- #1
cdummie
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- 5
Homework Statement
Very long thread, with constant longitudinal charge Q' is placed in a vacuum parallel to a very long conductive strip, whose width is a. thread is placed in the middle of the strip and it's a/2 away from it, if the surface density of charge of the strip is σ, find the longitudinal force to a thread.
Homework Equations
The Attempt at a Solution
since F=Q*E it means that F'=Q'*E so i have to find E to find F, i could think of strip as a very big number of thin lines of charge, then, knowing the value of E for every single line i could sum up all of the contributions from one to another end of the strip. Since the E of the single line is (it's dE for the whole system):
where d is distance from one charged line to a thread.
Because of the symmetry, x component of the vector E will be zero, which means there's only y component and it's
dEy=(σ*dl*cosθ)/(2πε0d)
Since dl, d, and cosθ are unknowns i need to express them with some values i know, since cosθ*dl=r*dθ it means that cosθ/r=dθ/dl i have:
dE=(σ*dl*dθ)/(2πε0dl)
dE=(σ*dθ)/(2πε0)
which means i have to integrate over angle,
i have the distance form middle of strip to a thread and it's a/2 and it's in the middle so i have the distance to both ends form middle equal a/2, which means i could find distance form first (and last) thin line using Pythagoras theorem and it's sqrt(2)*a/2, now i can easy find the angle I'm looking for since sinθ equals opposite over hypotenuse it's sqrt(2)/2 and it means that angle is π/3 so i have integral form -π/3 to π/3 for this expression
dE=(σ*dθ)/(2πε0)
when i solve it i get
E=σ/3ε0
and F'=Q'*σ/3ε0
Now, my question is. Is this good approach, because I'm not sure is this correct?