I normally don’t leave comments just b/c I don’t want to, but your videos on Gauss’ Law are saving me! My teacher did a very poor job explaining it (went quickly through a confusing PowerPoint) and I wish I’d seen these worked through examples earlier. You explain it slowly and thoroughly (unlike my teacher) and I appreciate the time you put into these videos to help students like me! Thanks and keep up your enthusiasm for teaching (something my teacher lacks).
For real. Anytime someone is struggling in class/within my group of friends, first thing I tell them to do is to go to google and type in michel van biezen + youtube + subject
the cylindrical imaging g of photocopier is to have an electric field just outside its surface of 1.40 x 10 raised to 5 newton charge a if drum has a surface area of 0.06 10 meters square the area of a 8 and 1/2 x 11 in sheet of paper what total quantity of a charge must reside on a surface of the drum be it the surface area of the drum is increased to 0.122 m squared so that larger sheet of paper can be used what total quantity of charge is required to produce the same 1.40 x 10 raised to 5 newton charge electric field just above the surface?
I was trying to help a friend about this topic but I have no background or whatsoever. But watching your videos, I even had more knowledge and understanding about this topic than them. I'm so thankful for yoU!!
I must say, I truly enjoy your videos and every time I look for a tutorial and see that red bowtie in the video image, I immediately click on it because I know it will be helpful. Your videos have been such a help for my physics classes so far and I just wanted to say thank you. :)
2:46 Prof., why didn't you write 2A on the RHS if we considered 2 times the area for both the sides of the cylinder since the charge is enclosed on both the sides? A lot of people have already asked this question before, as I've seen the comments but I didn't get it if you could elaborate it a little that would be much help. A response will be appreciated
At 1:57 you say that the electric field will be emanating through the ends on both sides. Does that mean that positive charges lie on both sides? (Just thinking that net flux is the total coming into the shape, if the other side was negative, wouldn't only one side have an electric field emanating outward) Thank you in advance to anyone who answers!
+Meg Foster The electric field emanates from any charge in all directions. When you have a flat sheet of charge, the effect of that charge acts exactly the same on both sides. If the object is a conductor, then the excess charge will be on the surface (on both sides).
Sir I am from Bangladesh. I love how you describe and feel us everything. I have a question of Electric field due to uniformly charged infinite plane sheet of insulator. The charges are one side of the sheet or plate for insulator. I mean the charges are in one surface of the plate not in two surfaces. Then why we consider electric field in two side of the plate or sheet for insulator in the proof ? Can a electric field line pass through inside of the sheet of insulator?Would you please explain it to me.
Sir, why wouldnt the two Electric fields from each side cancel out each other as they seems to emerge out from each end of the same Gaussian cylinder surface ? Thanks in advance..
Assume you have a small point charge. Will that cause there to be an electric field around that charge? Wouldn't there be an electric field of equal magnitude and the opposite side for each electric field line around the charge? How would they cancel each other out? For field line to cancel out they must be at the same location, have the same magnitude and be opposite in direction.
Why do we get 2piR² for A? Isn't it supposed to be the area of the circle belonging to the plane and contained in the Gaussian cylinder surface? Wouldn't it just be piR² then?
It all depends on what you assume. If the electric field exists on both sides of the sheet, then you have to take into account both ends of the cylinder. If the electric field only exists on one side, then you only need to take into account one side of the cylinder.
@@MichelvanBiezen Why would you have to take the electric field coming from the other side of the surface into consideration if you want to know the electric field from a distance ''D'' away from one of the sides of the surface. Assuming of course that both of the sides of the surface emits an electric field, how does the electric field of one side affect the electric field of the other side?
Hi Sir, first of all, your tutorials are great and easy to understand. It's noticed in some other tutorials, 2 EA surfaces are counted for insulating sheet but 1 EA surface is counted for conducting sheet. The explanation is that charge usually exsits on one side of the surface, but you're saying charges are evenly distributed on 2 sides of a conductive surface. Do you have any thoughts? What's the difference?
Charges are free to move on a conductor and thus they will try to move as far from one another as possible. Charges cannot move on a non-conducting material.
To find field due to any charged body the gaussian surface should enclose whole the charge body then why cylinder gaussian surface is used even though it does not enclose whole charge sheet
Professor, do you have a video of a right triangle laying down on its side, lets call this base a, this triangle being a wedge with a 60 degree angle. The flux being on a horizontal plain, entering the short side, lets call that x, and exiting the longest side, lets call that the hypotenuse. Thank you very much for all your work and that of Mrs. Professor for her awesome art work.
Hello professor, how does this apply to the case of a disk? I tried using the info provided in your video but it didn t work. Thanks in advance!! "34) A thin, circular disk of radius 30.0 cm is oriented in the yz-plane with its center at the origin. The disk carries a total charge of +3.00 μC distributed uniformly over its surface. Calculate the magnitude of the electric field due to the disk at the point x = 15.0 cm along the x-axis. "
iyad kobeissi You cannot use Gauss's law for a finite disk. I made a video just like that in the playlist on electric fields PHYSICS 36 (video 9 of 13)
Sir, wouldn't the electric field emerging from the plane sheet be equal and opposite and cancel each other out?? In my text book it's given that E vector is independent of x axis but I don't understand....
