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