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Problem 16.47 (Multistep)As shown in the diagrm_ three large_ thin_ uniformly charged plates are arranged s0 that there are two adjacent regions of uniform electric...

Question

Problem 16.47 (Multistep)As shown in the diagrm_ three large_ thin_ uniformly charged plates are arranged s0 that there are two adjacent regions of uniform electric field. The origin is at the center of the central plate: Location A is -0.4, 0, >m, and location B is 0.7, 0, 0>m. In region the electric field is E 1 = 725, 0, 0 Vlm In region the ectric field is Ez -450_ 0, 0 VIm_PartYour answer is correct:(a) Consider path fromAlong this path; what is the change in electric potential?AVthe t

Problem 16.47 (Multistep) As shown in the diagrm_ three large_ thin_ uniformly charged plates are arranged s0 that there are two adjacent regions of uniform electric field. The origin is at the center of the central plate: Location A is -0.4, 0, >m, and location B is 0.7, 0, 0>m. In region the electric field is E 1 = 725, 0, 0 Vlm In region the ectric field is Ez -450_ 0, 0 VIm_ Part Your answer is correct: (a) Consider path from Along this path; what is the change in electric potential? AV the tolerance is +/-2%/



Answers

SSM WWW A square metal plate of edge length 8.0 $\mathrm{cm}$ and
negligible thickness has a total charge of $6.0 \times 10^{-6} \mathrm{C}$ (a) Estimate
the magnitude $E$ of the electric field just off the center of the plate (at,
say, a distance of 0.50 $\mathrm{mm}$ from the center) by assuming that the
charge is spread uniformly over the two faces of the plate. (b)
Estimate $E$ at a distance of 30 $\mathrm{m}($ large relative to the plate size by assuming that the plate is a charged particle.

Hello, Prince. Here it is given Ah potential different stop. All people need established between the large letters stolen. So they say this is a plate, but the system relates to on very your potential different off for eight people between that, but really is that for people and other plate. Is that durable? The plates are separated apart. Why are the streets off 1.7? You know, sentiment discuss that you put into itself. Is that corresponding toe a sketch? You put interfaces responding toe do you don't? 1 20 for the 3. 60 and for world book for eight people in your escape. So that electric light that may So the airport so electric three lengths. This is that it keep its their face off. Do you know potential? This is the third place up. I want people. This is the surface off. She said she wouldn't. And this is a separate off for would be good electric free life is always perpendicular. Lee intercepted on There will be uniformed electric field between the plates. So these are the direction off electrically

In this problem on the topic of gases law, we're told that a square metal plate has a gland eight centimeters and negligible thickness and a total charge of six times 10 to the minus six school loans. We want to find the magnitude of the electric field just off center of the plate by assuming that the charges uniformly spread over the two surfaces of the plate. We then want to estimate the electric field at a distance of 30 m, which is large relative to the size of the plate. Now, using gas is low. We find the magnitude of the field E to be sigma over. Absolutely not recognize the area charge density for the surface just under the point, the charges distributed uniformly over both sides of the original plate, with half being on the side near the field point. And so sigma is Q. Over to A. So the surface charge density sigma is the total charge Q over two times the area of each phase, which is a charge of six times 10 to the minus six columns, divided by two times 0.0 eight m squared. So we get the surface charge density sigma to be four point 69 times 10 to the minus four columns per square meter. And so the magnitude of the field E. Is sigma over absolutely not. Which is 4.69 times 10 to the minus four columns per square meter, divided by eight 0.85 times 10 to the minus 12. Mhm. Cool. Um squared for newton meter squared. And so calculating you get the electric field to be 5.3 times 10 to the power seven newtons per column in magnitude. And the field is normal to the plate. And since the charge on the plate is positive, it points away from the plate. Yeah. Now, for part B at a point far away from the plate, the electric field is that of a point particle with the charge equal to the total charge on the plate and the magnitude of the field E. Okay, is equal to Q over four pi. Absolutely not. R squared. We are as the distance from the plate. And so this is one of the four pi absolute not, which is 8.99 times 10 to the nine newton meter squared per column squared times the total charge, which is six times 10 to the minus six columns, divided by a distance of 30 m squared, which gives the electric field strength at a large distance to be 60 newtons McCullum.

