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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)/ 24,2) 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 = (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 obtain E - (x0/ Evaluating the integral will lead us to Qxo 1 1 E= 4 TE,R? Xo (x3 + R²)=/2 For the case where in R is extremely bigger than x0. Without other substitutions, the equation above will reduce to E = Q/(

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)/ 24,2) 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 = (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 obtain E - (x0/ Evaluating the integral will lead us to Qxo 1 1 E= 4 TE,R? Xo (x3 + R²)=/2 For the case where in R is extremely bigger than x0. Without other substitutions, the equation above will reduce to E = Q/(

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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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 x0 distance from its center? (Consider that the surface of the plate lies in the yz plane)

Use the template in the attached pictures to solve the problem. 

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/ )(x0q/ 24,2) 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 = (1/ )(x0 %3D 24 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/ 2+ Evaluating the integral will lead us to Qxo 1 1 E= 4 nEGR? Xo (x3 + R²)/2 For the case where in R is extremely bigger than x0. Without other substitutions, the equation above will reduce to E = Q/(
Transcribed Image Text: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/ )(x0q/ 24,2) 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 = (1/ )(x0 %3D 24 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/ 2+ Evaluating the integral will lead us to Qxo 1 1 E= 4 nEGR? Xo (x3 + R²)/2 For the case where in R is extremely bigger than x0. Without other substitutions, the equation above will reduce to E = Q/(
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)/ 24,2) 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 = (1/ 24 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/ 2+ Evaluating the integral will lead us to QXo ( 1
Transcribed Image Text: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)/ 24,2) 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 = (1/ 24 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/ 2+ Evaluating the integral will lead us to QXo ( 1
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