# 8.02x – Lect 3 – Electric Flux, Gauss’ Law, Examples

Today we’re going to work on a whole new concept
and that is the concept of electric flux. We’ve come a long way. We started out with Coulomb’s law. We got electric field lines. And now we have electric flux. Suppose I have an electric field
which is like so– and I bring in that electric field a surface,
an open surface like a handkerchief
or a piece of paper. And so here it is. Something like that. And I carve this surface up in very small
surface elements, each with size dA, that’s the area, teeny weeny little area
and let this be the normal, N roof, the normal on that surface. So now the local electric field say at that
location would be for instance this. It’s a vector. The electric flux d phi that goes through
this little surface now– is defined as the dot product of E and the vector perpendicular to this element which has this, as a magnitude, dA. Now our book will always write for N dA simply
dA. So I will do that also although I don’t like
it but I will follow the notation of the book. So this vector dA is always perpendicular
to that little element dA and it has the magnitude dA. And so this since it is a dot product is the
magnitude of E times the area dA times the cosine of the angle between these two vectors, theta. It is a scalar. The number can be larger than zero, smaller
than zero, and it can be zero. And I can calculate the flux through the entire
surface by doing an integral over that whole surface. The unit of flux follows immediately from
the definition. That is newtons per coulombs, for the units
of this flux, is newtons per coulombs times square meters. But no one ever thinks of it that way. Just SI units. I can give you a, some intuition for this flux
by comparing it first with an airflow. These red arrows that you see there represent
the velocity of air and you see there a black rectangle three times. In the first case notice that the normal to
the surface of that area is parallel to the velocity vector of the air and so if you want to know now
what the amount of air is in terms of cubic meters per second
going through this rectangle it would be V times A, it’s very simple. However, if you rotate this rectangle ninety
degrees so that the normal to that rectangle is perpendicular to the velocity vector, nothing
goes through that rectangle and so it’s zero. So now the flux– the air flux is zero,
and if the angle is sixty degrees then it is of course V times A times the cosine of
sixty degrees. Now think of these red vectors as electric
fields. So now the electric flux going in the first case
through that surface is now simply E times A. In the second case it’s zero. And in the last case it is EA times the cosine
of sixty degrees. So you can sometimes think of this as airflows. We also saw that when we dealt with field
lines that can come in sometimes very handy. I now take a surface which is not open as
this one is, this is an open surface. Can come in from both sides. But now I choose one that is completely closed. Like a potato bag or a balloon. I’ll draw, put this line in here to give you
a feeling there’s a completely closed surface. So you can only get inside if you penetrate
that surface from the outside. And so now I can put up here and here these
normals, dA and there’s another normal here,
maybe in this direction, dA. In this case by convention the normal
to the surface, locally to the surface, is always from the inside of
the surface to the outside world. It’s uniquely determined because it’s a closed
surface. Here it was not uniquely determined. I arbitrarily chose this one but I could have
flipped it over a hundred eighty degrees since it’s an open surface, it’s ill-defined. Here it’s never ill-defined. So the normal is always chosen to go from
the inside to the outside. And now I can calculate the total flux going
through this closed surface. Locally multiplying E with dA,
dot product over the whole surface, out comes a certain number,
and that is now therefore the integral of E dot dA integrated over that closed surface and since it is a closed surface we put a
circle here, to remind us that it is a closed integral and here in this case it is
a closed surface. And this now is the total flux
through that surface. It could be larger than zero. It could be smaller than zero. It’s a scalar, it’s not a vector. It could be equal to zero. If it’s equal to zero then you can think of
it whatever flows in, if you think of it as air, also flows out. If more flows out than flows in
then it is positive. If more flows in than flows out
it is negative. So let’s now calculate the flux for a very
simple case where I have a point charge. So here I have a point charge and I’m going
to put a bag around this point charge and the bag is a sphere. It is a sphere and the sphere has radius capital
R. And let this charge be plus Q. Just for simplicity. Well, I pick a small element dA here. And at element dA is radially outward… dA… This is the normal to that surface so that
this radial… The electric field at that point is also radial. We have dealt with that before. So dA and E not only here but anywhere on
the surface of this sphere are parallel. For the cosine of the angle equals one. I can also introduce here the unit vector
R roof which is the unit vector going from capital Q to that element where I evaluate
the teeny weeny little amount of flux. So if now I want to know what the total flux
is through this sphere that’s very easy because since this is a sphere the E vector
in magnitude is everywhere the same because the radius is the same,
the same distance to this charge, and dA and E are parallel. So it’s simply the surface four pi R squared
of that sphere, times E. And so now I have that the total flux
through that closed surface is simply four pi R squared times E. Well, what is E?
The electric field at this distance R– equals Q divided by four pi epsilon zero R squared
times R roof. That gives me the direction. And so,
if I know that the flux is four pi R squared times E, I put the four pi R squared here,
I lose the four pi R squared and I find that the E vector,
at least the magnitude of the electric field– uh excuse me, that the flux phi,
that’s what I want to calculate, I multiply this by E,
equals Q divided by epsilon zero. And this is independent of the distance R. And that’s not so surprising because if you
think of it as air flowing out then all the air has to come out somehow
whether I make the sphere this big or whether I make the sphere this big. So the flux, being independent of the size
of my sphere, the flux is given by the charge which is right here
at the center divided by epsilon zero. Now if I had chosen some other shape,
not a sphere, but I have dented it like this, it’s clear that the air that flows out would
be exactly the same. And so I don’t have to take a sphere
to find this result. I could have taken any type of strange closed
surface around this point charge and I would have found exactly the same result. And if I put more than one charge inside this
potato bag then clearly since I know that electric fields from different charges can be added, should be added, vectorially, it is clear that the relation should also hold
for any collection of charges inside the bag and therefore we now arrive at our first milestone
