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## Physics library

### Course: Physics library > Unit 11

Lesson 2: Electric field- Electric field definition
- Electric field direction
- Magnitude of electric field created by a charge
- Net electric field from multiple charges in 1D
- Net electric field from multiple charges in 2D
- Electric field
- Proof: Field from infinite plate (part 1)
- Proof: Field from infinite plate (part 2)

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# Proof: Field from infinite plate (part 1)

Advanced proof of the formula for the electric field generated by a uniformly charged, infinite plate. Created by Sal Khan.

## Want to join the conversation?

- this is a really dumb question, but the parallel plates that aren't infinite are just an approximation of the infinite when charge and length of the plate are big enough compared to the distance right?(57 votes)
- Hi.

You are sort of correct. Khan actually mentions that at the end of the next video in this series: that for a finite plate where you are far away from the edge, this theory holds, so yes, you are correct. However, just remember that right near the edge, things fall apart.

Most theories are really good approximations; the only real values come when you test experimentally.

This isn't really a dumb question; I pondered in once myself, and most people will.(88 votes)

- At9:30, why is the area of the ring the circumference times dr? If dr were much larger (call it w), wouldn't the area be pi((r+w)^2-r^2) ? Like the difference between two circles.(24 votes)
- You are actually correct. If dr was larger, than it would be pi((r+w)^2-r^2) for the area. However, this is not the place to use that. We are assuming an infinitely small width between the two circles, we just assume that there is a central circle, right in the middle of the two circles, and we extend a length dr for the width. Because of the size, this average value is actually really accurate, so we stick to it, and that makes the math much more simple.

I hope that is not confusing; one of my teachers spent half a class teaching us why that is so, but his explanation was much better than mine.(18 votes)

- Around9:59, Sal says that the force generated by the ring is = kqq/d^2. However, how is the entire charge of the ring enclosed to one point? More specifically, how come he is treating the entire charge of the ring as enclosed to one point? Is he allowed to do that? Thanks.(14 votes)
- Since the ring is of uniform charge, and the charge density is sigma then Q (<-Charge of ring) = sigma (<-charge density) * (2 * pi * r * dr) (<-area of the ring). He is correct in doing this, but only in the case of a surface of uniform charge (The charge at every point of the surface is equal) and with the test charge (q) at the center of the ring so that all of the horizontal components of the electrostatic forces cancel, leaving only the vertical components.

Since forces are vector quantities and in the same direction, we can treat them as if they were all directly under the test charge and them sum them up to receive the total net force on the test charge from the ring.

Now if the ring was not of uniform charge and/or the test charge was not at the center of the ring, then we would have a much more difficult problem.

I hope this clears things up for you. ^^(12 votes)

- At9:18why is the width of the circle 'dr'; why does it share the same variable as the radius? Why not 'dx' or 'dw'? What allows us to use 'r' both times? Does it have to do with mstocksl@hotmail.com's question? Thanks!(12 votes)
- because dr is just a small change in radius

since he's talking a bout a circle with radius r it seems fitting to use dr

and because in the next video he's going to integrate across r

if he used dw he would have to rename r to w to be consistent(1 vote)

- Doesn't this video assume that the magnetic field will repel the particle, instead of attracting it?(7 votes)
- Yes it does. It was mentioned early in the video that the the chanrge on the plate is positive and the particle is positive as well so the force would be repulsive.(11 votes)

- Why does Sal take an infinite plane? I find it quite hard to imagine.(5 votes)
- If you pick a plane of finite size, you would have to compute the net effect of whatever you are studying at that boundary and include it in your final answer. In much of physics, an effect (gravitational potential or electric field for examples) are defined as zero at infinity, so you can ignore that 'boundary' value.

So think of it as 'choose a plane whose size is large enough that we can use infinity as the 'far' distance (for example, we could calculate (or integrate) from zero radius to infinity). If the effect has dropped to zero at infinity, that term becomes simpler to calculate - or may drop out altogether.(8 votes)

- At ~9:30, why didn't we take πr^2, since there are charged particles everywhere on the ring?(5 votes)
- We have to consider the point charges on the ring. So the total charge isn't the area because the "ring" is assumed to be hollow. So the charge is :-

(circumference) * (thickness) * (charge density)

Watch the next part and Sal will integrate this equation and obtain a general formula for this situation.(6 votes)

- He saying i need to learn about calculus first, can anybody tell me from where should i start learning it, i mean there are 3 different calculus topics ;Differential, multivariable, integral.Or any other type?(2 votes)
- For this specific video, you'd want to learn basic integration techniques, though learning most of differential and integral calculus (single-variable) would be a good idea so as to establish a thorough knowledgebase.(5 votes)

