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/ 242) 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/ (xo 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 E equals fraction numerator Q x subscript 0 over denominator 4 pi epsilon subscript OR squared end fraction open parentheses 1 over x subscript 0 minus 1 over open parentheses x subscript ó superscript 2 plus R squared close parentheses to the power of 1 divided by 2 end exponent close parentheses For the case where in R is extremely bigger than xO. Without other substitutions, the equation above will reduce to E = Q/( Repsilon subscript
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/ 242) 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/ (xo 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 E equals fraction numerator Q x subscript 0 over denominator 4 pi epsilon subscript OR squared end fraction open parentheses 1 over x subscript 0 minus 1 over open parentheses x subscript ó superscript 2 plus R squared close parentheses to the power of 1 divided by 2 end exponent close parentheses For the case where in R is extremely bigger than xO. Without other substitutions, the equation above will reduce to E = Q/( Repsilon subscript
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