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A slab of insulating material has a nonuniform positive charge density $ho=C x^{2}$, where $x$ is measured from the center of the slab as shown in Figure $mathrm{P}...

Question

A slab of insulating material has a nonuniform positive charge density $ho=C x^{2}$, where $x$ is measured from the center of the slab as shown in Figure $mathrm{P} 24.59$ and $C$ is a constant. The slab is infinite in the $y$ and $z$ directions. Derive expressions for the electric field in (a) the exterior regions and (b) the interior region of the slab $(-d / 2<$ $x<d / 2)$.

A slab of insulating material has a nonuniform positive charge density $ ho=C x^{2}$, where $x$ is measured from the center of the slab as shown in Figure $mathrm{P} 24.59$ and $C$ is a constant. The slab is infinite in the $y$ and $z$ directions. Derive expressions for the electric field in (a) the exterior regions and (b) the interior region of the slab $(-d / 2<$ $x<d / 2)$.



Answers

A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes $x = d$ and $x = -d$. The $y$- and $z$-dimensions of the slab are very large compared to $d$; treat them as essentially infinite. The slab has a uniform positive charge density $\rho$. (a) Explain why the electric field due to the slab is zero at the center of the slab ($x =$ 0). (b) Using Gauss's law, find the electric field due to the slab (magnitude and direction) at all points in space.

But if the problem we have following situation Well, here's the X axis and you're presentable. Ese Z axis. We have the charge slip. Here it is lying here. Char slave. Um, here's the origin. Oh, they're density off a charge. Function off X is given by roll nod Times X divided by D Hold square charge distribution is symmetric about X is equal to zero so e x is equal to minus key off X at X is equal to zero The field cannot be in a boat of osteo X axis in the 92 ex direction. So therefore, it must be zero. We can then right e off ex off X is equal to equal d x minus X and it X is equal to zero e off ex off zero is equal to minus T X over zero That implies e off X zero is equal to zero but to be off the problem But to be here Here we have a slab and, uh, the electric field lines can be drawn here. In this way, these are the electric field lines and, uh, the ex picks next more is greater than deed with the outside outside this live slept So it's somewhere here in here. The flocks electric flux is written is to 80 times a, um for the idea Here in the 80 a year we are asked to find a cure in closed, we can find it. The charge within the disk is then given by Dechaume Physical to charge them street terms of warning DVD. We can write them. This is a call to but all eight times a d x. At times he exits the volume. This is a call to Ronald Hey, divided by d square times ah d x Then the charging closed charging closed is given by two times off integral zero to D Oh, for D Q That is a to roll nod a divided by the square integral from zero to D this x square d x Oh, this integral equal rights then off into the cube or three. So the result is a two year old lord a d square d tree over three. So this d and the three will cancel out the square and we're left with the only Then the charging closed will be equal Thio to over three. We're all knowed eight tons of D From here, we know to e a physical to to roll nod Oh, a D or three times Sloan Onda We confined electric field with them. Uh, a little nord dans de So here's a little note Times a D divided by three weeks along X divided by X more That's an eye direction. Therefore, e electric field is directed away from away from ex physical zero What? Sea off the problem x more greater than the inside this layer. So this is less than the inside the slab So x more is here. You started to slip we can find in the flux. Electric flux will be cool too. Off e times, eh From here. Well, first, finding close charge From there I will find a electric field so enclosed charge is then given by two times off a tingle 02 x d Q to a lord a divided by the square from zero to x x squared the x and this integrity words into to your lord a divided by the square X tree divided by three. Then we use the relation That is it too e a Is it caused to tour or not extreme. Derided by three nor D Square Um, where flux is equal to the charge enclosed so flocks charging clothes divided by so long. We use this definition and right electric fields or electric field will be called to roll. Lord X three Divided by three. It's a long D Square. This is directed away from X is equal to zero, so our electric field use in ight direction.