The electric field always emanated away from a positive charge. Since it is directed in different directions on both sides o the sheet of charge, the electric field on one side cannot cancel the field on the other side.
The magnitude of the electric field must be the same all over the Gaussian surface and the direction of the electric field must always be perpendicular to the Gaussian surface. That would not be the case if you used a sphere here for the Gaussian surface.
excuse me sir, i dont understand why we didnt indicate that the charge enclosed would be sigma*2A which means sigma times two sides of the charged sheet? is it because the sheet is two thin that we can consider it as a one peace of one area?
There are different ways of expressing the charge. In this case it is an infinitely thin sheet. If you have a flat conductor with space between the 2 sides you can express the charge as residing on both sides.
you mean i can express the charge as residing on both sides for a case of a charged slab? but even though if this was not the case, gauss law considers all enclosed charges so why did we not consider both sides of the crossed sheet?
Hello Professor Biezen, I can't see how your approach would be different if the sheet were not infinite. What part of your math here incorporates that fact into finding the electric field because it seems to me that this would be equally valid for a very finite sheet of of charge. Thank you in advance if you get the chance to reply.
There are some examples in the playlist of how to calculate the electric field if it is not infinite. Then you can see how the answer will change to the infinite sheet answer when you plug in the correct parameters.
When we calculate Qin = segma*area, why we consider area is the area of the cylinder(gaussian surface) unlike the pervious videos and not calculate it as the area of the infinite plane ?
I had the same question but, i think, the only part of the surface, charge is passing through is the right face of cylinder, so we consider the area of that surface instead of area of the sheet
Professor, I understand why the sheet must be infinite to obtain this result using integration, but using Gauss’s law only, why is it important that the sheet is infinite? It didn’t seem that we used this “assumption” when applying Gauss’s law, the result naturally came, so where am I mistaken? Thanks in advance.
Note that the electric field would be the same regardless of how far you are from the sheet. It is a theoretical model for what happens in a capacitor. (there are no infinite sheets), but if you are close enough to a capacitor plate the plate acts like an infinite sheet and we can model the electric field using Gauss's law.
This is just a "fictituous" single sheet of charge emanating electric field on both sides. (In order to do so, the sheet would have to be infinitively thin). Normally for a conductor, half the charge would be on one side and the other half of the charge would have to be on the other side. The end result would be the same.
Why do we use area of a circle and not area of a square for this kind of problems? And why are we only multiplying by two only on one side of the equation and not on both because what I see is that the charge density is on both sides of the sheet?
Forgive me if it's simple but it is driving me nuts. If i need to find the Electric Field at a distance away from only one of the sheet's surfaces, will i apply Gauss' law on only one surface? So will E = σ/εo?
it depends on whether or not, the electric field emanates from one side of the sheet (like in a capacitor) or from both sides of the sheet. You need to be told which......
Why when estimating the charge inside you multiply charge density (sigma) times one A (I mean why not two times A if there is E both sides)? I assumed it is because the sheet is infinite but has no thickness...but I am not sure.
+Aníbal C. Ripoll R. It depends on how the problem is defined. In this case, there is a single sheet of charge (surface charge density). This is just a hypothetical problem to illustrate how Gauss' law works. It is a typical setup you will find in any physics text book. Look at the other videos with different charge distributions and you'll get a better understanding of Gauss' law.
That depends on several things, including how sigma is defined. But the electric field on one side is NOT affected by the charge on the other side of the sheet.
Hi Professor van Biezen, based on the solution you got for this, would the surface charge have to be half of what’s stated for each side? (Around 45 seconds you say that each side has that charge ) I ask that because I’m confused on why the charge enclosed inside isn’t 2* sigma instead of just sigma. (Times A) Hope to hear back from you and thank you for your videos!
It depends on how it is defined and what is "given" in the problem. In this example there is one sheet of charge and an electric field emanating from that charge on both sides. If the electric field was only emanating from one side only then the electric field would have twice the magnitude
Sir if the sheet is infinitely large the magnitude electric field is a constant term(σ/2ε). And the value is independent of the distance of the point considered. How can be the electric field be independent of distance from a practical point of view. Please explain
If the sheet is infinite, it doesn't matter how close or far away you are. In a practical sense, being close to a capacitor plate will make it appear that the plate is infinite in size.
Yes, it supports coulomb's law. You can take the contribution of every "point" on the infinite sheet and add them up. You'll get the same result. Take a look at the examples in the playlist.
sir I got a question, why do you multiply the area of the sheet by pi*r^2 instead of 2pi*r^2? I assume we do so because the area of 1 side of the sheet is calculated as whole. if so, is there a situation that we multiply by both sides of the sheet?
The Gaussian surface is used to find the electric field at a location (typically the surface of the Gaussian surface. When we calculate the electric field on one side, we only need to consider the area of the Gaussian surface for which the electric flux contributes to the electric field.
You could, but that would make it extremely hard to calculate. The shape of the Gaussian surface is chosen such that 1) the surface is perpendicular to the electric flux 2) The magnitude of the electric field is the same everywhere on the surface of the Gaussian surface
Are we assuming that the guassian cylinder encloses the entire sheet as opposed t just a segment of the sheet as you have drawn? Thanks so much for the videos too I find them very helpful as an aid for my textbook problems!
