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April 17, 2015

April 17, 2015

Posted by **Ashik** on Thursday, August 11, 2011 at 10:58am.

- Physics -
**Damon**, Thursday, August 11, 2011 at 6:32pmdQ is portion of Q along a segment ds of the ring

E = (1/4pieo) dQ/(x^2+a^2)

the component of this in the axial direction is

dEx = dE cos angle to x axis = dEx/sqrt(x^2+a^2)

(component in direction perpendicular to axis is adds to zero by symmetry)

so

dEx = (1/4pieo) x dQ/(x^2+a^2)^1.5

at a given constant x our integral is only over d Q and is Q

so

Ex = (1/4pieo)Q x/(x^2+a^2)^1.5

if x is >> a

then

Ex = (1/4pieo)Q x/(x^2 + 0)^1.5

=(1/4pieo)Q x/x^3

= (1/4pieo)Q/x^2 which is point charge

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