- So, a test charge could be light-years away and still feel the same force?(4 votes)
- If the plate were infinite, yes. Infinite is big even compared to light-years. But there's no such thing as an infinite plate, right?(2 votes)

- Why do you calculate the charge from the whole area, if youre only gonna use the points on the edge of the circle?(3 votes)
- since it is a ring and ring does not has a symmetry except at the ends of diameter, therefore he has taken two small charges 'dq' and then integrate it(2 votes)

## Video transcript

In this video, we're going to
study the electric field created by an infinite uniformly
charged plate. And why are we going
to do that? Well, one, because we'll learn
that the electric field is constant, which is neat by
itself, and then that's kind of an important thing to realize
later when we talk about parallel charged plates
and capacitors, because our physics book tells them that
the field is constant, but they never really prove it. So we will prove it here, and
the basis of all of that is to figure out what the electric
charge of an infinitely charged plate is. So let's take a side view of the
infinitely charged plate and get some intuition. Let's say that's the side view
of the plate-- and let's say that this plate has a charge
density of sigma. And what's charge density? It just says, well, that's
coulombs per area. Charge density is equal
to charge per area. That's all sigma is. So we're saying this has a
uniform charge density. So before we break into what may
be hard-core mathematics, and if you're watching this in
the calculus playlist, you might want to review some of
the electrostatics from the physics playlist, and
that'll probably be relatively easy for you. If you're watching this from the
physics playlist and you haven't done the calculus
playlist, you should not watch this video because you will
find it overwhelming. But anyway, let's proceed. So let's say that once again
this is my infinite so it goes off in every direction and it
even comes out of the video, where this is a side view. Let's say I have a point
charge up here Q. So let's think a little bit
about if I have a point-- let's say I have an area
here on my plate. Let's think a little bit about
what the net effect of it is going to be on this
point charge. Well, first of all, let's say
that this point charge is at a height h above the field. Let me draw that. This is a height h, and let's
say this is the point directly below the point charge, and
let's say that this distance right here is r. So first of all, what is the
distance between this part of our plate and our
point charge? What is this distance that
I'll draw in magenta? What is this distance? Well, the Pythagorean theorem. This is a right triangle, so
it's the square root of this side squared plus this
side squared. So this is going to be
the square root of h squared plus r squared. So that's the distance between
this area and our test charge. Now, let's get a little
bit of intuition. So if this is a positive test
charge and if this plate is positively charged, the force
from just this area on the charge is going to be radially
outward from this area, so it's going to be-- let me do it
in another color because I don't want to-- it's going to
go in that direction, right? But since this is an infinite
plate in every direction, there's going to be another
point on this plate that's essentially on the other side of
this point over here where its net force, its net
electrostatic force on the point charge, is going
to be like that. And as you can see, since we
have a uniform charge density and the plate is symmetric in
every direction, the x or the horizontal components of the
force are going to cancel out. And so that's true for really
any point along this plate. Because if you pick any point
along it, and we're looking at a side view, but if we took a
top view, if that's the top view and, of course, the plate
goes off in every direction forever and that's kind of where
our point charge is, if we said, oh, well, you know,
there's this point on the plate and it's going to have
some y-component that's on this top view coming out of the
video, but it'll have some x-component, this point's
x-component effect will cancel it out. You can always find another
point on the plate that's symmetrically opposite whose
x-component of electrostatic force will cancel out
with the first one. So given that, that's just a
long-winded way of saying that the net force on this point
charge will only be upwards. I think it should make sense
to you that all of the x-components or the horizontal
components of the electrostatic force all cancel
out, because they're infinite points to either side
of this test charge. So with that out of the way,
what do we need to focus on? Well, we just need to focus
on the y-components of the electrostatic force. So what's the y-component? So let's say that this point
right here-- and I'll keep switching colors. Let's say that this point-- and
once again, this is a side view-- is exerting-- its field
at that point is e1, and it's going to be going in
that direction. What is its y-component? What is the component
in that direction? And, of course, it's
pushing outwards if they're both positive. So what is the y-component? What is that? Well, if we knew theta, if
we knew this angle, the y-component, or the upwards
component is going to be the electric field times
cosine of theta. Cosine is adjacent over
hypotenuse, so hypotenuse times cosine of theta is
equal to the adjacent. So if we wanted the vertical
or the y-component of the electric field, we would just
multiply the magnitude of the electric field times the
cosine of theta. So how do we figure out theta? Well, that theta is also the
same as this theta from our basic trigonometry. And so what's cosine of theta? Cosine is adjacent