Michel van Biezen Sorry to bother as I'm sure you're very busy with a lot of engineering responsibilities but could you possibly make a couple videos or just one video on combining concepts. In other words a video that brings together multiple concepts with physics E and M. Thank you.
Evan Louder Evan, Not sure what you are asking for. There are over 500 videos on E&M concepts on this channel. Are there any specific topics that you are interested in that are not on the videos?
Yes, that is often a source of confusion. If all of the the electric field only points in one direction, then we use "sigma/e". But if the electric field emanates in both directions (left and right), then we sue "sigma / 2e" on each side.
@@MichelvanBiezen thankyou sir, I'm very grateful for such a quick response, but if you do not mind, i have another question..when exactly does the electric field emanates in both directions sir?
It is more of a "text book" situation. In real life, the electric field usually points in both directions. But in some cases (like on a capacitor plate) the electric field only exists on one side of the plate. The problem should make it clear what situation you are dealing with.
Gauss's law states that the electric flux emanating out of the Gaussian surface equals the charge inside divided by the permittivity of free space. Thus you must add the flux on both side of the surface.
Michel van Biezen But the flux is negative on other side, right? (Since cos180= -1) The normal to other surface is anti parallel to Electric field.So they should cancel each other out...
When it comes to flux, the sign is dictated by the determination if the flux goes INTO the Gaussian regions or COMES OUT of the Gaussian region. Since the flux leaves the Gaussian surface on both sides, the flux has the same sign. (good questions)
+jjjrrrrrrrrr The shape of the Gaussian surface is always chosen to make sure that the magnitude of the electric field at the surface is perpendicular to the surface. That is why cylinders and spheres are the usual shapes. In this case a cube would work as well.
Hello Michael going through the comments , I found one common query , while taking area of Gaussian surface we took twice the area , but while finding the charge enclosed in the surface we are multiplying sigma by single time A ( πr^2). Please elaborate.
It depends on the object containing the charge. If the charge is just a single "sheet" of charge, then you count the charges once. When the charge is situated on a metal sheet, it will reside on both sides of the sheet, so you have to count the charge double.
Excuse me , when calculating the charge enclosed , why didn't we multiply the area by 2 as you said that each side will have half of the charge ?[2Ao~]
+ahmed omran It depends on how the charge is quantified. In this example they give the charge in terms of a quantity of charge per square meter. There is nothing about that charge being on either side of the plane, simply the total amount of charge per square meter of plane.
i dont understan the idea of Charge, HOw can these positive and negative charge create electric field? when you talk about charge in your leasson which charge you are mention? postive charge or negative ? thank for your lesson
Andrew, It doesn't matter if the charge is positive or negative except for the direction of the electric field. With a positive charge the field is directed away from the charge. With a negative charge the electric field is directed towards the charge.
professor, as the direction of electric field due to a point charge is radially outwards in every direction.Similarly, what would be the direction of electric field due to a line charge
Think of it in terms of cylindrical coordinates. It would be radially outward in the rho direction away from the line charge. (No component in the z-direction).
Hello professor thx for the video~just one question. Why does the right side of the equation have the same area to the left side of the equation which could lead you to cancel the area on both side. I just wonder why can't we use the area of the sheet for the left side of the equation because that the place where the charges are enclosed in
honlolo That is a good observation. It all depends on how the charge density if defined. In this case the charge density is defined as the total charge (on both sides of the sheet) per unit area. (That is the traditional way of defining charge density on a sheet).
It looks to me that there will be electric field lines entering the cylinder through the sides. I guess because of symetry the normal component of the field entering the sides of the cylinder will cancel.Is this correct??
+Elang Zulfikar A is only the area of the portion of the infinite plane that is enclosed by the Gaussian surface, since you are only considering the charge within the Gaussian surface.
I had a question that came up in one of my previous exams similar to this principle. If there is a constant electric field given by E=k ĵ , what is the electric flux on a hemispherical guassian surface with its base on x-z plane? Now how do I solve a problem like that? Since At every point on the hemisphere the angle between the normal vector and electric feild changes.
By definition, the flux through the surface equals Q/Eo enclosed. Since you are only asked about the total flux, not the electric field on the surface, it is a fairly straightforward problem.
@@MichelvanBiezen Nevermind my question, i later realized that 2x10^-6 is the charge enclosed by the gaussian surface. Anyways, I respect that you took your time to reply my dumbass. Thank you.
The farther away you get from an infinite sheet, the more far away charges begin to have an effect on the electric field at that point as the sin(angle) increase with increasing angle.
@@MichelvanBiezen no no...i am saying if we take a veritical..which is divided by sheet into two parts..vertical in the sense means like a glass shape ..so that flux moves from lateral surface area
The electric field only emanates perpendicular to the plane. There are no electric field lines going through the sides of the Gaussian surface, only the ends
@@MichelvanBiezen sir 😅 why is it like that ... A charge emanates electric field in everywhere possible but when talking about sheet it just emanates perendicular to plane ?