in 802, which we call Gauss’s law. And Gauss’s law says that the flux,
the electric flux, going through a closed surface,
being the closed surface of E dot dA, is the sum of all charges Q which are inside
the bag, that you may choose at any time
you pick that bag, divided by epsilon zero. And this is the first of four equations of
Maxwell which are at the heart of this course. So the electric flux through any closed surface
is always the charge inside that closed surface divided by epsilon zero. And if that flux happens to be zero, it means
there is no net charge inside the bag. There could be positive, there could be negative
charges, but the net is zero. Gauss’s law always holds. No matter how weird the charge distribution
inside the bag. No matter how weird the shape of this bag. It always holds. But Gauss’s law won’t help you very much
if you don’t have a situation whereby the charges are distributed
in a very symmetric way. Gauss’s law holds but it doesn’t do you any
good if you want to calculate the electric field. In order to calculate successfully the electric
field you do need forms of symmetry and there are three forms of symmetry
that we will deal with in 802. One is of course spherical symmetry. Another one is cylindrical symmetry. And a third one is flat planes with uniformly
charged distributions. Then we also have situations of symmetry. And so now I would like to, as a first example,
use an application of Gauss’s law and I will start with a situation
of spherical symmetry. And I use a thin shell, a hollow sphere,
which is thin and so this radius is R and I put charge Q on here
but it is uniformly distributed. That’s crucial. If it’s not uniformly distributed I have no
symmetry, I can’t do the problem. So it’s uniformly distributed. We will learn later in the course,
that it’s very easy to do this, because any conductor of this shape,
if you bring charge on it, will automatically distribute itself uniformly. So we have the charge plus Q on there
uniformly distributed, that’s a must, and I would like to know now
what is the electric field here– at a distance R from the center and what is the electric field here
at a distance R from the center. In other words I want to know what is the
electric field everywhere in space. Just due to this charged, uniformly charged
sphere. And with Gauss’s law it just goes like that. You now have to choose your Gauss surface. And if you don’t choose it in a clever way
you get nowhere. In a case like this,
I would think it is rather obvious, that the Gauss surface that you would choose, are themselves spheres,
concentric spheres. If you want to know what the electric field
is at this point you choose a sphere with this radius R going through that point and
if you want to know what it here is you choose a sphere going through that point. All the way enclosed. It’s a concentric sphere. And now you have to use symmetry arguments. And the symmetry arguments are the following. Since this is spherically symmetric, this problem,
if you were here, whatever the electric field is here in magnitude
must be the same as it is there and it must be the same as it is there. Because of the symmetry of the problem,
it couldn’t be any larger here than it could be here. That’s obvious. That’s a symmetry argument because the charge
here is uniformly distributed. That’s symmetry argument number one. Now comes another symmetry argument
and that is– the electric field, if there is an electric field,
must be either radially pointing outwards or radially pointing inwards. So either it has to be like this or it has
to be like this and here the same. Either like this or like this. Because we already know if this is a positive
charge then it’s going to be pointing outward. It cannot go like this or like this
because nature could not decide, in this spherically symmetric problem,
to go like this or like this. It can only go radially. That’s the second symmetry argument. So now if we go to this sphere now
and we know that E is radially outwards apart from a plus or a minus sign,
apart from the fact that the angle between dA and E
could either be zero degrees or a hundred eighty degrees,
we know now that the surface area of that sphere,
which is four pi R squared, times the magnitude of the E vector right here, I can do that now because DA and E are either
parallel or antiparallel, that must be equal to Q inside divided by
epsilon zero. There is no Q inside, so E must be zero. That’s an amazing result. You say well uh there’s no charge inside. Still an amazing result. Because it means that anywhere inside here
no matter what radius you choose the electric field equals exactly zero. And it means there is some crazy conspiracy
of all these charges, that are uniformly distributed here, which each individually contribute to
the electric field inside through Coulomb’s law, that all those together can for through
a conspiracy make the E field everywhere inside zero. It’s a nontrivial result. All right. So now we know that the E field inside is
zero. So this is for r smaller than R. Let’s now go r larger than R. Everything I told you holds for the sphere
which is outside this hollow sphere. Everything holds. The E field here must be the same everywhere
on the surface. dA and E are either parallel or antiparallel. So I can write down again that four pi r squared
which is the surface area times the electric field vector must be the Q inside divided by epsilon zero. But this Q is that Q. It’s not zero. There is charge inside. And so now I know that the electric field E,
in terms of its magnitude, is Q divided by four pi r squared epsilon zero. And we know the direction
if it is positive of course it is radially outwards and if this is negative it’s radially inwards. And this is a nontrivial result. We have seen this before. If I had put all the charge right here at
the middle at the center we would have gotten exactly the same answer. We’ve seen that before. In other words whether the charge is uniformly
distributed over a sphere or whether the charge is, all of it,
exactly at the center of the sphere, that makes no difference for the E field
as long as you’re outside the sphere. If you plot the electric field as a function of R
and if here is capital R and if this is the — the field strength,
then you would get that the electric field is zero inside, jumps to a maximum value and this falls off
as one over R squared proportional to one over R squared. If I go back to the situation that the charge,
that the electric field inside is zero, you may say: “Isn’t that a little bit of a cheat?”. “Because yeah there is no charge inside.”. “But have you really used the charge outside.”. “And if you have used it how did you use it.”. Well I have used it. I use it through my symmetry arguments. The symmetry arguments take into account
that the charge is uniformly distributed. If the charge on the sphere had not been
uniformly distributed, I could not have used the symmetry argument, and therefore the electric field inside
would in fact not have been zero. If there is more charge on the sphere here
than there is there, the field inside the sphere is not zero. So I have used all that charge by using my
symmetry argument. Gauss’s law and Coulomb’s law in a way are
the same law. They both link the electric field with the
