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to this point, we've just discussed the electric force in terms of two particles. So if we have this blue particle here with the charge capital Q. And say we have a red particle over here with charge Q. One and it's displaced from the blue particle. Some vector are one, then using columns law, we know that the force on the blue particle from the red particle, which we call F one simply K capital Q Q one over our one squared in the R one direction. But now let's add green particle over here. It's got a charge of Q two, and it's got a displacement to the blue particle of our two. Well, how does this affect Herbal Particle? We don't know what's going on with the red particle, but we knew we know if the red particle wasn't there in the green particle will just import in part a force of F to like Who loans Law of K Capital que que two over our two squared in the art to direction, and similarly, we can add another particle here with charge Q three at some displacement are three, and we know the force, ignoring the red and the green particles on the blue particle would be simply this. But what is happening to the blue particle from all three of these four forces happening at the same time? How do we figure out what the Net forces and its direction? Well, for this we use The principle of superposition on with superposition tells us, is that the net force on our blue particle would charge. Q. It's simply the vector sum of the individual forces created by each of the three charged particles in the area in superposition will help us deal with more complex problems where we have many charges or we actually have charges distributions, not just point charges of we, as we've been dealing with so far, we'll see how powerful that can be in a few examples. So our first demonstration of superposition will use this very basic example. We have three charges on a line. We'll call the direction of this line the extraction with positive to the right in the middle. We have a particle in blue with the charge of capital Q on the left of particle in red, with the charge of Q one and on the right of particle of in red with the charge of Q two, and each of the red particles is the distance X from the blue particle. So what is the total net force on our blue particle with charge? Capital Que? While superposition tells us that is just the vector sum of the individual forces from Q one and Q two, and so we can calculate those first. And so the red particle on the left well apart. Some force F one of K capital Q Q one over X squared in the X direction. Now we'll know that Q two is on the opposite side, so it's forceful in part of this way. And so f two, we'll give us, um, negative value of K Q Que two over X squared in the extraction. And so, using superposition for the Net force on our blue particle, we know this is simply the some vector some of F one and F two, which will give us K U Q one or X squared minus K Q Q two over X squared, all in the X direction, which we could simplify a little bit more by factoring out the common terms and arriving at this final answer. So the first thing we'll note is that the direction positive or negative off this force will be determined by the relative values and signs of Q one and Q two here in the center term. Also, if Q one and Q two have the same sign and magnitude, this center term goes to zero and our net force on cue. Our blue particle is zero. Next, we look at an example of superposition in two dimensions. Here we have to read particles, each with charge. Lower case Q. On our X axis on our Y axis above the X axis, we have a blue particle charge. Capital que. The red particles are each a distance are from the blue particle, and the separation doctor makes an angle fada for each other. Red particles with the horizontal. So what is our net force on our blue particle? Well, to use superposition first, we calculate the individual forces from each red particle. The particle on the left will give us some force in this direction That will call F one, and its magnitude is simply K capital. Q. Lower case Q. Over our square particle on the right will in part of force and part of force that we call f two and this is gonna have the exact same magnitude. Of course, the difference will be the direction of these two. For this will want to break each of these force vectors into their components in the X and Y direction and out of those components up. So F one we know looks like this with an angle theater to the horizontal and so we can calculate using that angle the X and Y components for F one s o f one oops, F one of X. It's simply equal to F one times CoSine data in the X direction. Similarly, f one of y It's simply the magnitude of F one time sign of data in the Y Direction F two. Well, look like this with an angle theater. And so when we break it into components, it will look like that. So f two of X. We'll notice points in the negative X direction, and so we'll give it a negative sign. It is a magnitude of F to cosign data in the X direction. If one of why still too, of why still points up. And so it has the exact same form as F one of why we'll note that F one and F two are actually just the magnitudes which are identical. So are net force on our blue particle is simply the vector sum of these components. So for our X component, we have F one cosine theta minus F two cosine theta, and our white component is F one sign data plus F to sign data right off the bat. Since F one and F two are equal in magnitude, will notice this component becomes zero. And so this reduces to once we've substitute in for F one and F two zero in the X Direction and K Capital Q. Lower case Q over R squared times to since there two of them adding up in the Y direction or alternatively written to K que que over r squared. Why

University of California, Berkeley

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