you indicate. But with a finite length line of charge that is not the case and you have to solve the problem as indicated.
Actually only the very middle point will have the electric field go straight out. For all other points on the rod, the electric field will NOT go out perpendicular to the rod.
okk , now i want to know one thing more is that the electric field emitting from point charges or line charges are always perpendicular to the surface at point of contact ryt?
It depends on the shape and the distribution of the charges. (For example, when the charges are on the surface of a spherical conductor, or along an infinitely long straight conductor) the field will emanate perpendicular to the surface everywhere on the surface. I know what the "text books" explain, but they sometimes explain the "simplified" version of things and don't always look at a more realistic scenario.
Is x =0 in the middle of the rod? So if I put the point charge at L/8, would it be correct to integrate from L/8 to L/2 and then add that with the integral of the electric field from L/8 to -L/2?
If you place a charge that is not directly across the middle of the rod, then the vertical component of the electric field will not be zero and you'll have to integrate as you indicated for both components of the electric field.
Wouldn't shifting the problem upwards make it so that the y components don't cancel anymore? I think you made a small mistake at the end. Good video otherwise!
+Michel van Biezen it was minor but at the end he mentions how if the point of consideration was not at the middle of the rod then all you had to do was change the limits of the integral. However, shifting the point anywhere other than the middle makes it so that not all the y components cancel.
Is it possible to find the electric field if I assume a cylinder of radius R with length of the wire and take a small area da on the cylinder and calculate using gauss law?
Only if the cylinder is long enough and the point is close enough so that we can ignore end effects. If the cylinder is short we cannot use Gauss's law.
That is actually a very common integral in E&M. The integral equals: (-1) / sqrt(x^2+y^2) assuming that x is the variable and y is a constant (or the other way around)
The field lines coming from a single point source (or a singe point on a line charge) do not cross one another. But field lines coming from multiple sources, of coming from different points on a line charge do indeed cross one another.
If the point of interest is picked at the half way point between the 2 ends of the charged rod, the electric field will not have a component in the y - direction.
Gauss Law can be used for an infinite line segment. For a finite line segment like this example, it would make it much more difficult to use the Gauss's law technique to find the answer. It is possible, just very difficult.
Apologies...When we first started our TH-cam educational channel, we had ABSOLUTELY NONE, ZERO experience in video making. We were math/physics not film-making majors. So we had absolutely zero clue about lighting and focus. Fortunately we do have some sort of learning curve. So our “less old” videos have better lighting and focus.
Sir my lecturer gave similar question but I was given relative permittivity and relative permittivity of free space pls which do I use and I was asked to calculate for the force acting on the electron which formula should I use pls i need urgent reply thank you ...i have quiz tomorrow
If the region around the line of charge is space (or air) you use the permitivity of free space. If it is another material, then you use the permitivity of that material.
sir in this video u r integrate with limits in length , if used to integrate with limits in angle , then the final electric field comes in terms of angle which is differ from ur calculation . is i m doing ryt?/
Too much advertise. I am trying to focus and then comes another ad. I understand it is to make some money, but lost concentration quite a few times. :(
I used your videos in A level, now that I am a first year engineer at Oxford I am still using your help!!!! thank you so much!
Glad to be of help. Keep working hard at your studies.
Update?
@@trexbattle Got a job
And in India This is just a part of the vast syllabus for Engineering entrance exams!😂
These things r in our fingertips😎🔥
@@philomath4747 JEE ne gand phaad di😢
Michel van Biezen or should I start calling you professor? I'd just like to thank you and your assistants for providing this educational content on youtube. You have multiple videos with in depth examples and it just about covers everything i'm learning in my first year engineering class at Mcmaster University. Sir, I just got my results back, and I scored an 87% on my midterm. Although I worked my ass off, you made it incredibly easy for me to learn what I needed to know for the test. So thank you once again. I'm going to be asking a lot more questions as we progress through this semester aha. Next up! Circuits!