charge Q. Key is the fact that the electric force falls
off as one over R square. If the electric field strength did not fall
off as one over R square Gauss’s law would not even hold. And the electric field inside this uniformly
charged sphere would not be zero. So it is the immediate consequence of the fact that electric forces fall off as one over R squared. Gravitational forces also fall off as one
over R squared. Therefore if you take a planet if it existed
which is a hollow spherically spherical planet but hollow inside it means there would be
no gravitational field inside that hollow planet. So if you were there, there would be no
gravitational force on you. If it is spherical. If that planet were a cubical planet then
the gravitational field inside would not be zero. You say well big deal, with 801 we
always take a planet and then it’s not as far as we’re outside the planet we put all the mass
and we consider it as a point. Yeah indeed. It’s not a big deal for you
and it is not a a big deal for me but it was a big deal for Newton. He intuitively sensed that it was correct
that if you have a planet of uniform mass distribution that you can
consider it as a point mass as long as you are outside the planet. But it took him twenty years to prove it
and he finally published his results. It would take us now thirty seconds. He didn’t have access to Gauss’s law. Came about a hundred years later. But the net result is that you see here in
front of you that if you have uniformly charge distribution
and you can draw the parallel with gravity, that it’s you get the same electric field outside that you would have gotten if
all the charge is at one location. At the center. This is spherical symmetry, number one. That’s the easiest symmetry that we have in
802. Now I will present you with a second form
of symmetry which is a flat horizontal plane. And I want you to work out most of it but
I’ll help you a little bit to set it up. Suppose we have a plane which is very very
large. Think of it for now as infinitely large. That doesn’t exist of course infinitely large. And I put on this plane charge. And I put a certain amount of charge density
which I call sigma. Sigma is an amount of charge Q per area A. So it is a certain number of coulombs per
square meter. And it’s uniformly distributed,
so the whole plane everywhere has the same number of coulombs
per square meter. Or microcoulombs or nanocoulombs,
whatever you prefer. And this is a plane which is huge
and you are being asked: “What is the electric field anywhere in space?” Just like we before, we asked,
what is the electric field anywhere inside the sphere and anywhere outside the sphere? Now I want to know what it is anywhere in
the vicinity of this plane. If now you pick a clever Gauss surface
the answer pops out very quickly. If you would choose a sphere as a Gauss surface
you’re dead in the waters, you get nowhere because there’s no spherical symmetry. I will define for you that Gauss surface
but I want you to work out at home how you get the electric field. Suppose I want to know what the electric field
is at a distance d above the plane. What I do now is I choose this as my Gauss
surface. Watch me closely. This is the intersection with the plane. This is my Gauss surface. It is a closed surface. Three conditions have to be met for you to
be able to calculate what the E vector is at that location D. The first one is that this is a flat plane
here and this is the same flat plane. Must be parallel to this plane. That’s a must. If you don’t do that you can’t use Gauss’s
law. The second one is that these vertical walls,
that you have here, are indeed perpendicular to that plane. In other words these are parallel and these
are ver- exactly vertical. If you don’t make them vertical if you do
this you’re dead in the waters. Can’t use Gauss’s law very effectively. And then the third argument
which is very important, that this flat surface is a distance d
above the plane and that this flat surface is exactly
the same distance below the plane. And you can already smell why that is important. Because if you ever want to use a symmetry
argument if this plane is uniformly charged the electric field vector here in terms of
magnitude obviously must be the same as there in terms of magnitude,
maybe not in terms of direction, as long as this d is the same as that d. So that’s why it’s important that the two
d-s are the same. And the only charge that you have inside
when you apply Gauss’s law is the charge which is of course here. That’s the only charge inside that closed
box. If you work this out at home you will find
an amazing result. You will find that the electric flux through
these vertical walls is zero. Nothing comes out through the vertical walls. Think about it. Why that is. Use symmetry arguments. But something comes out here or comes in here
if it is a negative charge and something goes out here. And so you only have two contributions from
those two end plates. You’ll work on that and you will find perhaps
to your amazing result that the electric field equals sigma divided by two epsilon zero and
that it is independent of how far you are from that plane. Whether you’re very far away or whether you’re
close it’s the same. So if this is that plane
and if the plane is positively charged then E would be like this here
and E would be like this here and it would be independent of distance
and if it is negatively charged E would be like so and it would be like,
like so pointing towards the plane and in all cases would the magnitude be
sigma divided by two epsilon zero. Does it mean if I go very far away from that
plane that it is still independent of the distance? Yeah, if that plane is infinitely large. But if the plane is only as large as the lecture
hall here then clearly it would hold very accurately as long as I stay relatively close
to the plane. In other words if my distance to the plane
is small compared to the linear size of the plane. But if I go miles away, well of course then
that plane is charged looks like a point charge if I’m five miles away
from 26-100. if the plane is only as large as this lecture hall then it looks like a point charge
and obviously the electric field will then fall off as one over R squared. So when I say the E field doesn’t change
with distance, it means of course that you have to be
relatively close to the surface, relative to the linear size of that surface. So you are going to prove this
and I’m going to use this now to calculate for you
a much more complicated configuration of two charged planes. But I use that result. That’s very important. And suppose I have here a, a plate,
very large, nothing is infinitely large of course, and it has a surface charge density plus sigma
and I have here a plate which has surface charge density minus sigma and the separation between these
two plates happens to be d. And the question now is what is the electric
field anywhere in space. Here, here and here. And we’ll think of them as being infinitely
large, each plate. And I now use the superposition principle. I say to myself aha! This plate alone, forget this one, this plate
alone would give me an E vector, oh, stick to my colors,
give me an E vector like so and that is sigma divided by two epsilon zero, this one is also pointing away from this,