If the field is not infinite, then the electric field will diminish with increasing distance. In the real world the plane will act as if it is infinite if you are close enough to the surface.
+Kabir Manrai Not a silly question at all. We pick the shape of the Gaussian surface depending on how the charge is distributed. The goal is to make sure that the electric field will emanate through the Gaussian surface perpendicular to the surface. (If not, it becomes very difficult to solve).
Mathematically speaking, Gauss’s theorem uses a dot product of electric field and surface normal vectors and since those are parallel for the curved surface cos(0°)=0 so there’s no flux.
In this video, the charges are given as a charge distribution per unit area. To find the total charges in a particular area, you must multiply the charge density with the area and you get charge. The units usually help in figuring that out. Note that the m^2 cancels out and you are left with Coulombs.
Savitha Mysore When you are close to the sheet of charge, the portions far away have very little contribution to the E field. When you are far away from the sheet of charge, many more portions of the sheet contribute to the E field. Notice that the contributing component to the electric field from any portion of the sheet depends on sine or cosine of the angle (depending on how you look at it)
I normally don’t leave comments just b/c I don’t want to, but your videos on Gauss’ Law are saving me! My teacher did a very poor job explaining it (went quickly through a confusing PowerPoint) and I wish I’d seen these worked through examples earlier. You explain it slowly and thoroughly (unlike my teacher) and I appreciate the time you put into these videos to help students like me! Thanks and keep up your enthusiasm for teaching (something my teacher lacks).
I know you are doing this for free but I feel kinda guilty for watching this without paying. You're saving the asses of so many students
We are glad to give something back to the world.
For real. Anytime someone is struggling in class/within my group of friends, first thing I tell them to do is to go to google and type in michel van biezen + youtube + subject
He getting money by views so issokay
@@Savvytoogood he never uses ads on his videos so he won't be getting much if anything.
At the end of the video an ad just poped up for me hmmmmmm V:
I know that this was posted 8 years ago, but just had to pop in and say thank you and that you are a very good teacher
Thank you. We appreciate the comment. 🙂
Anonymous poster
There is no electric flux through the sides of the Gaussian cylinder.
So we don't need to integrate across the sides.
That is correct.
the cylindrical imaging g of photocopier is to have an electric field just outside its surface of 1.40 x 10 raised to 5 newton charge a if drum has a surface area of 0.06 10 meters square the area of a 8 and 1/2 x 11 in sheet of paper what total quantity of a charge must reside on a surface of the drum be it the surface area of the drum is increased to 0.122 m squared so that larger sheet of paper can be used what total quantity of charge is required to produce the same 1.40 x 10 raised to 5 newton charge electric field just above the surface?
@@mylenececista4977 hey do it on your own! Or ask any specific conceptual doubt on stackexchange
You are an absolute legend.I've already lost count of the days you saved.
you explain things so well, a true saint
I was trying to help a friend about this topic but I have no background or whatsoever. But watching your videos, I even had more knowledge and understanding about this topic than them. I'm so thankful for yoU!!
Glad I could help!
I must say, I truly enjoy your videos and every time I look for a tutorial and see that red bowtie in the video image, I immediately click on it because I know it will be helpful. Your videos have been such a help for my physics classes so far and I just wanted to say thank you. :)
Great to know. We appreciate the feedback.
2:46 Prof., why didn't you write 2A on the RHS if we considered 2 times the area for both the sides of the cylinder since the charge is enclosed on both the sides? A lot of people have already asked this question before, as I've seen the comments but I didn't get it if you could elaborate it a little that would be much help.
A response will be appreciated
At 1:57 you say that the electric field will be emanating through the ends on both sides. Does that mean that positive charges lie on both sides? (Just thinking that net flux is the total coming into the shape, if the other side was negative, wouldn't only one side have an electric field emanating outward) Thank you in advance to anyone who answers!
+Meg Foster
The electric field emanates from any charge in all directions. When you have a flat sheet of charge, the effect of that charge acts exactly the same on both sides. If the object is a conductor, then the excess charge will be on the surface (on both sides).
+Michel van Biezen Thanks for the help, I appreciate it!
Sir I am from Bangladesh. I love how you describe and feel us everything. I have a question of Electric field due to uniformly charged infinite plane sheet of insulator. The charges are one side of the sheet or plate for insulator. I mean the charges are in one surface of the plate not in two surfaces. Then why we consider electric field in two side of the plate or sheet for insulator in the proof ? Can a electric field line pass through inside of the sheet of insulator?Would you please explain it to me.
Sir, why wouldnt the two Electric fields from each side cancel out each other as they seems to emerge out from each end of the same Gaussian cylinder surface ? Thanks in advance..
Assume you have a small point charge. Will that cause there to be an electric field around that charge? Wouldn't there be an electric field of equal magnitude and the opposite side for each electric field line around the charge? How would they cancel each other out? For field line to cancel out they must be at the same location, have the same magnitude and be opposite in direction.
Dear Professor,
I'm from Bangladesh. It was really awesome!!!... Thank you.
Welcome to the channel!
Love frm🇧🇩🇧🇩🇧🇩🇧🇩🇧🇩
Why do we get 2piR² for A? Isn't it supposed to be the area of the circle belonging to the plane and contained in the Gaussian cylinder surface? Wouldn't it just be piR² then?