Tomorrow I am writing my first physics semester test in 2nd semester and I am feeling pretty confident since I have been making use of your assistance. Thank you so very much. We are all grateful.
i tried five different explanations before yours: my professor, my TA, chatgpt, my textbook and a book full of solved problems for reference. this is the only one that gave me steps to follow. thanks to you i can do a single question on my homework (after the last four hours alone of trying to apply the previous five explanations to it). you are saving my life.
Professor, many thanks for this excellent explanation. I totally didn't understand this from the textbook. I guess you already know this, but I will say it again, you can make anyone learn physics.
Thank you for your kind words. 🙂
you are covering exactly what I am seeing in class tghank you sooo much! I shoud have found you before !!! my exam is tomorow and I hope for a miracle!!!!
I wish all profs were like this thanks sir ur helping a lot people out here
Mark Yu,
That makes no difference.
Just turn the vector around at the end.
(Or solve the problem with pointing all vectors in the other direction)
7:57 any further explanation on that 'common integral' ?
Just perfoem u-substutution
Why am i like this?? Why do i even start reading the comments of a educational video like why in the hell my fking brain just orders me to read random comments when i'm sitting here tryna learn some electric field
Is anyone else like me or i'm the only soul left to posses such immense power!!!
I think somebody once said bro was the professor Leonard of physics which is high praise I agree lol
Thank you.
Allah razı olsun hocam.
Thank you. God bless you as well.
Thank you so much for this video, it was very very helpful
You are welcome. Glad it helped.
You save my life! Thank you sincerely.
Happy to help!
Regarding toward the end, when you discuss the situation where the test charge is NOT equidistant from the ends of the rod: Wouldn't the calculations get more complex, because the Y-components of the field vectors in THIS CASE do NOT cancel each other out. ie. the y-component in one direction will be more than the Y-component in the opposite direction??
Yes, in that case you would also have to calculate the y-component of the electric field.
Thank you❤❤
Thank you you're the best teacher Ive never had I love you !
At the very end of the video you discuss what would change if the electric field wasn't at the center of the rod. You said we would simply need to change the bounds. However I think there would be some Y components that would otherwise have canceled out. Since the angle from the top of the rod to the electric field would be a different angle from the bottom of the rod to the electric field. Therefore I think that the Y components would not cancel and it would be more complicated. Am I correct in this assumption? Please resond
+Ely Fialkoff
Indeed, you would have a vertical and horizontal component. The technique would be the same with one integration for the vertical component and one for the horizontal component.
Ok thank you very much. Your videos are very helpful. Thank you.
Simply Awesome! Glad to see you
Isn't integral of 1/(x^2+a^2)^3/2 =
(1/(x^2+a^2)^0.5)*(arctan(x/a)/a).
What happened to arctan..is the angle small?
thank you so much can do if the charge is negative
If the charge is negative you can simply turn the final result around 180 degrees in direction and the magnitude would be exactly the same.
Can we integrate it from 0 to L....
I always thought thumbnails were supposed to be attractive but meh. I can definitely welcome myself to my reality. Thanks for the video
Thank you so much 😊
You're welcome 😊
professor, how do we prove that the dEy is zero? what would be the equation for that? thanks!
It is not so much a question of an equation, but of logic. In this case due to symmetry the component in the y direction caused by the upper half of the charged line, will be cancelled out by the component in the y-direction caused by the lower half of the charged line, which will be directed in the opposite direction.
Michel van Biezen,
I noticed that you chose to integrate from 0 to L/2 and then multiplying the integral by 2 to get the total charge on the rod. I, however, chose to integrate from 0 to L and believing that I would get the same final answer as you, but I didn't. My answer was: E= [K*(lambda)*L] / [a*(L^2 + a^2)^(1/2)]. The only difference between our answers is that in your denominator, the L is divided by 2, while mine is not. Why am I wrong? Shouldn't we get the same answer? Thank you.
Muhammad,
I find it more practical to look at the problem and decide what makes the most sense.
