Electronic field calculation of 2 wires

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SUMMARY

The discussion focuses on calculating the electric field produced by two nonconducting wires, each carrying equal and opposite charges of ±2.5 × 10-6 C, positioned at a right angle. The electric field at point P, located 0.06 m from each wire, is derived using the equation |E| = |kQ/r2|. A key point raised is the incorrect approach of integrating the expression (kQ/2r)dr over symmetric limits, which fails to account for the vector nature of the electric field.

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ThiagoG
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Homework Statement


2 1.2m nonconducting wires meet at a right angle. One segment carries +2.5*10-6 C of charge distributed uniformly along its length, and the other carries -2.5*10-6 C distributed uniformly along it. Find the magnitude and direction of the electric field these wires produce at point P, which is .06 m from each wire.

Homework Equations



|E|=|kQ/r2|

The Attempt at a Solution



I have the solutions manual so I don't care for knowing how to solve it. I just want to know as to why I can't take the integral of (kQ/2r)dr with limits from -.08485(the distance from P to the tip of the wire) to +.08485.
 
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ThiagoG said:
I have the solutions manual so I don't care for knowing how to solve it. I just want to know as to why I can't take the integral of (kQ/2r)dr with limits from -.08485(the distance from P to the tip of the wire) to +.08485.
Because that is not how to solve it - which you don't care to know.
Hint: E is a vector.
 

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