sigma divided by two epsilon zero and here it’s also sigma divided by two epsilon zero because it’s independent of the distance to this plate. What is the negative charge doing? Well, the negative charge has
E vectors pointing towards it. So here I have an E vector which is sigma
divided by two epsilon zero. Here I have one that is sigma divided by
two epsilon zero and I have one that is pointing towards the plate,
which is sigma divided by two epsilon zero. I use the superposition principle,
I can add electric vectors, and when I do that I find that
these two cancel each other out. So the electric field here is zero. The electric field here is sigma
divided by epsilon zero. The two support each other. They are both in the same direction. And the electric field here is again zero. And that is an amazing result. Of course it’s only accurate if these plates
are extraordinarily large and so I have to draw the field lines
in the situation like this then the field lines would be like so. If the upper plate is positive and the field
in here would be the same everywhere, would be outside zero and outside zero here. Now clearly this cannot be true
if you get into this area here where you are near the end
of these plates. That is not possible. Why not? Well you can’t use your symmetry
arguments so Gauss’s law is not going to help you if you get anywhere near this area. And it is very difficult to calculate the
electric field configuration when you are near the edges,
which we call the… the fringe field. Maxwell of course was a clever man
and he knew how to do that. Today we can also do that very easily
with computers. But I’ll show you from Maxwell’s original
publications that in a situation like that,
he was already perfectly capable of calculating these electric field lines
and you have these two horizontal plates, which one is plus and which one is minus
doesn’t matter, he doesn’t put arrows in there
and what you see is an extremely strong field inside the two plates
and remember that the density of field lines tells you something about the very strong field
but when you get near the edge the field is not really zero. The field strength drops very rapidly
because look the density is very low. But it is not zero. And the electric field is not zero here either
and is not zero there. In our assumption, in our simplification,
we have however assumed that the plate is so large that we don’t have to worry about any end effects and in that case the electric field is only existent
in between the plates but not anywhere else. I now want to demonstrate to you some of the
things that we have learned today. And the first thing that I want to demonstrate
is that the electric field outside a large plane is more or less constant. Doesn’t matter how far away you are. Now the way I’m going to do that is of course
I don’t have an infinite large plane, the plane that you’re going to see only a few
square meters in size. And so with only something like one by one
meter, then it would only be true that the electric field is very close to constant
if I stay very close to that plane. The moment that I go out as far as a meter
of course it’s no longer true. So it’s very qualitative, what I’m going to
show you. But you’re going to see very shortly there
a very large plane. I’m going to get it in a few minutes. And let’s assume that we look at that plane
edge on. So here is that plane. Look at it from edge on,
it will be put here. It will block your view,
that’s why we don’t have it up now. And what I will do now is I will connect that
with the VandeGraaff which is behind it. If you wait a few minutes then class will
pay attention to me and not to you. Uh here is the VandeGraaff,
we’re going to attach it to the VandeGraaff and then we use this interesting fishing rod
which is a small Mylar balloon which we will charge with the same charge
as the VandeGraaff, the same charge as the plate
and we will hold that in front of the plane. And then of course there will be a force. So here is my glass rod. This is the vertical. And because there will be a repelling force
on this air-filled balloon, there will be an angle. There’s an electric force on it because the
two have the same charge. And this is the angle theta that I will show
you projected on that wall. And when I move this away from this plane
you will see that the angle theta becomes smaller. Yes of course because look how small that
plane is. No matter what I do if I go from twenty to
forty centimeters you can’t really say that the plane is infinitely large compared to
forty centimeters. But you will see that the angle of theta will
change very slowly. And then we will remove that plane and then
we will do exactly the same experiment but we will use only the VandeGraaff
which produces now an electric field. And that electric field now falls off as one
over R squared. It’s not constant as a function of distance
but it falls off as one over R squared. This is a hollow sphere. So you can think of it as all the charge right
at the center. As we demonstrated, it’s on the blackboard
still here. You know, you get that amazing result. And so now if I place this,
if I place this fishing rod, this balloon, near the spherical VandeGraaff
you will see that this angle theta drops very fast when I start moving my hand away. Extraordinarily fast. If I double the distance to the center,
the force on that little object will become four times smaller. It’s inverse R square. So let’s first do the plane and then we’ll
try to do the… the single VandeGraaff. And we’ll try to optimize the light conditions. We have a projection here. There’s a carbon arc. Which will hopefully produce some light in
that direction. If the carbon arc works. [chuckles] Marcos, oh I forgot to turn on the
power. Thank you. So this carbon arc is coming on now and you’ll
see there the shadows on that wall. See my hand, here is that plane, and it is
far from infinitely large, that plane. If I were this far away from it, four centimeters,
very close approximation, it would be infinitely large. But if I’m here and there and that’s where
I will be, of course it is not infinitely large anymore. So let’s start the VandeGraaff. You can see that I turned it on. It’s rotating now. I have to put charge on here so I’ll touch
it with the VandeGraaff and so this is now charged. It has the same charge as the plane. The plane is being charged. And here you see the angle. Try to remember that angle. It’s hard to estimate, maybe fifteen degrees. You see the vertical and if now I–
it’s about um thirty centimeters away from the plane. And if now I go back to fifty centimeters,
which is where I am now, you see the angle hasn’t changed
very much. If I go further out, to sixty centimeters,
yeah, the angle goes down a little. Of course it does. But not very much. And if I go far away,
all the way to Mass Avenue, of course the force on this little object would be inversely R squared because then the whole plane would behave
like a point source. So I’ve shown you that very close to this
plane the electric field stays approximately constant. So if now we remove this, Marcos if you can
yeah, you’ll have to take this also off. Thank you very much. So now we have the VandeGraaff alone. So now we know that the electric field falls
off as one over R squared. It’s a very good approximation now. We can think of the charge as being right