Notice that the point of interest is on the perpendicular bisector, which means that the magnitude of the electric field caused by the upper portion is exactly the same as the electric field caused by the lower portion, due to its symmetry. Therefore is is easier to integrate from 0 to L/2 and then double the result.
The way you did it places the point of interest at the bottom of the charged rod, which is not what the problem was asking for.
Hey professor, in the end of the video you said if we changed the place of the charge instead of being in the middle to be any where else we would change the integral boundaries and you didn't mention the electric field in the y-axis
Is the y-axis field not affected really?
Only if you consider a point directly across the middle of the line charge distribution will the y-component of the electric field cancel from both halves and therefore will be zero. That is because of the symmetric distribution of the charge. At any other point you don't have that symmetry and there will then be a y-component of the electric field.
Thank you so much...this was so helpful 🙏
Why do we have to multiply the integral by a factor of 2? for a infinite line charge we directly integrate from negative infinity to positive infinity so wehy cant we integrate this integral directly from 0 to L?
Integrating from zero to L gives you a different geometric situation then placing the point of interest directly across from the middle of the line charge as shown in the video.
We can also integrate from -L/2 to L/2 correct? And disregard multiplying by 2
+Jomel Sagsagat
Yes you can.
If the Y-components of the Electric Field were cancelling out, wouldn't the integral go from (-L/2) to (L/2) and not 0 to (L/2)?
Yes, but it is easier to have limits from 0 to L/2 and just double the result.
If and only if the professor taught like this
Hi Michel, if the origin and the centre of the rod do not coincide, how does the integral produced in the video account for the newly introduced net y component? You said dE_x = E_total, which makes sense at the centre of the rod due to symmetry arguments, however if it is assymetric, won't the problem of calculating net electric field be more complicated than simply changing the limits of integration?
After some further thought, won't it make sense to produce two integrals - one for x-components (as produced in your video), and one for y-components (similar to one produced in your video, except using sin(theta) instead of cos(theta)), then taking the vector sum both?
Where the origin is placed is completely arbitrary. It is better to place the origin at the center of the bar to make the math easier.
you said that when you move the point that your testing far from the middle of the rod you have to change only the limits of the integral but as i know the Ey component will taking into consideration since the R will not be equal (not symmetrical ) so can you clarify more
Yes, when the point of interest is taken across from the middle of the rod, then the y-components will cancel and you will only have an x-component of the electric field. If the point of interest is not in the middle, then the y-components will not cancel out. (We probably should do such an example)
Is such an example already done?
Like it, but I have a doubt here: would anyone explain me why don't we double the final solution since we have two x axis components of E field? Thanks in advance.
The integral is from 0 to L/2, which gives you half the electric field (in the x-direction). So you need to double the answer to get the whole electric field. In the y-direction, the field cancels out.
Simply amazing content
Glad you liked it!
I didn't understand where did we get 'a squared' in the dominator after integrating the equation ??
Actually it was a square root ((x^2 + a^2) ^ (1/2) (or are you referring to the a^2 (these are integration techniques)
If the point is not at the middle, are the y components still cancel each other?
No, if the point is not in the middle, you will also have a y-component for the electric field
When i take the limit L-->infinity then E = kλ/2a. If the rod was infinite then why the E depend on a? For a infinite rod the E shouldn't be constant? Like the infinite plane?
Unlike the infinite charged plane where the electric field is independent from the distance from the infinite plane, as the mathematics show, that is NOT the case with an infinite line. With an infinite line of charge the magntitude of the electric field does depend on the distance from the infinite line of charge.
@@MichelvanBiezen thank you so much for the immediate response!!
What's the point of taking the small segment there in the beginning? I know it has something to do with Integral Calculus but I never got the concept of it.
If the force or the electric field (or the quantity you are interested in) changes with position (as is the case in this example), you find the electric field caused by a very small piece and then do it again and again with the adjacent pieces and sum them all up. That is the same as integrating them.
