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

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

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)...

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Problem Using the method of integration, what is the electric field of a uniformly charged thin circular plate (with radius R and 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/ )(xOq)/( 2472) 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 = (1/ )(x0 24 We wish to obtain the complete electric field contribution from the above equation, so we integrate it from 0 to R to…
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find the electric field strength everywhere. please explain each step
A 2D annulus (thick ring) has an inner radius Ri and outer radius Ro, and charge Q non-uniformlydistributed over its surface. The 2D charge density varies with radius r by η(r)=Cr 4 for Ri ≤ r ≤ Roand η=0 for all other r. C is a constant. Answer the following in terms of the variables given above. Note: Gauss' Law will not be useful here.a) Find an expression for C such that the total charge of the annulus is Q. Include the SI units for C next to your answer.Do the units make sense?b) Draw a clear picture and use it to set up the integral to calculate the E-field at a point on the axis of the annulus (this axis is perpendicular to the plane of the annulus) a distance z from the center of the annulus. *** You must complete all the steps short of computing the integral (i.e. your eventual answer must be an integral with only ONE variable of integration and all other variables constant.) Show that your answer has the correct SI units for the electric field.
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