at the center. I will give it a little bit of charge. Oh, it is already charged. [chuckles] OK. So look at the projection. The eh, the balloon is now uh oh maybe thirty
centimeters away from the center, maybe forty. Boy, the angle is almost forty-five degrees. And now I go, I double the distance,
I go to about ninety centimeters and look at that angle theta. The angle theta is now down to oh maybe ten
degrees. I will go back where I was. This angle is about forty degrees. And now it’s very small and when I go here
which is about a meter-and-a-half you can hardly see that there is any angle. It’s only a few degrees. And so I’ve shown you only qualitatively that
the electric field falls off very rapidly. In the vicinity of a hollow uniformly charged
sphere. And that it doesn’t fall off very fast
if you are in the near vicinity of a plane. The second thing I want to show you
has to do with the fact that the electric field inside a uniformly charged sphere is zero. Here I have a sphere which is not entirely
closed. I can’t make it closed because I want to demonstrate
to you that there is no electric field inside when I charge it uniformly. And since I have to get inside I need an opening. There’s nothing I can do about it. Since there is an opening, the electric field
is not exactly zero inside. It’s only true if this is a complete closed
surface and if the charge is uniformly distributed. But it’s a good approximation. The opening is quite small. And what I’m going to do is I’m going to charge
this sphere. I’m putting charge outside. I use a device that we have not used before,
but that’s not so important, but here is now that hollow sphere. I’m going to put charge on there. Let’s suppose it is positive charge. So this will be positively charged. Since it is a — a conductor, as we will learn
I think the next lecture or at least this week, that the charge will automatically distribute uniformly, only does that on a conductor and now to demonstrate to you that there is
an electric field here, I will use induction. I have two Ping Pong balls
painted with conducting paint. They touch each other. Under influence of this electric field
this one will become negative and this one will become positive,
we have discussed that last time, you create a dipole. Not important that it is a dipole. I separate the two. I have negative charge here
and positive charge there. I will touch any one of these two balls,
it doesn’t matter which one, with the electroscope and you will see
that there is charge there. So I have demonstrated then that there is
an electric field outside that sphere. Now I will do exactly the same demonstration,
but now I put these two conducting balls inside, so here they are. I touch them, you just have to trust me
that I really will touch them and then I will take them out. And if I didn’t make a mistake,
if I didn’t touch the rim by accident, then I will show you that there is
no electric field inside, it means there is no induction,
so these balls did not pick up charge. I show you with the electroscope that indeed
there is no charge on it. So that is the way I want to do this. So there is the electroscope. Here is the sphere. And the way I’m going to charge it,
has a nice name, it’s called electrophorus,
hard to pronounce. I first rub a glass plate with cat fur. Then I take a metal plate. I put on top. And I touch it with my finger. And now I transfer charge
and you think about it why that is. I put it on here again
touch it again with my finger. I’m again charging it. Put it on top, touch it again with my finger. I want a little bit more charge. So I’m rubbing this again. Put this on top. Touch it with my finger. Every time I do that I feel a little shock. Put it on there. Touch it with my finger. OK, let’s hope that’s enough. So now comes demonstration number one. These two spheres conducting completely discharged,
I bring them close to this sphere. There they are. I separate them. And now they must have picked up charge. Shall I use this one or this one to touch
the electroscope? The same to me. My right hand or my left hand? Who wants right? Who wants left? The right ones have it. And there’s the charge. So I’ve shown you that there is an electric
field there. Through induction I have created
charge on here. Now I’ll do the same inside. It’s always tricky because if I hit–
if I hit the rim then it’s not zero. This one has to go in first
because the opening is too small. Then the second one has to come in. Now I have to touch them
and I really do. I wouldn’t cheat on you. Not this time. They are now in contact with each other. And now I take one out. And I take the other out. Which one shall I touch it, there shouldn’t
be any charge on either one of them. We had left before or we had right before?
Well let’s do this one. This one? Who is for left? Who is for right? The lefts have it. Nothing. Maybe a teeny weeny little bit,
well, the electric field inside is not necessarily exactly zero. But it’s extremely close. The last thing I want to show you
has to do with the fringe field that we have seen here. I have here two parallel plates
which I’m going to charge with an instrument
that we have not seen before which is called the uh
— a Wimshurst. If I rotate this crank I can produce positive
and negative charge. And this plate becomes positively charged
and the other plate automaticly becomes
negatively charged. And I’m going to show this to you
right there. That’s the idea. Yeah, we will make it uh… So there you see these two plates. And you see a Ping Pong ball. And this Ping Pong ball is a conductor,
we put conducting paint on it. And remember when I did the demonstration
with the balloon which bounced between my head
and the VandeGraaff and everytime that it bounced
on the VandeGraaff it took all the charge of the VandeGraaff
and when it bounced on my head it took my charge,
and so it went back and forth and that is what I want to show you now. That this Ping Pong ball will start to probe
that field first outside the capacitor, or I shouldn’t use the word capacitor but these
plates and then I will bring the Ping Pong ball inside and then you will see that the
field is much stronger there. So let’s first get some charge on there. And listen to the sounds. Every time that it hits it bangs. So it’s following almost those field lines. And in doing that it’s actually transferring
charge every time from one plate to the other. It’s nicely going around in an arc
the way you see it there. So it’s clear that there is an electric field
outside. I’ve proven that to you. Otherwise it would never do what it’s doing. So the electric field outside is not exactly
zero, of course not. This plate is not infinitely large. And now I will bring this Ping Pong ball inside,
I have to open up the — the gap a little and I will bring it inside. And you see the field is much stronger. Now it’s going back and forth between those
very high-density field lines, very strong electric field, going back
and forth each time that it hits the plate, it changes polarity and this is not too different from
the experiment I did with the balloon when I bounced it back from the VandeGraaff
to my head and back to the VandeGraaff. OK. Start working on that assignment. It’s not an easy assignment this week. See you Wednesday.

