Electric field on the axis of a ring-shaped charged conductor

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The discussion centers on the calculation of the electric field on the axis of a ring-shaped charged conductor. A participant questions why the horizontal component of the electric field is not multiplied by two, given that the y-components cancel out. Another contributor clarifies that the integration bounds from 0 to 2π account for the entire ring, eliminating the need for additional multiplication. If the integration were limited to 0 to π, then a multiplication by two would be necessary. The conversation emphasizes the importance of understanding integration limits in electric field calculations.
yashboi123
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Homework Statement
A ring-shaped conductor with radius a = 2.20 cm has a total positive charge Q = 0.130 nC uniformly distributed around it
Relevant Equations
E = F/q
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Hello. I was wondering why do we not multiply cos(alpha) by 2. I believe we should do this since the y-components of the electric field cancel out, meaning there would be 2 x-components of the electric field(at least I think so). Currently, this derivation/answer only considers one horizontal component, not the other half.
 
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How does it only consider one horizontal component when the integration is from ##0## to ##2 \pi##?

The integration bounds take care of this “doubling up effect”.

If you did it from ##0## to ##\pi## you would certainly have to multiply by 2 at the end.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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