Calculating Electric Field & Proton Acceleration

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The discussion focuses on calculating the electric field and proton acceleration due to a charged wire. The user attempts to find the electric field at a specific distance from the wire using the formula E = K*q/r² but finds their result differs from the expected answer. Suggestions include using charge density to express dq and adjusting the integral to account for the geometry of the problem. Additionally, the importance of considering the vertical components of the electric field vectors is emphasized, along with the recommendation to create a diagram for clarity. The conversation highlights the need for detailed calculations and proper setup in physics problems.
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Homework Statement


A total of 3.22 106 electrons are placed on an initially uncharged wire of length 1.29 m.

(a) What is the magnitude of the electric field a perpendicular distance of 0.396 m away from the midpoint of the wire?


(b) What is the magnitude of the acceleration of a proton placed at that point in space?


The Attempt at a Solution



A total of 2.82 * 10^6 electrons are placed on an initially uncharged wire of length 1.34 m.
(a) What is the magnitude of the electric field a perpendicular distance of 0.404 m away from the midpoint of the wire?

so i did
2.82 X 10^6 electrons X 1.6 X 10^-19 C = 4.512 * 10 ^-13 = q

then i used E = K*q / r^2

E = 2 integral( k*q / r^2, min .404, max .78 (the hypot))
and got 0.00969 but the answer is 0.0128 n/C

mmm a little offf any suggestions?


(b) What is the magnitude of the acceleration of a proton placed at that point in space?
Newtons famous law Force = mass*accel
Look up the mass of a proton
a = F/m
does this look right?
 
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Your part (a) is missing a lot of detail; I can't tell if it is right without doing all the work! In Integral k*dq/r², what did you replace dq with? Say you use λ = q/1.34 as the charge density along the wire. Then dq = λ*dx would work. But your integral is now over x, so r² has to be expressed in terms of x. Also, the horizontal components of the dE vectors cancel out, so you must put in a cosine or sine to take only the vertical component. I think a diagram is required to make sense of what is vertical and what is horizontal. Did you do all that?
 
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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