## 100 thoughts on “8.02x – Lect 3 – Electric Flux, Gauss’ Law, Examples”

• ### master thou Post author

But you are talking about uniform surfaces if I take a cuboid then what or a aircraft wing

• ### Shivankar Singh Post author

Sir, In March 2018 you suggest me One movie before my exam "Charlie and the chocolate factory" when asked which chocolate should I carry with me on Exam Center. Sir, I watched it yesterday and It make my Day, It was amazing Movie. I live Charlie Bucket Character in Willy Wonka Chocolate Factory. Thank you Sir

• ### videoviewer viewer Post author

Thank You Sir

• ### Erbaz Khan Post author

Great lecture! There is one thing that I didn't understand, during the lecture you said, "If the electric field strength did not fall off as 1/r^2, Gauss's law would not even hold." What did you mean by that? And, why wouldn't Gauss's law hold if electric field strength were to fall off as 1/r^3 or simply 1/r as in the case of a line charge?

• ### sushant sareen Post author

41:29 Is this guy is his son……???

• ### Romir Post author

Loved your lectures sir, they did made me love physics and now I really now want to discover the science behind everything . Sir can you please also tell me where i can get lectures of chemistry and mathematics by teachers like you who make students see the science and feel it. 🙂

• ### Fahim Mumand Post author

28:09 I ended up with E = Sigma / (4 epsilon0)

• ### Suryanil Banerjee Post author

Sir, I would like to know that if flux for a closed sphere is independent of the radius (R), but we also know that if the radius is huge, say approaching infinity, the electric field (E) also approaches zero. So as we know, if there would be zero Electric field, then there wont be any flux. So how can flux in this case not be dependent on the radius?

• ### Thomas Offenbecher Post author

Such an excellent lecture. Thank you!

• ### HyperChannel Post author

Something to think about: The Gravitation inside a sphearical Planet is never sero. And the density of Matter in a non hollow Planet is not in the middle of a Planet the highest. Gravitation creates no dipoles. Am I right ?

• ### Aashish Kumar Ambasta Post author

What does it mean about the open surface here.?

• ### imran zakir Post author

I am in 2nd year college pre-medical currently we are studying electrostatics in class which was kinda boring until i found your lectures your passion to teach physics is amazing the way you teach with complete dedication and love.Teacher's like you prove to student's like us physics is indeed beautiful thank you sir!.

• ### Samruddhi Deode Post author

At 28:47 won't the two electric field vectors cancel each other ?

• ### Samruddhi Deode Post author

My textbook says that electric lines of induction don't depend on the medium they are in…it also later mentions
ø= q/*EPSILON*…how ?

• ### Shuru10 Post author

this lecture gets 11/10 bagels

• ### Sidhu Bhalla Post author

Does field lines come in contact when two charges attract or repel

• ### Ayush Soni Post author

It does look like what teacher teaches to student preparing for JEE exam

• ### 江泽民主席万岁 Post author

I love this professor!!!!!

• ### AKRAM A. A. AL-KHAZZAR Post author

I watched the three lectures consecutively and thank you very much sir, is the last demo is the basic concept of electric motor of course with magnetic field instead

• ### Me Post author

What was the Logic behind dipoles induction when they were brought near to charged sphere?

• ### Mostafa Nouh Post author

Sir, I don't understand why you touched the metal plate with your finger at electrophorus

• ### DJ Toddles Post author

21:50 gravity/electric field

• ### Javier Sutil Post author

Amazing experiments. Thanks,

• ### Ahindra Biswas Post author

Sir are the resnick halliday problems sufficient to clear concepts regarding the basics of electrodynamics?

• ### __________ Post author

Sir, Will the negatively charged shell be electric field proof ?
Because my teacher told me that shells are electrically shielded!!
But what about the field lines going inward?😫

• ### Shivam Kakkar Post author

Sir, I think you are touching your finger at the time of charging solid sphere because of when you approach metal plate to glass plate it induce some charge but when you touch with it with glass plate it acquire same charge as that of glass plate so there is neccessary to remove that induce charge hence by touching your finger.. you acquire that induce charge.

• ### Incognito LeWIN Post author

How long will it take to complete just all videos of 8.02 course If I get to give it 3 hours per day. I have physics exam in January.

• ### Sourabh Mushre Post author

Sir, in case, if we have two plates parallel to each other with same polarity, the elec. Field must be zero in between, but it still experiences repulsive force. what makes this force generate and How this force can be calculated?

• ### Real Ale Homebrewer Post author

These are great. Thank you for your dedication to teaching and making physics fun and understandable

• ### Bry WHsbg Post author

why touching metal plate, doesn't that action decreases the charge of it because when you touch it, you almost ground it completely?

