Engineering Calculating Q such that E(0,0,0)=0

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The discussion focuses on calculating the electric field at the origin due to a half ring of charge. The contributor presents an integral expression for the electric field's magnitude, which involves parameters such as the radius of the half ring and the line density of charge. They note that the radial components cancel out, leaving only the vertical component directed along the negative y-axis. The contributor seeks confirmation of their calculations and understanding of the electric field's behavior at that point. The discussion emphasizes the importance of correctly applying integral calculus in electrostatics.
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Homework Statement
Attached below.
Relevant Equations
Attached below.
Question

1636907568790.png

Relevant equation;
gif.gif

My attempt:
%5C%5C%20Solve%5C%3Bfor%5C%3BQ%20%5C%5C%5C%5CQ%3D2.gif


Could someone please confirm my answer?
 
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Electric field at the Origin contributed from half ring is of direction - y and its magnitude is expressed by integral
\frac{1}{4 \pi \epsilon_0} \int_0^\pi \frac{\rho r \sin\theta d\theta }{r^2}
where r is radius of half ring, ##\rho## is line density of its charge.
 
The component radial is cancel out, because of all direction of component radial around circle center
 

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