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this video will explore how charges spread over a conductor
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. And in the first case here we have,
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ah, conducting shell. It's got an inner radius
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of a and an outer radius of B, and
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the cavity is centered at the center of the shell
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. It's also got a total charge of Q.
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So first we want to know what does the chart
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. What does the electric field look like everywhere in
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space? Well, we've already discussed that inside a
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conductor electric field is always zero. And so right
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away we can say e for are greater than a
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and are less than B is simply zero. Now
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how about inside the cavity? Well, we could
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draw Gulshan surface and we see that Phi equals integral
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of e dot d A, which is equal to
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Q and closed over Epsilon. Not and there's no
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charge in there. So this has to be zero
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, which means our electric field is zero or more
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appropriately when our is less than a equals zero.
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So now we draw Argh! Ocean surface on the
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outside and I'll leave the work to you. But
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we know that again by this formula here, the
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total charge enclosed is the charge on the conducting shell
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, which is Q and so we get that e
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dot d a. Is equal to Q over Epsilon
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. Not and again, The electric field is perpendicular
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to the area of our gout to the surface area
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of Goshen surface, and it's always pointing Radio Lee
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outward. So this is simply e times Thean a
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Gral of D A. On the radius of the
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sphere is four pi r squared, and we get
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back to our familiar formula for the electric field of
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a point charge when we're outside of the conducting sphere
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. Spherical shell of Q over four pi Excellent,
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not R squared. But now the more important question
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becomes, How is this charge distributed in this conductor
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? It's got two surfaces and a total charge.
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Q. So how was that charge? Q Located
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. So we'll clean this up a little bit and
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consider this on each case. We know that inside
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the cavity there is no charge in the electric field
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. Is zero further? We know that inside the
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conductor itself in this guy ocean surface, there is
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no charge because the field there is zero. So
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if there's no charge in the cavity. There can't
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be any charge on this inner wall. So we
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can say that our surface charge density that the surface
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were at the inner cavity where R equals a zero
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, and that extends all the way out to the
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edge of the outer cavity, since there can't be
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any charge inside, otherwise we'd have an electric field
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inside the conductor. So we conclude that all our
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charge has to be distributed over this outer surface.
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And we already knew that in a conductor charge will
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distribute to the surfaces. But in this case,
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we've proven that it's gotta be the outer surface,
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and I'm stressed that it's Onley. In this case
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, we're not only in this case, but its
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particular to this case in this manner, so we
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can write that are surface charge density at R equals
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B is the total charge over the area, which
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is Q over four pi B squared. But now
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let's consider a slightly different case. We have the
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exact same conducting shell. It's got an inner radius
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of a and outrageous of be spherical in shape.
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So this time, instead of charging it, we
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put a point charge of Q at the center of
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its cavity, so at the center of the entire
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show. First question. What is the electric field
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everywhere in space? Well, again, without having
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to do any work, we know that inside the
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conductor, the electric field has to be zero.
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Now. Inside the shell, we have Argh!
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Ocean surface here of some radius R. And the
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charging close is simply Q. And we know we'll
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get the familiar expression for a point. Charge electric
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field as such. And if we go outside well
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again, the total in charges Q. At the
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center of this Goshen surface. And so we know
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this applies for inside the cavity and outside of the
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entire conductor, and you can work through that.
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But it's a situation we've seen many times. More
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, interestingly, is now how is charged distributed in
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this conductor. Well, we know that the electric
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field on some Gulshan surface inside the conductor has to
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be zero, which means the net charge inside has
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to be zero from the simple fact that the flux
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, which is equal t e d. A equal
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to Q enclosed or epsilon, not there is no
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charging closed because there is no field inside a conductor
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. That means we have toe have some kind of
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charge inside the conductor or inside of this guy ocean
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surface of negative que. And as we know,
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charging a conductor always goes to the surfaces and spreads
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out evenly over symmetric surfaces. And so we get
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little charge around the inner cavity of this shell.
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The total charge of Q. So we can say
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the surface charge density on the inner shell is negative
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. Que over four pi. Hey, squared charge
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per unit area. Now what happens to a situation
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where we go outside of the entire thing? Well
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, we've just shown that there's gotta be a charge
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of negative Q on the inner shell, which means
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right now we have zero total net charge. Also
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, we know this entire shall This entire conducting shell
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is uncharged. And right now we have a total
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charge of negative cue. So that means that somewhere
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out here we need a charge of plus cute to
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counter act that negative que and again in a conductor
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. These charges will spread out all over the surface
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like so and so we see that the surface charge
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density at the outer shell is positive. Que over
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four pi b squared. And just to double check
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our total charge, well, we have our point
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charge. Q. We have the charge on the
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outer part of the shell, which is also que
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, and we have a charge on the inner part
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of the shell, which is negative, que?
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Which gives us a total charge of Q as stated
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in the problem. So we can see that when
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you place a charge inside the shell of a cavity
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of a conductor, the charges within the conductor will
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move around, which they can do very easily because
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it is a conductor in order to make sure that
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the electric field within the connector remain zero. And
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so there's no there's no charge on the conductor.
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There is net charge on the surfaces of it.
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Clean things up a little bit because I want to
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consider a special case, particularly what happens at our
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boundaries. We know the charge on the inner shallow
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, this conductor or the inner surface becomes Q over
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four pi epsilon doubt a squared, and at the
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outer surface we get something very similar but would be
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squared instead of a squared. So let's consider this
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first. Well, I can actually rewrite this.
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And sorry, let me put in the vector directions
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because that will matter in a moment. I can
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actually write this Sigma of r equals B over Epsilon
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. Not conversely, over here, I can also
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write this as Sigma of r equals a over epsilon
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, not. And let's put in the radio directions
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and you'll notice Epsilon, not Epsilon. So articles
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A is negative, which points us in the right
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direction. It's pulling us toward the surface, which
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is Radio Lee outward that way. So the negative
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sign accounts for the direction of the electric field.
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And so we see one question is always well,
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what happens right at the border of a charge conductor
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or the charge conducting surface and free space. What
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happens to the electric field? Well, at that
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point, the component is simply the perpendicular component of
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the electric field is simply the charge density on that
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surface over epsilon Not, and this will always be
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able to. This is something will always be able
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to check by checking our boundary conditions. Uh,
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whenever we do a problem like this,