It is all about calculus. How can you sum infinitely small changes between discrete values? It is valid for any measurement: we use this time for small changes of electric field if we move on the wire, because every little piece of it creates a little electric field dE. When we do this, we can see we have the derivative of what we wanted. Yet, its integral requires tan. subs. i guess.
What would be the general electric field solution for when then point is not half way between the line?
The vertical components would not cancel, and the limits on the vertical axis would change and would not be symmetrical. The process would be exactly the same.
Great video. Thanks for posting it.
Sir what about if we want to take the limits in angle form...
sir i think two field lines should never intersect but in your case your dE field lines is intersecting
Two field lines from the same source should not intersect. In this case the field lines come from two different sections (2 different sources) of the line of charge.
tnk sir
I have the same problem except the rod is horizontal and the point is on top of the rob. I looked up how to do is and it says that dEy=dE*cos(theta). Why would is not by dE*sin(theta) since in finding the Y component
Don't automatically associate x with the cos and y with the sin. It depends on the geometry of the problem. Note that if you were to rotate the drawing of this problem by 90 degrees you get exactly the problem you have and it would still be the cos(theta).
I think the infinite length line problems are easier than this one because we can use Gauss Law for infitinite length lines. Thank you anyway
They are indeed easier, but we also need to know how to solve this type of problem where we cannot use Gauss's law.
I have a doubt..... For every point upwards there is a symmetrical point downwards.. Till a limited length ( eg if upper part is 5cm and lower part is 3cm and we cut a dl at 3 cm in upper portion.... Then won't all the sin components btw -3 & 3 cancel?.... And we will only consider sin components from 3 to 5 cm in upper portion.?
Yes, if the point of consideration is not exactly at the middle (and there is no symmetry) the electric field in the positive direction will not cancel the electric field in the down direction.
@@MichelvanBiezen Then.. What will be the final result?
So is a "test charge" always positive unless otherwise specified?
A test charge can be both positive or negative. Typically we use a positive test charge such that it experiences a force in the same direction as the electric field.
Thank you for your great videos. Perhaps the question has been asked but I just can't spot it. The direction of the net electric field on the point should be in the + i hat direction? Perhaps your x hat notation meant this also but since you used x as the vertical direction I am a bit confused. The point itself also seems to be labeled x (as opposed to e.g. point P)
Yes, x-hat and i-hat can be used interchangeably.
Infinity divided by infinity does not equal 1!! That is a indeterminant form! How would you evaluate that integral??
Take the limit as you approach infinity and you'll see that it converges to 1
@@MichelvanBiezen Hey nice, thanks!
I am trying to find the force of a rod like this one on an identical rod standing a distance 'a' away. I know force is F=qE. Would I just do the process you did here and then integrate that over another length of L? I have found lots of resources explaining the force of a distribution on a point charge but not of a distribution on another distribution.
Is it possible to do this using gauss' law, via a vertical gaussian cylinder?
No, since it is of finite length.
@@MichelvanBiezen thank you! Are you saying it would work for the infinite line of charge?
Prof Mickel can you tell me how you solved this integration to get this result (i mean in details how will we solve it u substitution , by parts ??
This is done with trig substitution: let u = a tan (theta) where du = a sec^2(theta) d(theta).
@@MichelvanBiezen how Thank you so much! will you change the limit
What happens if the point was in the Y axis ? I mean the same axis as the charges
There is an example like that in the playlist.
Oh alright! Thank you so much professor!
Why do you integrate the (x^2+a^2)^3/2 in this example and not in the next example with the ring? Confused about this if anyone could clarify. Thanks
In a ring, the distance remains constant (when you are at the center of the ring). That is not the case here.
Ok thank you!
thanks a lot i understand all of that. KEEP GOING
can the limit of integration also be -L/2 to +L/2?
Yes it can.
