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:

Thank you. We appreciate the comment. 🙂

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

Great to know. We appreciate the feedback.

Glad I could help!

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.

Welcome to the channel!

Love frm🇧🇩🇧🇩🇧🇩🇧🇩🇧🇩

No, we don't have one like that.

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.

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.

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)

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......

Glad that the video helped in explaining Gauss's law.

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.

+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.

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?

+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!

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.

In order to answer you question, I'll need to know uniform what and conducting what?

Michel van Biezen spheres please

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.

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!

A is the area of the Gaussian surface (perpendicular to the direction of the electric field)

+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

Not necessary since the area of the end of the cylinder is proportional to the amount of charge encompassed by the Gaussian surface.

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.

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!

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).

If the field lines are uniform and parallel, then the intensity of the field remains constant

Thankyou soo much sir..very helpful for me

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.

Yes, there are a number of examples in the electric field playlist.

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.

You are welcome. 🙂

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?

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

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.

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 ?

+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.

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

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.

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

+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.

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.

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

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.

When you are close to a capacitor plate, it looks like an infinite sheet.

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!

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.

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 🙏

+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.

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

We are only looking at the electric field on one side of the plane.

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.

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)

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.

For simplicity we are assuming a single "sheet" of charge.

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?

The Gaussian surface has 2 ends, each of them with an area of pi * r^2

Then charge would reside only on the surface.

Michel van Biezen both sides of the surface of the conductor? (excessive charge)??

+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

If the sheet is small you cannot use Gauss's law, but you must calculate the electric field as shown in the other playlist.

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..

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 ?

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?

Thank you.

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.

Sigma = Q / A. If the sheet is infinitely thin, the electric field will emanate in both directions.

thanks for replying sir

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

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