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Let the infinitesimal dL=Rdθ and dq=λdL=λ0Rsinθdθ, then: dE = dq/4ωωR? =(λ0/4ωωR)sinθdθ

Its x component: dEx=dEcosθ=? (λ0/4ωωR)sinθcosθdθ=？ (λ0/8ωωR)sin 2θdθ

Its y component: dEy = desinθ=(λ0/4ωR)sin? θdθ

So: Ex=∫dEx=? (λ0/8ωωR)∫sin 2θdθ

Substituting the upper limit π/2 and the lower limit 0, we can get the integral: ex = λ 0/8 π ε r.

Ey=？ (λ0/8ωωR)∫sin？ θdθ=(λ0/8ωεR)(π/4)=λ0/32εR

Then: E=Ex i +Ey j= ......................

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