It all depends on what you assume. If the electric field exists on both sides of the sheet, then you have to take into account both ends of the cylinder. If the electric field only exists on one side, then you only need to take into account one side of the cylinder.
@@MichelvanBiezen Why would you have to take the electric field coming from the other side of the surface into consideration if you want to know the electric field from a distance ''D'' away from one of the sides of the surface. Assuming of course that both of the sides of the surface emits an electric field, how does the electric field of one side affect the electric field of the other side?
what is the difference between this and previous example?
Hi Sir, first of all, your tutorials are great and easy to understand. It's noticed in some other tutorials, 2 EA surfaces are counted for insulating sheet but 1 EA surface is counted for conducting sheet. The explanation is that charge usually exsits on one side of the surface, but you're saying charges are evenly distributed on 2 sides of a conductive surface. Do you have any thoughts? What's the difference?
Charges are free to move on a conductor and thus they will try to move as far from one another as possible. Charges cannot move on a non-conducting material.
Awesome you ,,, thanks a lot ,,, you’re so organized and give the info striate way to us
At 3:13, why is A= Pi * R^2 not 2Pi *R^2 ??
We are only looking at the electric field on one side of the plane.
To find field due to any charged body the gaussian surface should enclose whole the charge body then why cylinder gaussian surface is used even though it does not enclose whole charge sheet
Gauss's law does not require ALL of the charge to be included. The electric flux will be proportional to the charge ENCLOSED.
@@MichelvanBiezen thanks
Professor, do you have a video of a right triangle laying down on its side, lets call this base a, this triangle being a wedge with a 60 degree angle. The flux being on a horizontal plain, entering the short side, lets call that x, and exiting the longest side, lets call that the hypotenuse. Thank you very much for all your work and that of Mrs. Professor for her awesome art work.
No, we don't have one like that.
Hello professor, how does this apply to the case of a disk?
I tried using the info provided in your video but it didn t work.
Thanks in advance!!
"34) A thin, circular disk of radius 30.0 cm is oriented in the yz-plane with its center at the origin. The disk carries a total charge of +3.00 μC distributed uniformly over its surface. Calculate the magnitude of the electric field due to the disk at the point x = 15.0 cm along the x-axis. "
iyad kobeissi
You cannot use Gauss's law for a finite disk.
I made a video just like that in the playlist on electric fields
PHYSICS 36 (video 9 of 13)
You are saviour thank you so much.....You explain each and every thing in simple and understandable manner.....Thank-you......
Sir, wouldn't the electric field emerging from the plane sheet be equal and opposite and cancel each other out?? In my text book it's given that E vector is independent of x axis but I don't understand....
The electric field always emanated away from a positive charge. Since it is directed in different directions on both sides o the sheet of charge, the electric field on one side cannot cancel the field on the other side.
Why can't we use a sphere to find electric field where answer comes to around Sigma /4*epsilon
The magnitude of the electric field must be the same all over the Gaussian surface and the direction of the electric field must always be perpendicular to the Gaussian surface. That would not be the case if you used a sphere here for the Gaussian surface.
Yes therefore curved surface in the cylinder cancels to zero... btw thanks 🙏
excuse me sir, i dont understand why we didnt indicate that the charge enclosed would be sigma*2A which means sigma times two sides of the charged sheet? is it because the sheet is two thin that we can consider it as a one peace of one area?
There are different ways of expressing the charge. In this case it is an infinitely thin sheet. If you have a flat conductor with space between the 2 sides you can express the charge as residing on both sides.
you mean i can express the charge as residing on both sides for a case of a charged slab? but even though if this was not the case, gauss law considers all enclosed charges so why did we not consider both sides of the crossed sheet?
There is only one layer of charge in this example as defined in the problem.
Hello Professor Biezen, I can't see how your approach would be different if the sheet were not infinite. What part of your math here incorporates that fact into finding the electric field because it seems to me that this would be equally valid for a very finite sheet of of charge. Thank you in advance if you get the chance to reply.
There are some examples in the playlist of how to calculate the electric field if it is not infinite. Then you can see how the answer will change to the infinite sheet answer when you plug in the correct parameters.
When we calculate Qin = segma*area, why we consider area is the area of the cylinder(gaussian surface) unlike the pervious videos and not calculate it as the area of the infinite plane ?
I had the same question but, i think, the only part of the surface, charge is passing through is the right face of cylinder, so we consider the area of that surface instead of area of the sheet
Sondos Hassan you can take the area of a rectangular plane and it won't matter. The area cancels anyway.
it's not the area of the cylinder, it's the area enclosed by the cylinder.
Sir I have another question..does the intensity of the field remains constant in case the field lines are uniform?
If the field lines are uniform and parallel, then the intensity of the field remains constant
Thankyou soo much sir..very helpful for me
Thank you so much for these videos! They make everything a lot clearer! :)
Professor, I understand why the sheet must be infinite to obtain this result using integration, but using Gauss’s law only, why is it important that the sheet is infinite? It didn’t seem that we used this “assumption” when applying Gauss’s law, the result naturally came, so where am I mistaken? Thanks in advance.