Can the limit of integration be 0 to L? Without multiplying it for 2
HD MM yea it can it’s the same as integrating from -L/2 to L/2. Your limits of integration should be where on the rod it is charged. Studied for physics midterm that’s today :D
Can any one give some comments according to the below question: Does the the periodic boundary condition interfere with the application of a finite electric field?
Thanks in advance
I wouldn't know why a periodic boundary condition is applied to this problem as that is not where a periodic boundary condition technique would be applied. The application is more suited for atoms packed in a repeatable patteren, not the electric field caused by a charge distribution which can be easily calculated with traditional methods.
couldn't you have just integrated from -(L/2) to L/2 instead of multiplying it by 2 because of symmetry?
Yes, it is faster as shown in the video.
here it should be dy*lambda not dx because x is horizontal axe the charge is on y-axe or not ????
It would have been better if I had used the y variable here instead of the x variable to indicate the line charge. However that said, it works fine regardless of what variable we use.
So nice thanks sir
Most welcome
The integral of our dEx function should be a^3, rather than the a^2 written on the board. I check with integral calculators and I'm not sure which one to use now. Can anyone verify?
The equation is correct.
Professor, I have a similar problem I'm trying to solve but the rod is positively charged 0 upwards and negatively charged 0 downwards on the y axis. I realize that in that case, the x components will cancel each other out rather than the y ones, and the total field will be in the y direction, downwards. But I keep getting the answer wrong and I'm not sure what I'm doing wrong. Could you shed some light please? Thank you so much!
That is an interesting problem. I will go ahead and make a video of it. Your assumptions appear correct. What is the answer you get and what should the answer be? That will help me understand where you may have gone wrong.
Thank you for your reply! I unfortunately don't actually know what the correct answer is, it's an online problem portal where we're supposed to type in our answer and submit. I got an "incorrect" but they don't send any feedback on what I might be doing wrong. I got a 55.24x10^4 N/C. The problem states: "A charge per unit length λ = +7.00 μC/m is uniformly distributed along the positive y-axis from y = 0 to y = +a = +0.500 m. A charge per unit length λ = -7.00 μC/m, is uniformly distributed along the negative y-axis from y = 0 to y = -a = -0.500 m. What is the magnitude of the electric field at a point on the x-axis a distance x = 0.311 m from the origin?" I really appreciate the attention!
I'll take a look at this within the next few days.
I worked it out quickly and I got 190,906 N/C (1.909 x 10^5 N/C) or (19.09 x 10^4 N/C)
Thank you, Professor! That seems to be the correct answer. I'm going to go back and review my work to see if I can catch any mistakes I've made. And I'll try to get it to the right answer. Thank you very much!
sir , electric field are parallel to each other because of repulsion , but here u draw electric field in the another direction , i m get u , help me to clear my doubt
The electric field of any particular piece of the total charge, emanates in all directions, but if we have an infinitely long line of charge, the components to the left and right cancel out and the electric field lines will be parallel as you indicate. But with a finite length line of charge that is not the case and you have to solve the problem as indicated.
but sir only at the 'ends ' the electric field should spread , not in the middle of the rod ,
Actually only the very middle point will have the electric field go straight out. For all other points on the rod, the electric field will NOT go out perpendicular to the rod.
okk , now i want to know one thing more is that the electric field emitting from point charges or line charges are always perpendicular to the surface at point of contact ryt?
It depends on the shape and the distribution of the charges. (For example, when the charges are on the surface of a spherical conductor, or along an infinitely long straight conductor) the field will emanate perpendicular to the surface everywhere on the surface. I know what the "text books" explain, but they sometimes explain the "simplified" version of things and don't always look at a more realistic scenario.
Is x =0 in the middle of the rod? So if I put the point charge at L/8, would it be correct to integrate from L/8 to L/2 and then add that with the integral of the electric field from L/8 to -L/2?
If you place a charge that is not directly across the middle of the rod, then the vertical component of the electric field will not be zero and you'll have to integrate as you indicated for both components of the electric field.
Wouldn't shifting the problem upwards make it so that the y components don't cancel anymore? I think you made a small mistake at the end. Good video otherwise!