over hypotenuse from SOHCAHTOA, right? Cosine of theta is equal to
adjacent over hypotenuse. So when we're looking at this
angle, which is the same as that one, what's adjacent
over hypotenuse? This is adjacent, that
is the hypotenuse. So what do we get? We get that the y-component of
the electric field due to just this little chunk of our plate,
the electric field in the y-component, let's just call
that sub 1 because this is just a little small
part of the plate. It is equal to the electric
field generally, the magnitude of the electric field from this
point, times cosine of theta, which equals the electric
field times the adjacent-- times height-- over
the hypotenuse-- over the square root of h squared
plus r squared. Fair enough. So now let's see if we can
figure out what the magnitude of the electric field is, and
then we can put it back into this and we'll figure out the
y-component from this point. And actually, we're not just
going to figure out the electric field just from that
point, we're going to figure out the electric field from a
ring that's surrounding this. So let me give you a little bit
of perspective or draw it with a little bit
of perspective. So this is my infinite
plate again. I'll draw it in yellow
again since I originally drew it in yellow. This is my infinite plate. It goes in every direction. And then I have my charge
floating above this plate someplace at height of h. And this point here, this could
have been right here maybe, but what I'm going to do
is I'm going to draw a ring that's of an equal radius around
this point right here. So this is r. Let's draw a ring, because all
of these points are going to be the same distance from
our test charge, right? They all are exactly like this
one point that I drew here. You could almost view this as
a cross-section of this ring that I'm drawing. So let's figure out what the
y-component of the electric force from this ring is
on our point charge. So to do that, we just have to
figure out the area of this ring, multiply it times our
charge density, and we'll have the total charge from that
ring, and then we can use Coulomb's Law to figure out its
force or the field at that point, and then we could use
this formula, which we just figured out, to figure
out the y-component. I know it's involved, but it'll
all be worth it, because you'll know that we have a
constant electric field. So let's do that. So first of all, Coulomb's Law
tells us-- well, first of all, let's figure out the charge
from this ring. So Q of the ring,
it equals what? It equals the circumference
of the ring times the width of the ring. So let's say the circumference
is 2 pi r, and let's say it's a really skinny ring. It's really skinny. It's dr. Infinitesimally
skinny. So it's width is dr. So that's
the area of the ring, and so what's its charge going to be? It's area times the charge
density, so times sigma. That is the charge
of the ring. And then what is the electric
field generated by the ring at this point here where
our test charge is? Well, Coulomb's Law tells us
that the force generated by the ring is going to be equal
to Coulomb's constant times the charge of the ring times our
test charge divided by the distance squared, right? Well, what's the distance
between really any point on the ring and our test charge? Well, this could be one of the
points on the ring and this could be another one, right? And this is like a
cross-section. So the distance at any point,
this distance right here, is once again by the Pythagorean
theorem because this is also r. This distance is the
square root of h squared plus r squared. It's the same thing as that. So it's the distance squared
and that's equal to k times the charge in the ring times
our test charge divided by distance squared. Well, distance is the square
root of h squared plus r squared, so if we square
that, it just becomes h squared plus r squared. And if we want to know the
electric field created by that ring, the electric field is
just the force per test charge, so if we divide both
sides by Q, we learned that the electric field of the ring
is equal to Coulomb's constant times the charge in the
ring divided by h squared plus r squared. And now what is the
y-component of the charge in the ring? Well, it's going to
be this, right? What we just figured out is the
magnitude of essentially this vector, right? But we want its y-component,
because all of the x-components just cancel out,
so it's going to be times cosine of theta, and we figured
out that cosine of theta is essentially this, so
we multiply it times that. So the field from the ring in
the y-direction is going to be equal to its magnitude times
cosine of theta, which we figured out was h over
the square root of h squared plus r squared. We could simplify this
a little bit. The denominator becomes what? h squared plus r squared
to the 3/2 power. And what's the numerator? Let's see, we have kh and then
the charge in the ring, which we solved up here. So that's 2 pi sigma r-- make
sure I didn't lose anything-- dr. So we have just calculated
the y-component, the vertical component, of the electric
field at h units above the plate. And not from the entire plate,
just the electric field generated by a ring of radius r
from the base of where we're taking this height. And so I've already gone 12
minutes into this video, and just to give you a break and
myself a break, I will continue in the next. But you can imagine what
we're going to do now. We just figured out the electric
field created by just this ring, right? So now we can integrate across
the entire plane. We can solve all the rings of
radius infinity all the way down to zero, and that'll give
us the sum of all of the electric fields and essentially
the net electric field h units above the
surface of the plate. See you in the next video.