WEBVTT
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So here we have a familiar picture. Well,
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blue particle with charge Capital que No red particle with
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a charge Q one that's at some displacement vector are
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from the blue particle, and we know by columns
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law that the red particle will feel a force from
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the blue particle of K capital. Q Q one
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over R squared in the direction of our hat.
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Let's ask ourselves, what happens if we remove Q
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one? Well, there's no particle there for the
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blue particle to act on, so this force has
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to go away. So what's happening here at this
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point? Well, we can put Q one back
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, and we know that force will come back.
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Alternatively, we can put a different particle there,
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say, with the charge of Q two, and
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we know we'll have a force in the same direction
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this time with Q two in its magnitude instead of
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Q one. We also know we can move Q
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one. Stay down here at some displacement are to
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well, we know this will feel a force from
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the little particle of F K Capital Q Q.
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One over our two squared in the direction of our
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two in fact, we know we can place charged
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particles anywhere in the vicinity of the blue particle,
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and each one will feel some force from that blue
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particle. And so just the mere presence of the
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blue particle is doing something to the space around it
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. And we call that something the electric field,
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and it gives us a new way to think about
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electric force. We've always said that Q two feels
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a force from Q. But a better way to
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think about that is que creates an electric field.
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Q. Two feels the force from that electric field
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, and so we say that this blue particle charge
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capital Q grades an electric field in space. But
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how do we quantify that field was considered this yellow
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particle here, which we know feels this force from
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the electric field created by the blue Particle? I'm
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going to rewrite this a little more suggestively like that
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. Recall are Red Particle was originally at this point
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, and it felt this force in that direction,
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and we could say that this particle here has a
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charge of Q three and move it there, and
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we know by cool arms law the force it feels
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would be equal to that, and right away we'll
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note that all of these forces have that in common
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. So part of the electric force is determined by
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the charge of the particle feeling that force from the
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field. But the rest is determined by the charge
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that is the source of the electric field, its
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distance from that source and, of course, the
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direction from that source. And so we can rewrite
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the electric force on a charge. Q. As
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Q times the electric field. At that point,
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we're now e of our It's simply the force divided
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bike or cake over R squared our had Remember this
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. Q is the source of the electric field,
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and this is true for all point charges. Cool
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. Now what's interesting to note is that, yes
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, that is the value of the electric field here
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. But our is arbitrary. We just called that
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are we could call this our and we'll have a
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value there for the electric field. We can call
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this our and we have an infinite amount of ours
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because space is infinite. And so there are unfinished
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amount of Decter values possible from the electric field and
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So we called the electric field of Vector Field.
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We'll know two things about the electric field. The
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first is its units. This is the expression we
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used to derive an expression for the electric field.
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Of course, ah force has units of Newton's and
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a charge has Newtons of units of cool arms.
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So the electric field has units of Newtons per kula
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, and this makes sense. Um, electric field
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applies a force upon a charge. The second thing
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will note is that the electric field obeys the principle
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of superposition. So say we have a space that
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has a bunch of test charges. Q one through
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Q I and we're interested in a point here that
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will call point P well if we put a test
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charge there at point P with charge. Q.
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We know that by the principle of superposition the net
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force on that charge, it's simply the some of
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the vector forces on it from the individual charges around
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it. What if we take you out and just
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let focus on playing P? We know we can
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divide this side by that charge, which will give
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us a vector some of electric fields and so the
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electric field at that point P is given by the
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vector sum of the electric fields from all the charges
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affecting Point P to demonstrate the superposition of electric fields
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will look at a very common example, and that's
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the electric dipole on electric. Dipole is simply a
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system of two particles with equal and opposite charges separated
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by some distance. So here we have a particle
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in blue with charge, plus Q and a particle
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in reading might with charge minus que They're both on
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the X axis, and they're both separated from the
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origin by a distance D. And for this example
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, we want to find what the field is at
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some point p that we can arbitrarily place on the
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Y axis. Since it's arbitrary, we could just
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call this distance. Why? Well, the field
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created at that point by the blue particle, which
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will call E plus simply equal to cake over R
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squared in the art direction. Where are is this
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displacement vector? But from the geometry of the problem
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, we have a right triangle, and we know
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R squared is equal to D squared, plus y
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squared for the red particle. Our electric field is
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equal to cake over this displacement our prime. But
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from symmetry Lino, our prime has the same magnitude
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as our. And so our prime squared is also
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equal to D squared plus y squared. And this
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is the our prime direction. Now here you'll know
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I've drawn the electric field vector for the blue particle
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away from the particle and for the red particle toward
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the particle. This is very common. Let's consider
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what happens if we put a particle with charge at
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Point P. Well, we know that force equals
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Q times, the electric field vector. So if
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we put a positive particle at point P, these
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force vectors will point in the same direction as the
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field. Factors in a positive particle will be repulsed
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by a positive particle and attracted to a negative particle
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, which we expect. However, if we point
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put, a negatively charged particle appointee will flip the
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direction for the field vector for the four. Specter's
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and a negative particle will be attracted to the positively
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charged particle and repulsed by the negatively charged particle.
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And that's what we expect. So we always draw
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field vectors as pointing away from a positively charged source
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and toward a negatively charged source. And now it's
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simply a matter of finding our components and adding those
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together. And what will note is that these feel
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vectors have the same magnitude. We'll also note that
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they make the same angle with the horizontal, but
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in opposite directions. So by symmetry, we know
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that the white components are going to be equal and
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cancel each other off. So the net in the
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Y direction it's simply zero. Similarly, we know
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they're going to have the same X component. And
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so we can say that E. Net X is
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simply this same as two times e plus X or
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two times E minus X, and we only have
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to calculate one. So what is the X component
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? Well, e minus X or e plus X
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. It's simply equal to cake over de squared,
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plus y squared times, the cosine of that angle
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theta and from our triangle, we know the coastline
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of data is simply the adjacent side over the opposite
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side, the adjacent side of being D over the
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high pot news and the high Patna is being squared
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of d squared plus y squared and so we see
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that plus X equals K Q D over de squared
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plus y squared to the three halfs. And,
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of course, that's in the X direction. So
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we met at that point. P is two k
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Q d over de squared plus y squared to the
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three halves hold racks. We'll notice that as why
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goes to zero. So as we move this point
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down to the origin, the Nina simply goes to
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K two cake over de squared in the extraction,
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which we've seen before, is the correct expression for
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a point between two equally charged particles at its center
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.