What is the formula for the electric field above a plane?

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SUMMARY

The formula for the electric field above a uniformly charged wire of length L at a perpendicular distance x from its midpoint is given by E = (λ/2πε₀) * (L/x√(L² + 4x²)). The discussion highlights the integration process involved in deriving this formula, specifically using the relationship between cos(θ) and the geometry of the problem. The correct approach involves substituting variables effectively to simplify the integral, ultimately leading to the desired result. The participants confirm that using r = cos(θ)/x yields the correct outcome, while alternative methods may lead to complications.

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Homework Statement


This is problem 47(chapter 21) in the textbook - Physics for engineers and scientists (Giancoli)
Uniformly charged wire has length L, where point 0 is the mid point. Show that the field at P, perpendicular distance x from 0 is

E= (lambda/2*pi*epsilon_0) *(L/x*sqrt(L^2+4x^2)

Homework Equations

The Attempt at a Solution



I tried solving it, I got E = - (lambda/2pi*epsilon_0)*[1/sqrt(L^2+4x^2)-1/2x)
Is something wrong with my integration?

My attempt is correct until

E= lamda/r^2*4pi*epsilon_0 * integration (cos theta)dl

After this, I use cos theta = x/r (r is the hypotenuse = sqrt (L^2+4x^2))

I am trying to do this instead of taking r=x cos theta.
 
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Looks like something wrong with the "(cos theta)dl."
I'll use A instead of theta. And z in place of your l, running from -L/2 to L/2.
I get some constants times integral of cos(A)/(x^2 + z^2)*dz
Since tan(A) = z/x, I can use z = x*tan(A) to simplify the integral.
And dz = x*sec^2(A) dA
After the dust settles on this change of variable, I get integral of cos(A)dA.

Not the difference from your integral: I have dA where you have dl
I end up with the given answer.
 
If I use r=cos theta/x, then i get the correct result.
I am trying to get there without that substitution and just cos theta = x/sqrt(x^2+(l/2)^2)
huh. Doesn't work if I use the integral but does work if I take l/2 as y initially. The whole term becomes (x^2+y^2) and then apply limit to get answer.
 

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