Okay, so we're in chapter 22 problems. 43. So a very large assume Infinite flat slab of non conducting. Near material has a thickness D and a uniform volume charged Density Positive Road E. We want to show that a uniform electric field exists outside of the slab and determine its magnitude and direction. So first we can choose a a cylindrical galaxy in surface with the flat ends parallel to an equal distance from the slab. By symmetry, the fuel must point per particularly away from the slab, resulting in no flux through the curved part of the galaxy and cylinder. And by symmetry, the flux through each end of the cylinder must be equal with the electric field constant across the surface. So what that is saying is the electric field darted into the surface in a girl of our galaxy and surface is given by two times magnitude times the area. Awesome. So now we have to have this set this equal to the charge enclosed over. I'm not ask out his losses. So what is the charge enclosed. So this is given as our surface area, our surface charged density. The volume doesn't volume charged density row e times How much volume we're closing? Well, the volume we're enclosing is yeah or e a the area times the thickness of the smiled, which is deep. So if we plug that into our equation appear we consult for what eat is and see that this comes out to be a row e do you over to Absalon? Not in the areas cancel out. So it doesn't really matter how we draw our galaxy and surface, so that's very cool. And also, the electric field is independent from the distance from the slab. Just also interested. Okay, Part B. Now this is the more challenging part. Says has shown the figure This lab is now aligned so that one of its surface is lie on the line y equals X. That's what show me here, line y equals X. In that time, T equals zero. A point like particle of mass M in charge positive was located at Why not point? Why not here? Cool. And as a philosophy of being the heart in a positive ex directory, we want to show that the particle will collide with the slab. If the knot is greater than the square root of route too. Q? Why not roti over or d over up slow? Not. And that's all in the scammer square, Rick. If he not greater than that, then it will collide and we're ignoring gravity. Okay, so this is gonna be tricky because we have lots of coordinate system stuff to think about. So when we when the coordinate system of the problem is changed two axes parallel and perpendicular to the slab, it can easily be seen in the particle or hit the slab if the initial particular velocity is sufficient for the particle to reach this lab before. Okay, so what we're gonna do is we're gonna rotate our axes. So now this is our Y axis are new. I access, which I'll call y prime. And this is the new X Axis X prize. So these get rotated an angle of 45 degrees. Okay, so in the new coordinate system, we can do some transitions here. So initially, all right and black for the order coordinate system we have are not factor is why not? J and v not doctor is vey not Oops, you know I So now let's right out our Becker's in the new cornet system we have our are not in the new court system to be. Why not co sign of 45 degrees or huh? Plus, why not sign of 45 degrees Z, huh? Okay. So we can also transform the velocity vector to this r Z basis with our being the perpendicular basis and Z being the parallel basis to our slab. So RV not thus becomes negative. Veena sign 45 degrees or huh, plus the not co side 45 degrees see her. And lastly, our acceleration vector since in this basis is very easy because we know it's all radio and the force is given by Q E and then to revert to make her acceleration. We defined by mass. And this is his radio. So in the art direction. So now if we want to find just the perpendicular components that is given as the knot or and what we can do here is insert this into equation to find it says equation to 12 sees this is old mechanic stuff. So we have zero equaling r v r squared What it is to a over our minus are not this equals Veena squared over to minus to Q over in wrote E d over to absolutely not just plugging in what her e is for acceleration here and now we just plug in the differences in our vectors and this is why not over group to minus zero. Cool. So just as a set again. So this is the initial velocity in the art direction squared minus this and this equals the final velocity. But we're sending the final velocity to zero because we want to see if it will even reach the slap. So to see if the velocity will reach the slab, we then solve this equation for the knot. So let's rewrite the equation down. Professor zero equaling Dean up square over to harnessed to Q over in. And now we write what he is. Is Rodi d over to? Absolutely not. Then we have our our vector. Our vector is why not over experience to minus zero. So now we can solve this for the not squared in this over two equals two. Q over. M wrote e d over to its long Why not over route cool. So now Veena squared equals Cancel these guys out. This equals to over group, too. Q. Over end Road E D. Why not over absolutely cook. And that's the same as read, too. Q row e d line Hot over. Absolutely not in all squared equals, and that's exactly what we wanted to show. So this is the exact the initial velocity it would take to stop exactly on the slab. So if the velocity is anything greater or equal groups, if the velocity and is greater than equal this value, it will contact the slab.