Note that the electric field would be the same regardless of how far you are from the sheet. It is a theoretical model for what happens in a capacitor. (there are no infinite sheets), but if you are close enough to a capacitor plate the plate acts like an infinite sheet and we can model the electric field using Gauss's law.
Is this sheet a conductor or an insulator?
This is just a "fictituous" single sheet of charge emanating electric field on both sides. (In order to do so, the sheet would have to be infinitively thin). Normally for a conductor, half the charge would be on one side and the other half of the charge would have to be on the other side. The end result would be the same.
Why do we use area of a circle and not area of a square for this kind of problems? And why are we only multiplying by two only on one side of the equation and not on both because what I see is that the charge density is on both sides of the sheet?
Forgive me if it's simple but it is driving me nuts. If i need to find the Electric Field at a distance away from only one of the sheet's surfaces, will i apply Gauss' law on only one surface? So will E = σ/εo?
it depends on whether or not, the electric field emanates from one side of the sheet (like in a capacitor) or from both sides of the sheet. You need to be told which......
why is the cylinder symmetrical about the sheet? thanks in advance
Why when estimating the charge inside you multiply charge density (sigma) times one A (I mean why not two times A if there is E both sides)? I assumed it is because the sheet is infinite but has no thickness...but I am not sure.
+Aníbal C. Ripoll R.
It depends on how the problem is defined. In this case, there is a single sheet of charge (surface charge density). This is just a hypothetical problem to illustrate how Gauss' law works. It is a typical setup you will find in any physics text book. Look at the other videos with different charge distributions and you'll get a better understanding of Gauss' law.
u said the sheet is conduction therefore charge will be residing on both side of it so wouldn't we multiply sigma by 2 ??
That depends on several things, including how sigma is defined. But the electric field on one side is NOT affected by the charge on the other side of the sheet.
Hi Professor van Biezen, based on the solution you got for this, would the surface charge have to be half of what’s stated for each side? (Around 45 seconds you say that each side has that charge ) I ask that because I’m confused on why the charge enclosed inside isn’t 2* sigma instead of just sigma. (Times A) Hope to hear back from you and thank you for your videos!
It depends on how it is defined and what is "given" in the problem. In this example there is one sheet of charge and an electric field emanating from that charge on both sides. If the electric field was only emanating from one side only then the electric field would have twice the magnitude
Michel van Biezen really appreciate it!
Sir if the sheet is infinitely large the magnitude electric field is a constant term(σ/2ε). And the value is independent of the distance of the point considered. How can be the electric field be independent of distance from a practical point of view. Please explain
If the sheet is infinite, it doesn't matter how close or far away you are. In a practical sense, being close to a capacitor plate will make it appear that the plate is infinite in size.
Sir how does it support the coulombs law? (as it says that for point charge electric field is inversely proportional to square of distance)
Yes, it supports coulomb's law. You can take the contribution of every "point" on the infinite sheet and add them up. You'll get the same result. Take a look at the examples in the playlist.
sir I got a question, why do you multiply the area of the sheet by pi*r^2 instead of 2pi*r^2? I assume we do so because the area of 1 side of the sheet is calculated as whole. if so, is there a situation that we multiply by both sides of the sheet?
The Gaussian surface is used to find the electric field at a location (typically the surface of the Gaussian surface. When we calculate the electric field on one side, we only need to consider the area of the Gaussian surface for which the electric flux contributes to the electric field.
Michel van Biezen thank you sir for your speedy replies.
Is A of the Gaussian surface equal to the A of the charged object because the area of the infinite sheet enclosed is also pi*R^2?
A is the area of the Gaussian surface (perpendicular to the direction of the electric field)
Can't we consider a vertical gaussian surface so that the flux is perpendicular to the curved area of the cylinder?
You could, but that would make it extremely hard to calculate. The shape of the Gaussian surface is chosen such that 1) the surface is perpendicular to the electric flux 2) The magnitude of the electric field is the same everywhere on the surface of the Gaussian surface
@@MichelvanBiezen Thank you very much
If it is a thick sheet then surface charge density would be 2@(sigma),if not,then why?
Do you have a video, that talks about the difference of uniform, non uniform, conducting and non conducting?
In order to answer you question, I'll need to know uniform what and conducting what?
Michel van Biezen spheres please
Are we assuming that the guassian cylinder encloses the entire sheet as opposed t just a segment of the sheet as you have drawn? Thanks so much for the videos too I find them very helpful as an aid for my textbook problems!
Evan Louder
The Gaussian surface drawn in the video assumes there is an electric field on both sides of the sheet,
Michel van Biezen Sorry to bother as I'm sure you're very busy with a lot of engineering responsibilities but could you possibly make a couple videos or just one video on combining concepts. In other words a video that brings together multiple concepts with physics E and M. Thank you.
Evan Louder
Evan,
Not sure what you are asking for. There are over 500 videos on E&M concepts on this channel. Are there any specific topics that you are interested in that are not on the videos?
when is the electric field sigma/e?
i get confused cuz some cases its sigma/e and in some its sigma/2e
Yes, that is often a source of confusion. If all of the the electric field only points in one direction, then we use "sigma/e". But if the electric field emanates in both directions (left and right), then we sue "sigma / 2e" on each side.
