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Proof (Advanced): Field from infinite plate (part 1) : Electric field generated by a uniformly charged, infinite plate
相關課程
- 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.
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