So we have a very large love. Let's say this is our slob with the thickness D because the length and the birth of the slabs are very large. You can think of this as an influence lab on Because of symmetry, the field must be perpendicular to the plane off the slap. Now let's say our role is positive, which means our electric field will be away from this lab. So now, because this is our origin that's going on, we want to find it. Feel a distance X considera cylinder For this, which looks something like this. This distance is ex. Obviously on this distance is Steve. I do no. We want to find the flux through this Carson cylinder. Now the flux through the seven circular surface will be zero because they feel is simply perpendicular to the surface. They'll be. But they wouldn't be flux from the two. So close their faces on either side. So our flocks will be. There's some off the two flexes so they will be electric will be here. But because off symmetry, the l A. People on the other side will also be equal that say it is e so That's e times theeighty off the surface. That's it. The radius is some art. So in 24 by our square, I'm sorry into pi r squared or the upper surface on. Even for the lower surface, it's into four i r Square. Notice that the electric field is up upwards in at the top on Downward City pardon, which means the flux in both the cases will be positive. So that's Ian Do by a square. Now this must be equal. Do charging closely. Very big exit or not. No charging clothes to simply wrote times the volume. The volume of the slender is I. R Square finds two. X because the length of the cylinder is to X and the idea of this slanderous pi r squared, we know that the volume of a cylinder is area times the length of this murder. You had a very excellent not this most people do to buy our square. Hence, from this, you can see that he's equal. Do we'll cancel off a few similar numbers? You and I are square, giving us easy will do through X, but it still or not as long as X is less than be delighted. Notice that if X is negative, your e will be negative, which is in in line with our expectation. When X is Nick, do we expect you to be down or sister off upwards? Now that we have this, what about distances which are beyond? Do I do it is what if this is Alex then executed that day? We do. So we do the same thing again. We know that the flux, it's simply two times electric bill times The dest area which is pipe are square. This is deeply divided. Now this is equal to charge in close. To be ready by Solano of charging closed will simply be the charge within this volume. So even though the cylinder extends beyond this lab, only the cylinder with the art of cylinder that is within this lab will have any charge. So we only considered the charge here, So that will be road gangs. I r squared times be because the part of cylinder that is inside the slab will have length off me. Bye. Incident. That's it. I asked what he excuse us. Easy. Koldo canceled pi r squared on Get Brody bye to cancel or not. This is the electric field for any point beyond this love. It is for Marx, greater than divided. That is it.

Okay, so we don't Chapter 22 problems 47 here. And in this problem, we have a flat slab of non conducting material with thickness to d go back to mark You just small compared to its height and breath. Define the X axis to be along the direction of the slabs thickness with the origin at the center of the slab. That's what shown is here. And if the slab carries a volume charged density of P e x B negative, he not for for negative t x less and zero and positive p not for zero x Austin deep. Okay, so we now want to determine the electric field as a function of X in the regions outside the slab and and in the left partner in the right part. Okay, So, first of all, the electric field from a flat, infinite slab is given by Road D over to Absalom. So this means we can take the total as D from the left plus e from the right. But this now becomes d over to Absalom. Not over row left, close road, right? And that's just the over to absolutely not. Our little row on the left from the left slab is negative. Peanut, peanut, and that's just zero. So outside of this lab, no matter where we are, there's gonna be no charge because the electric fuel from an infinite slab does not depend on D. It does not depend on the distance away from this lab has been on this deal of thickness but does not depend on the distance away from the slough. So now for part B, we want to look at the electric field for ex Lister equal a d. So in the right part of this lab. So to find this, we know we're gonna need to use gas Is law here where we have e dot d A. So if we draw our galaxy and surface here to be a cylinder with cross sectional area A, then we see that the electric field is going to be perpendicular to the surface at all points and only in parallel to the surface on the curve parts of the cylinder. So if it's always perpendicular, we can. He composed this integral to just being e dot ta or e Times Day. And this equals, he charged, enclosed over absolute up. So Now we re righted the electric field in the right slab is written as Q enclosed and you write slab over Absalom, not a So now we've solved this problem. Except we got to find what Q and close This So you enclosed is given as the sir the volume charged density. So now we need to multiply this by the the cross sectional area of our galaxy and surface, which is a and how thicknesses. And that's gonna be given by D minus x. Okay, So if this is our cue in close as a function of X and we know that our e r as a function of X for the right side you, then this is given as row, not D I miss X over absolute nut. Awesome. That's it. So let's move on to part. See here on Adam Page part See, we're gonna be doing the same thing, but the left side now we're gonna be doing the other region where X is less than zero and greater than negative. D. So now we have the same equation R e a equals charge and closed over. Absolutely not. But now our card enclosed is negative peanut D plus X Yeah, terms in the area across Xperia, this is all a perhaps learn knots are A's cancel out and we're left with an e of negative peanut D plus X all over. Absolutely awesome. Um, also, we know that this is in the left area, so this is pointing to the left so we can rewrite this as a vector as negative peanut e plus x all over slow not. And we know this is in the negative. I direction also so small backed apart because we didn't actually do the direction. They're so in part B. This was what our magnitude is. A function of X is. So let's write it as a vector. And this is now a positive peanut d minus X over. Absolutely not. And we know that the field is proving to be left, so this has got to be a negative. I awesome


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