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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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
(xOq)/
2+r2)
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
2+
= (1/
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
E equals fraction numerator Q x subscript 0 over denominator 4 pi epsilon subscript 0 R squared end fraction open parentheses 1 over x subscript 0 minus 1 over open parentheses x
subscript 0 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 x0. Without other substitutions, the equation above will reduce to
E = Q/(
Depsilon subscript o'
Transcribed Image Text: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 (xOq)/ 2+r2) 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 2+ = (1/ 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 E equals fraction numerator Q x subscript 0 over denominator 4 pi epsilon subscript 0 R squared end fraction open parentheses 1 over x subscript 0 minus 1 over open parentheses x subscript 0 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 x0. Without other substitutions, the equation above will reduce to E = Q/( Depsilon subscript o'
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