+Alf a If you place the point at a different location (upward or downward), the vertical component would not cancel out. What was the mistake?
+Michel van Biezen it was minor but at the end he mentions how if the point of consideration was not at the middle of the rod then all you had to do was change the limits of the integral. However, shifting the point anywhere other than the middle makes it so that not all the y components cancel.
Is it possible to find the electric field if I assume a cylinder of radius R with length of the wire and take a small area da on the cylinder and calculate using gauss law?
Only if the cylinder is long enough and the point is close enough so that we can ignore end effects. If the cylinder is short we cannot use Gauss's law.
@@MichelvanBiezen What are the effects of using gauss law on a short cylinder?
sir how we do it by using guass law
This cannot be done by Gauss's law. (only if the line of charge is infinitely long, or effectively so).
THANK YOU SO MUCH FOR YOUR VIDEOS!!!
Thank you sir
You're welcome!
hii I really need your help. what if the term u need to integrate is xdy/ (x^2 + y^2) ^3/2 from -a to +a?
That is actually a very common integral in E&M. The integral equals: (-1) / sqrt(x^2+y^2) assuming that x is the variable and y is a constant (or the other way around)
Plz answeer i am confused electric field lines never cross each other then why at the point concern they are crossing each other??
The field lines coming from a single point source (or a singe point on a line charge) do not cross one another. But field lines coming from multiple sources, of coming from different points on a line charge do indeed cross one another.
@@MichelvanBiezen thank you .you are a great help for me always 😊😊😊
will the formula boxed in the end be the same if the question is asking for the dy component.
If the point of interest is picked at the half way point between the 2 ends of the charged rod, the electric field will not have a component in the y - direction.
no I mean if the rod is horizontal on x axis and I am being asked to Calc the e field at y = 4
The approach would be exactly the same. Replace y with x and x with y.
What if the rod is negatively charged?
If the rod is negatively charged then the direction of the field changes, that is the X component of field will go in negative X- axis
Teachers can interpret the integral in more detail.
Please i donot understand y
Can explain the integration in detail
Sir
How if we do integral form 0 to L not from 0 to L/2 times 2 ?
The limits should then be: from - L/2 to + L/2
Only if my teacher taught like you
Would Gauss Law still work on this problem?
Gauss Law can be used for an infinite line segment.
For a finite line segment like this example, it would make it much more difficult to use the Gauss's law technique to find the answer.
It is possible, just very difficult.
Thank you a lot :)
You're welcome!
Camera focus problem in the video
Apologies...When we first started our TH-cam educational channel, we had ABSOLUTELY NONE, ZERO experience in video making. We were math/physics not film-making majors. So we had absolutely zero clue about lighting and focus. Fortunately we do have some sort of learning curve. So our “less old” videos have better lighting and focus.
Sir my lecturer gave similar question but I was given relative permittivity and relative permittivity of free space pls which do I use and I was asked to calculate for the force acting on the electron which formula should I use pls i need urgent reply thank you ...i have quiz tomorrow
If the region around the line of charge is space (or air) you use the permitivity of free space. If it is another material, then you use the permitivity of that material.
thanks a lot
sir in this video u r integrate with limits in length , if used to integrate with limits in angle , then the final electric field comes in terms of angle which is differ from ur calculation . is i m doing ryt?/
8:22 hmm how is that integration is equal to that 💔🙂
You can always check by differentiating the result and you will get the integrand again.
5:36 what was that?hey hey XD
can we use dL as the small segment of the electric charge other than dx?
sure, you can use any variable you like, but it has to match up with your sketch.
Thanks sir
thanks you sir!!!!!
Too much advertise. I am trying to focus and then comes another ad. I understand it is to make some money, but lost concentration quite a few times. :(
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omg thank you!!!!
👊👊
Thanks 🤗
thaaaaank yooooooouuuuu
lambda = q/l/2 ?
or lambda=q/l
Q/L
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