• ### NEHA SAKKA Post author

2019?
In last his greatness was wearing an egg, now a doughnut, intuitive.

• ### Cumar Axmed Post author

He look like Einstein explaining physics lectures

• ### Cumar Axmed Post author

Ii am from Somalia thank u sir

• ### mohammad saqqa Post author

Hello Prof. Lewin
I have a little question, can a college student succeed in Physics by watching your lectures and solving problems without reading the book line by line? I know this seems a bit ridiculous but I have been thinking about it for a while.

• ### Hexect Post author

Wow at 49:36 you can see a purple spark. But in the previous video when professor beat the kid with cat fur it was a red spark do anyone know why the colour changes? Is it because purple light have more energy in it?

• ### Hamza Tahir Post author

Asslamoalikum sir I m from Pakistan and I really likes your video keep it up sir

• ### FUTUREISTIC S Post author

At 17:15 you said that magnetic field can be either radially inward or radially outward but then why did you consider qinside as 0
Please tell why q inside is zero

is this MIT

• ### David Valle Post author

Is the reason why it's 2EA because there are 2 gaussian surfaces?

• ### Anand p Post author

could you confirm that even the charge was distributed uniformly inside a closed surface area, the net electric field is zero. And my observation was that every charge(suppose positive) have radially outwards field and that's going to be cancelled due to their surrounding charges.

• ### Sourav Das Post author

Sir,
The first thing I want to know that in the last experiment, where a ball oscillates in between the plates, if, the neutral ball was held at first in between the charged plates, without touching any of them, and then allowed to move, would that remain at rest or somehow will start to oscillate? Does Earnshaw's theorem have any application here?

The second thing, while touching the above end of Electrophorus to transfer the negative charge to the earth, were you wearing shoes? Then how do the charges move to the ground?

I will be pleased if you answer these doubts. Respect and Love from India.

• ### Chris Molinared Post author

Can we all agree that this man is an absolutely amazing teacher?

• ### Huma Parveen Post author

I really like your videos sir i dnt knw about your videos but the sir from lighthouse channel has suggested in almost his every lecture about your videos and i just loved it….Thank you so mch for being an amazing teacher

• ### GOVINDARAJ S Post author

Reference 44.46,when rubbing cat fur on glass plate it Gains – ve charge, keeping metal plate on glass plate result in induction – ve top surface, on touching the metal plate some electron flows into your body, make metal plate +ve charge, ON touching metal plate on sphere makes it also +ve charge, This continues
IF ABOVE EXPLANATION IS WRONG, CAN YOU PLEASE EXPLAIN HOW YOU CHARGED THE SPHERE

• ### Rama Pandey Post author

Great sir # best lecture ever😊😊👍👍

• ### A MAN'S MACHINE Post author

How can we choose a Gaussian surface , I mean what factor should we keep in mind while choosing a Gaussian surface.

• ### Rahul Kumar Post author

आपका vidio बहुत अच्छा लगा।
👍👍👍👍👍👍👍👍

• ### Watch Momo Post author

43:10 I lost it here

• ### Smackimation Post author

Why did he touch the metal with his finger at the end of the lecture? I know it has to do with grounding, but wouldnt he not want to ground anything?>

• ### surya kant Pathak Post author

Hello Sir, what would be the position of charges if a closed conductor having a cavity is placed in uniform electric field?

• ### Sandhya Pathak Post author

Hello Sir, is electric field inside a conductor always zero or is it conditioned by the fact that it would only be zero when it is placed in an external uniform electric field?

• ### Sagartirtha Sengupta Post author

Hello sir…..in the question of infinitely large plane you've taken charge of the end surfaces of cylinder…..It is not enclosed charge so why've you taken it?

• ### FreedomFighter Post author

geeze man this 480p is rough

• ### 1paper 1pen Post author

Professor, I love the way you teach us by using demonstrations. I got a bit confusion in the air flow analogy. It is analogous but what is the proof that this analogy holds in each case, for example, using this analogy, flux is flow amount (which depends upon source and) is obviously shape independent but how do you prove by using the main definition of flux that flux is shape independent, if the shape is closed. You can ask me again if I didn't explain my question correctly.
Thanks for all the videos.

• ### Irfan Ullah Post author

i am from pakistan sir so nice

You are a great teacher for all students.

• ### Aryavansh Saraf Post author

Sir I am a jee aspirant plz can u tell me a reference book for the subject

• ### John Smith Post author

Don't we need a double integral to integrate over a surface?

• ### Channaveeraiah Hiremath Post author

Sir from which college you're

• ### sushree satapathy Post author

Why cant a charge be distributed symetricaly in cube

• ### Thet Aung Post author

In the case of cylinder, the flux through curved surface of the cylinder is zero because normal to that surface is radially outward to the surface ,thus perpendicular to the electric field.
Instead of cylinder if it were some other shape, like a box, is the flux through the surface the same or different? My intuition is that it'll be different since Sigma depends on the area, but I really want your confirmation.

• ### Karlyann Laboy Post author

I am studying for my Physics 2 final and your lectures had help me all over the semester. Thanks! Such an inspiration. Thanks from University of Puerto Rico at Mayagüez, from a Chemical Engineering student ☺️

• ### pkdude junction Post author

wow you are amazing

• ### Mimic Nature -The Natural Way of Farming Post author

Sir, Thank you.

• ### Amit Kumar Post author

Dear sir …You are na great teacher ..really love Ur videos … I am from India …We study without any experiments but Ur lectures are outstanding… Want to go into MIT … Thank you sir

• ### sashank Mathamsetty Post author

I am from India and never seen such teacher!! (though we were taught this in 11 grade)

• ### Chahat Sharma Post author

SIR DO UR LECTURES HELPFUL FOR AIIMS UG WHICH IS HELD IN INDIA??????