@@MichelvanBiezen thankyou sir, I'm very grateful for such a quick response, but if you do not mind, i have another question..when exactly does the electric field emanates in both directions sir?
It is more of a "text book" situation. In real life, the electric field usually points in both directions. But in some cases (like on a capacitor plate) the electric field only exists on one side of the plate. The problem should make it clear what situation you are dealing with.
ohh..thankyou sir! thankyou very much, it means a lot to learn from you
Why does the direction electric field -x(other side) and +x add up? (-x)E+(x)E = 0 right.. why 2E
Gauss's law states that the electric flux emanating out of the Gaussian surface equals the charge inside divided by the permittivity of free space. Thus you must add the flux on both side of the surface.
Michel van Biezen But the flux is negative on other side, right? (Since cos180= -1) The normal to other surface is anti parallel to Electric field.So they should cancel each other out...
When it comes to flux, the sign is dictated by the determination if the flux goes INTO the Gaussian regions or COMES OUT of the Gaussian region. Since the flux leaves the Gaussian surface on both sides, the flux has the same sign. (good questions)
Michel van Biezen Thanks, You Are The Best!
Why is it that in the cases of infinite plane sheets like this one,
professors always like to use cylinders the way you did? Why not a cube?
+jjjrrrrrrrrr
The shape of the Gaussian surface is always chosen to make sure that the magnitude of the electric field at the surface is perpendicular to the surface. That is why cylinders and spheres are the usual shapes. In this case a cube would work as well.
Hello Michael going through the comments , I found one common query , while taking area of Gaussian surface we took twice the area , but while finding the charge enclosed in the surface we are multiplying sigma by single time A ( πr^2). Please elaborate.
It depends on the object containing the charge. If the charge is just a single "sheet" of charge, then you count the charges once. When the charge is situated on a metal sheet, it will reside on both sides of the sheet, so you have to count the charge double.
@@MichelvanBiezen kindly specify in this case for infinitely long sheet is charge only on the one side ?
Excuse me ,
when calculating the charge enclosed , why didn't we multiply the area by 2 as you said that each side will have half of the charge ?[2Ao~]
+ahmed omran It depends on how the charge is quantified. In this example they give the charge in terms of a quantity of charge per square meter. There is nothing about that charge being on either side of the plane, simply the total amount of charge per square meter of plane.
Area of the charge enclosed in the guassian surface is only one surface while the areas where the electric field emenates are 2 surfaces I think.
You make everything so easy to understand 😘
Beautiful video explanation!
i dont understan the idea of Charge, HOw can these positive and negative charge create electric field? when you talk about charge in your leasson which charge you are mention? postive charge or negative ? thank for your lesson
Andrew,
It doesn't matter if the charge is positive or negative except for the direction of the electric field. With a positive charge the field is directed away from the charge. With a negative charge the electric field is directed towards the charge.
Why didnt we multiply area by 2 when we found Q inside?
professor, as the direction of electric field due to a point charge is radially outwards in every direction.Similarly, what would be the direction of electric field due to a line charge
Think of it in terms of cylindrical coordinates. It would be radially outward in the rho direction away from the line charge. (No component in the z-direction).
what if it wasn't an infinite charge what we will do?
will we have to do some integration?
Yes, there are a number of examples in the electric field playlist.
Is it correct to assume that the cylinder will be encompassing the whole sheet? since the E will be result of the whole charge on the sheet.
Not necessary since the area of the end of the cylinder is proportional to the amount of charge encompassed by the Gaussian surface.
Thank you very much! I found this very helpful!
Wow, that was very well explained
Hello professor thx for the video~just one question. Why does the right side of the equation have the same area to the left side of the equation which could lead you to cancel the area on both side. I just wonder why can't we use the area of the sheet for the left side of the equation because that the place where the charges are enclosed in
honlolo
That is a good observation. It all depends on how the charge density if defined.
In this case the charge density is defined as the total charge (on both sides of the sheet) per unit area. (That is the traditional way of defining charge density on a sheet).
Michel van Biezen Oh thank you once again and god bless you ..THANKs
It looks to me that there will be electric field lines entering the cylinder through the sides. I guess because of symetry the normal component of the field entering the sides of the
cylinder will cancel.Is this correct??
+Benjulie Benjulie i see i'm wrong. The electric field at any point in space above or below the surface is pointing perpendicular to the surface
professor, why isnt the integral of dA is the area of the infinite plank?
+Elang Zulfikar A is only the area of the portion of the infinite plane that is enclosed by the Gaussian surface, since you are only considering the charge within the Gaussian surface.
I had a question that came up in one of my previous exams similar to this principle. If there is a constant electric field given by E=k ĵ , what is the electric flux on a hemispherical guassian surface with its base on x-z plane?
Now how do I solve a problem like that? Since At every point on the hemisphere the angle between the normal vector and electric feild changes.
By definition, the flux through the surface equals Q/Eo enclosed. Since you are only asked about the total flux, not the electric field on the surface, it is a fairly straightforward problem.
Michel van Biezen Oh I get it now. Thanks!
