Problem Using the method of integration, what is the electric field of a uniformly charged thin circular plate (with radius Rand total charge Q) at xo distance from its center? (Consider that the surface of the plate lies in the yz plane) Solution A perfect approach to this is to first obtain the E-field produced by an infinitesimal charge component of the charge Q. There will be several approaches to do this, but the most familiar to us is to obtain a very small shape that could easily represent our circular plane. That shape would be a ring. So for a ring whose charge is q, we recall that the electric field it produces at distance x0 is given by E- (1/ 2,2) Since, the actual ring (whose charge is dq) we will be dealing with is an infinitesimal part of the circular plane, then, its infinitesimal electric field contribution is expressed as 2. = (1/ Xx0 We wish to obtain the complete electric field contribution from the above equation, so we integrate it from 0 to R to obtain .R E- (x0/ Evaluating the integral will lead us to Qxo E=- 4 TEGR? Xo (x3 + R?)=/2 1 1 For the case where in R is extremely bigger than x0. Without other substitutions, the equation above will reduce to E= Q/ Eg)

Problem Using the method of integration, what is the electric field of a uniformly charged thin circular plate (with radius Rand total charge Q) at xo distance from its center? (Consider that the surface of the plate lies in the yz plane) Solution A perfect approach to this is to first obtain the E-field produced by an infinitesimal charge component of the charge Q. There will be several approaches to do this, but the most familiar to us is to obtain a very small shape that could easily represent our circular plane. That shape would be a ring. So for a ring whose charge is q, we recall that the electric field it produces at distance x0 is given by E- (1/ 2,2) Since, the actual ring (whose charge is dq) we will be dealing with is an infinitesimal part of the circular plane, then, its infinitesimal electric field contribution is expressed as 2. = (1/ Xx0 We wish to obtain the complete electric field contribution from the above equation, so we integrate it from 0 to R to obtain .R E- (x0/ Evaluating the integral will lead us to Qxo E=- 4 TEGR? Xo (x3 + R?)=/2 1 1 For the case where in R is extremely bigger than x0. Without other substitutions, the equation above will reduce to E= Q/ Eg)

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

We have calculated the electric field due to a uniformly charged disk of radius R, along its axis. Note that the final result does not contain the integration variable r: R. Q/A 2€0 Edisk (x² +R*)* Edisk perpendicular to the center of the disk Uniform Q over area A (A=RR²) Show that at a perpendicular distance R from the center of a uniformly negatively charged disk of CA and is directed toward the disk: Q/A radius R, the electric field is 0.3- 2€0 4.4.1b
As shown in the figure, inner radius R1 outer radius R2homogeneously distributed over the metal sphere shell, total positiveelectric charge is given as + 4?. If in the center of the sphere shellThere is a point electrical charge with a total amount of 2?.Answer the following questions using the Gaussian flux law. a) Find the electric field strength in the region r <1. b) Find the electric field strength in the region 1 <r <?2. ) c) Find the electric field strength in the region r> ?2. d) Express the electric field strengths you find as vectors.
Assume a uniformly charged ring of radius R and charge Q produces an electric field E at a point Pon its axis, at distance x away from the center of the ring as in Figure a. Now the same charge Q is spread uniformly over the circular area the ring encloses, forming a flat disk of charge with the same radius as in Figure a. How does the field Eick produced by the disk at P compare with the field produced by the ring at the same point? O O Ek Ering O impossible to determine
A thin rod has uniform charge per length z. The distance between point A and point B is 5W and the distance betweer point A and point C is 8W and the distance between point Cand point P is 2W. We introduce an integration variable h with h = 0 chosen to be at point B and the th direction to the left. The small r segment has length dh and charge dq. We want to find the electric field at point P. Draw it out--label the all the lengths and the integration variable! A B C P
Fill in the blanks.
Good morning could you help me to solve the following problem?Thanks in advanceA ring of radius a carries a uniformly distributed positive total charge. uniformly distributed. Calculate the electric field due to the ring at a point P which is at a distance x from its center, along the central axis perpendicular to the plane of the ring. Use fig. a The fig.b shows the electric field contributionsof two segments on opposite sides of the ring.
Please don't provide handwritten solution .....
Needs Complete solution with 100 % accuracy.
FILL IN THE BLANKS
Problem An electric charge Q is distributed uniformly along a thick and enormously long conducting wire with radius R and length L. Using Gauss's law, what is the electric field at distance r perpendicular to the wire? (Consider the cases inside and outside the wire) Solution To find the electric field inside at r distance from the wire we will use the Gauss's law which is expressed as PĒ - dÀ= We will choose a symmetric Gaussian surface, which is the surface a cylinder excluding its ends, then evaluate the dot product to obtain A = (Equation 1) Case 1: Inside the wire Since, r falls inside the wire, then all the enclosed charge must be: denc = On the other hand, the Gaussian surface inside the wire is given by A = Using Equation 1, the electric field in simplified form is E = Case 2: Outside the wire Since, r falls outside the wire, then, all the charge must be enclosed, thus denc = On the other hand, the Gaussian surface outside the wire is given by A = Using Equation 1, the electric…
Ggg
Problem 4: who is right? Two parallel, infinite planes of charge have charge densities +20 and -o as shown in Fig. 1 They are at a distance L, from each other. Three students are asked to determine the electric field at points A and B, at a distance L above the top plane, and at a distance L below the bottom plane, respectively, as shown in the figure. Each student comes up with a different answer and a different explanation. Read carefully their statements below. The students are reminded that the electric field magnitude of an infinite plane of charge with charge density a is E = . +26 3+26 Figure 1: Two cquivalent views of the two infinite parallel planes • Student 1 draws the sketch in Fig. 2 [top] and says: "I think from the outside we can think of the two planes as a single infinite plane with charge density +20 - o =0, so then we know the electric field must be a/(2) pointing upwards at point A and also o/(2c) pointing doumwards at B". • Student 2 draws the sketch in Fig. 2…