Hi. In the given problem, these are the two large copper plates. This is first copper plate than in parallel. With this. There is another copper plate which are having the charge is spread over them as positive and negative. 4.0. Cool. Um, 4 m squared. This is not a charge. The sister These are the charge densities. Sigma, the surface charge Density is over. Both off. These copper plates are given us having the same magnitude but opposite nature. The first plate is having a positive charge and the second plate is having a negative charge. So the electric field between the players will be like this away from the positive charge and two words the negative charge. So the electric field between these plates will be given by cause the serum using the concept, of course. The serum it will be given by Sigma by Absolutely not. So here it becomes 4.0 Coolum for meter square. The magnet, you know, Charge density divided by permitted Vitti having a value off 8.854 into 10 days. Par minus 12 Coolum Square, Part Newton Meter square, which comes out Toby, 4.52 in tow. 10 days for 11 Newton per column. Now, in the first part of the problem, we have to find the electric flux passing through a rectangular area, which is kept just between these plates. And in the first case, this is skeptic Carol toe the place. So the electric plugs passing through this rectangular area, probably given by e dot e or we can see this is E a cost. Tita. We're theater is the angle between electric field vector and area vector. And, as we know, area Vector is given by an arrow drawn perpendicular to the area. So here we can consider this to be a bar, and this is a bar already so we can say in the first case angle between electric field and area. Vector is zero degree, so plugging in unknown values here, electric flags comes out Toby 4.52 in tow. 10 Dish, Part 11 Newton for column multiplied by area off this rectangular area, the angular plane whose dimensions are given as three centimeter by four centimeters so we can write it like three centimeter or 3 500 m multiplied by 4, 500 meter so Finally, this answer here comes out Toby. Five point for two into 10 days, the Part eight Newton meters Squire for Coolum, and it becomes the answer for the first part off the problem. Now, in the second part of the problem, they are saying, if the area this rectangular area is stilted at an angle of 30 degree from the parallel direction means in this case, if we re draw the diagram, these are the two copper plates kept parallel to each other. This is the direction off electric field. This was the original orientation off the rectangular area parallel to the plate, and now it is said that it is still did through an angle off 30 degree from its original position. So it is clear that the area vector will also get tilted by the same angle. So this time the angle between electric field, an area vector, will be 30 degree. So you can see in this case in second part of the problem, the flux will be given by E cost Tita, or we can say this is 4.52 in tow. 10 dish for 11 Newton per column for area. This is three by 100 into four by 100. Meet early square into cause sign off 30 degrees. So finally this answer comes out, Toby. 4.7 into 10 days to the part eight Newton meters squared for Coolum. And here it becomes the answer for the second part. Off top problem. Thank you.

Hi. In the given problems, there are two very large metallic charged plates, which are kept parallel to each other. Having a separation off be is equal to 2.0 centimeter between them and, ah, potential difference off we is equal to 12 10 volt, such that the first plate is at a potential off. 12. Old in the second plate is at a potential off zero world, so we can say electric field between the plates from first play to second plate. It's given asked the potential creating int the magnitude of this electric field. Hence it will come out Toby, 12 old divided by 210 centimeter and hence we can say electric field between the plates is six world per centimeter. No. In the first part of the problem, we have to draw few equal potential surfaces at the given distances. And for the first distance, the potential difference has been given, as we want is equal to zero world. So if we put this we even as zero world first of all, we will have to find the position where this potential difference will take place, for which we will use an expression for electric field as we by d. So using that the expression for position from South Toby we buy it Or we can say this is we There we is the potential difference by E which is six world per centimeter. And here in this given part off this problem we won has been given a zero world. So be one we'll come out Toby zero by six Bolt per centimeter means zero. So clearly this point is at the position off first played only hands We can say the first equal potential surface is actor location off first late on Lee here which is shown in red No For the second part off this problem we to has been given us four world so toe find our location off this potential to friends again using the formula be visible to be by a here this d two will come out Toby four world divided by six volt per centimeter This Walt get canceled and hence it will remain as two by three centimeter or 0.67 centimeter means the second week we potential surface having a potential difference off for world from the first plate is at a distance off 0.67 centimeter. Hence we can draw it. Yeah, So this second red line will be representing the second Agree potential surface. Now for the third evening potential surface. The potential difference has been given us eight world, so its position will come out. Toby eight world divided by six world per centimeter again and hands. It will come out to be four by three centimeters or we can say 1.33 centimeter from the position off the first place. So this will be the third equal potential surface. Yeah, so this distance waas 0.67 centimeter. This distance is 1.33 centimeter. Now Finally, for the third equal potential surface, the potential difference has been given us overall 12 old means the gap will be equal toe. A total gap between the plates means 2.0 centimeter hands. The fourth equal potential surface is after location off the second metallic plate only like this. So the second metallic plate itself will be behaving like the equi potential surface. Now we have to draw in the second part of the problem. We have to draw a few equally few electric field lines. So we know a concept to draw the electrical lines, and that concepts is electric. Field lines are always perpendicular to equally potential surfaces. So the electric three lines, which we even sure here are these This is First Electric field line, this ISS second electively line and the third and the fourth, and many more electrical lines we can show on all of these electric field lines should be perpendicular toe every I agree potential surface. We should keep this concept in mind, and even we can prove it also for which to justify this concept. First of all, as we know, Electric Field is given us the negative off potential ingredient. So if we consider and equally potential surface potential is constant over it, potentially same at each and every point over the equal potential surface means we is constant soul. It's Grady int, or we can say its differentiation with respect to position comes out Toby zero, which means there is no electric field and long and equally potential surface. So to show that electrical lines are always perpendicular to equal potential surface, we use ah contradictory method. Suppose this is the equi potential surface and electrical line is not perpendicular to, but making an angle Tita with the perpendicular drawn toe that equip potential surface So this electrical line will be having two components. E cost Tita and e scientific equals. Sit up perpendicular to the surface and e sign theta along the surface. While this is and equally potential surface, so out of these two components is signed, Tita must be along the given equal potential surface. While they know there is no electric field along the equal potential surface hands, we can see this e sign. Tita should also be zero. So the conclusion is that electrical lines are only perpendicular to equal potential surface. They cannot be at any other angle because electric field at any other angle will give ah component off electrical along the equal potential surface, which is impossible. Thank you.


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