May I have a practical example of such a sheet?
When you are close to a capacitor plate, it looks like an infinite sheet.
Is this the total electric field?
Not sure what you mean by the "total" electric field.
Michel van Biezen the electric field of both sides, not the electric field at a point at one side of the sheet
Michel van Biezen or I should say, where does the E-field becomes (sigma) / (absalon not) and what does it mean?
Is 2x10^-6 Coloumb the magnitude of a single charge?
The magnitude of a single charge = 1.602 x 10^ -19 C
@@MichelvanBiezen Nevermind my question, i later realized that 2x10^-6 is the charge enclosed by the gaussian surface. Anyways, I respect that you took your time to reply my dumbass. Thank you.
thanks for the video i just can't get why there is no dependency on the distance?
The farther away you get from an infinite sheet, the more far away charges begin to have an effect on the electric field at that point as the sin(angle) increase with increasing angle.
Sir plz answer why dont we take a vertical cylinder..and how to solve using verticsl cylinder so that flux moves out of lateral surface area
You want to draw your Gaussian surface such that the the flux of the electric field passes through the surface, perpendicular to the surface.
@@MichelvanBiezen no no...i am saying if we take a veritical..which is divided by sheet into two parts..vertical in the sense means like a glass shape ..so that flux moves from lateral surface area
What would happen if the sheet wasn't infinite?
If the sheet is small you cannot use Gauss's law, but you must calculate the electric field as shown in the other playlist.
Why are we taking it perpendicular to infinte sheet why not in its plane ??
The electric field only emanates perpendicular to the plane. There are no electric field lines going through the sides of the Gaussian surface, only the ends
@@MichelvanBiezen ohk thanks sir
@@MichelvanBiezen sir 😅 why is it like that ... A charge emanates electric field in everywhere possible but when talking about sheet it just emanates perendicular to plane ?
thank you sm life savior
Glad that the video helped in explaining Gauss's law.
Thank you, it is very helpful
Why the plane must be infinite?
If the field is not infinite, then the electric field will diminish with increasing distance. In the real world the plane will act as if it is infinite if you are close enough to the surface.
why isnt the area on the right side of the equation 2pir^2
The area on the right side of the equation represents the area of the sheet of charge enclosed by the Gaussian surface.
@@MichelvanBiezen But there are two sides of the sheet front and back so charge enclosed should be counted on both the sides. Isn't it?
sir what if the sheet is conducting
Then charge would reside only on the surface.
Michel van Biezen both sides of the surface of the conductor? (excessive charge)??
Sir..wouldn't the electric field depend on distance?
Not in the case of an infinite sheet (or a physical situation, like a capacitor plate, where the sheet of charge acts like an infinite sheet.
Thankyou sir..
why we took cylinder as the Gaussian surface, I know it's a silly question
+Kabir Manrai
Not a silly question at all. We pick the shape of the Gaussian surface depending on how the charge is distributed. The goal is to make sure that the electric field will emanate through the Gaussian surface perpendicular to the surface. (If not, it becomes very difficult to solve).
Michel van Biezen thank you sir, really helped me
very well explained
Say the total charge on the sheet in Q and the Area is A. Then, will σ be Q/A or Q/2A?
Sigma = Q / A. If the sheet is infinitely thin, the electric field will emanate in both directions.
thanks for replying sir
why there isn't flux through the sides of the cylinder? thanks for the great video though
Because the electric field emanates away from the surface in a direction perpendicular to the surface.
Mathematically speaking, Gauss’s theorem uses a dot product of electric field and surface normal vectors and since those are parallel for the curved surface cos(0°)=0 so there’s no flux.
sir ..plz tel me how to take charges whn we r finding out E due to a thick infinite plane sheet....
In this video, the charges are given as a charge distribution per unit area. To find the total charges in a particular area, you must multiply the charge density with the area and you get charge. The units usually help in figuring that out. Note that the m^2 cancels out and you are left with Coulombs.
can u please explain once again , how E is independent of distance?
Savitha Mysore When you are close to the sheet of charge, the portions far away have very little contribution to the E field. When you are far away from the sheet of charge, many more portions of the sheet contribute to the E field. Notice that the contributing component to the electric field from any portion of the sheet depends on sine or cosine of the angle (depending on how you look at it)
is the sheet non conductor?
For simplicity we are assuming a single "sheet" of charge.
I like you teaching
why is the Area on the left side just pir^2 instead of 2pir^2 like the left side?
The Gaussian surface has 2 ends, each of them with an area of pi * r^2
Hello ; please why haven't we taken a CUBE to be our gaussian surface instead of a cylinder
It makes no difference, you can do the same problem with a cube and you will get the exact same answer
@@MichelvanBiezen Oh the two A will cancel each other
Thank you sir SO MUCH god bless you
nice bow tie, bow ties are cool
Thank you.
Thank you
You are welcome. 🙂
I am iit jee inspiration
thank u it is very helpful
thank u sir
شكرا
Amazing
thanks a lot
Great
omg I might actually pass my physics 2 class
I like your bowtie 👍 that was also super helpful.
Sir draw graph
👍
Be hindi mein bolna..
its funny how a lot of indians are here......P.S I am one