• ### Chad Giovanni Post author

No wonder he teaches at MIT! He’s a great man!

• ### Beautlin Stanley Post author

Ok can anyone explain very easily what's the physical meaning of flux

• ### OrangeC7 Post author

I absolutely loved that demo with the ping pong ball showing the strong electric field between the plates. You could even see it sparking sometimes!

• ### harshit krishnatre Post author

Sir, I want ask which is the most fascinating physics equation of all time except Maxwell equations In your mind

• ### kanchan bala Post author

Sir which instrument you have named at 47:15

• ### Chirag Taneja Post author

Sir why are you touching the metallic plate with your finger while using the electrophore for charging the sphere ?

• ### Ungalil Oruvan Post author

Sir how electric field is equal to
Q/4π£°R2
In previous lecture you said
Electric field = F/q
Sir Please explain how you arrive at that equation.. Thankyou

• ### Andre Caldas Post author

Wonderful lectures! I would like to take the opportunity to share something that has bothered me since I was a graduate student of electrical engineering.

Like at 25:50, I have always found very annoying when the integral surfaces crosses the surface where the charge is distributed, or when it crosses boundaries of two different materials. I wish that when stating the law, professors would comment about this. After all, the "zero" in "epsilon zero" is there for a reason… 🙂

• ### Mohammmad Nayem Post author

Teacher,You defined electric flux as a mathematical quantity defined as integral of E.dA.My little knowledge says it should be so.But,in google,some of famous books like resnick haliday,and various well known youtube channels are saying that flux is the amount of field line😰 passes through a surface.Even,they are saying that if the charge inside a close surface doubled the force line should be doubled that completely make no sense to me(Following you I believe that force line should be infinite regardless how small the charge is)Teacher,help me a little bit to learn some actual physics

• ### TBoy205 Post author

What is the deal with the bagel?

• ### Manoj Kumar Post author

You are a wondful person to taught me physics nicely

Very good!

• ### Anamika Sinha Post author

Does his lectures sufficient for class 12 grade student cbse

• ### Cf Mh Post author

Can someone please help me letting me know the name of the machine Walter is cranking at 48:43 ?

walter i think u are the only real teacher rest are virtual

• ### Mahmoud Desokey Post author

Sir, is there a mathematical proof for zero electric flux through a closed surface having a charge outside it ?

• ### Mahmoud Desokey Post author

Sir, you mention at 21.39 that the electric field inside a uniformly charged sphere would not be zero if the electric field falled of as 1/r^2.why is that?

• ### Raja Sarkar Post author

I m a physics student teacher…As i love physics..therefore I do love ur teaching styles..Thank You sir..
Some one Said"If u hate Physics (though it shouldn't be) dan U've probably learnt it from a Wrong teacher" rightly said.

• ### Mahmoud Desokey Post author

Sir , you mentioned that the electric field inside a cube of uniform charge distribution will not be zero . but my intuition tells me that every face of the cube will cancel the effect of the other face .so the E field should be zero .

• ### Messi H Post author

Professor Lewin, around minute 46:54, you demonstrated that the Electric field inside the large sphere is (almost) zero, but my question is, why don’t the 2 small charges (which you inserted into the sphere) retain the positive and negative charge that you gave them through induction? Shouldn’t they retain their charges if the electric field inside the sphere is Zero? How did they lose the charge?

• ### Soham Sen Post author

Professor, I had a question regarding the calculation of Electric Field using Gauss’ Theorem. It goes as follows:

If we consider that two identical charges ‘+q’ (same magnitude & polarity) are placed at coordinates (-a,0,0) & (a,0,0) respectively then Electric field at origin should be 0, using Coulomb’s Law & then superposing the forces.
However if we construct a spherical Gaussian surface of radius ‘a’ with centre at origin then Φ= q/ε =surface integral of E & dS . Since Φ is non zero the integral will be non zero signifying that E has some non zero value.

I wanted to know how are we supposed to take into consideration the other charge for the field E to be 0.
Thank you.

• ### Alex Sinek Post author

I may have missed it, but does anyone know what is the book he refers to during the lectures? Anyway I am loving the lectures so far!

• ### Ahmad Eldesokey Post author

Sir, will the electric field inside a hollow charged metalic cube be zero ?

• ### Ahmad Eldesokey Post author

Sir, is there is any mathematical derivation other than Gauss law for the case of zero E- field inside hollow conducting sphere ?

• ### Ahmad Eldesokey Post author

Sir , for the case of a hollow charged conducting sphere with an opening i understand intuitively that the E-field will not be zero inside the sphere but is there is a quantitative proof for such case ?

• ### Ahmad Eldesokey Post author

Sir, will the electric field be zero inside a thin insulating spherical shell that is uniformly charged over the surface ? I am talking about the E-field in the hollow part .thanks

• ### Ahmad Eldesokey Post author

Sir , is the E-field zero inside the hollow part of a cylindrical shell that is opened at the end faces ?

• ### Hidden Temple Post author

Can you give an explanation why the E field inside a closed spherical surface is 0 without using a Gauss's theorem (maybe using the expresion of E derived by Coulomb's law)?

• ### Vinod Srivastava Post author

Before this lecture: I think I should leave physics

After: I should do research on physics

• ### Manish Bishnoi Post author

Love from India . Your are such a great legend of teaching physics. Any one will fall 😍 in physics# sir WL.🙋🙋

• ### Chad Sweeney Post author

Came here to get away from politics, and I see now I can't escape the "left vs right" battle