Electric field from a line of charge

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SUMMARY

The discussion focuses on calculating the electric field produced by a uniformly charged nonconducting rod with a positive charge of 7.81 pC and a length of 14.5 cm, at a point 6.00 cm away from the rod along its perpendicular bisector. The electric field (E) is derived using the formula E = kQ/(R^2 + L^2/4), where k is Coulomb's constant, Q is the total charge, R is the distance from the rod, and L is the length of the rod. The solution emphasizes the importance of considering only the x-components of the electric field due to symmetry, as the y-components cancel out.

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Homework Statement


Positive charge q= 7.81pC is spread uniformly along a thin nonconducting rod of length L=14.5cm. What are the magnitude and direction (relative to the direction of the x axis) of the E-field produce at pt P, at distance R = 6.00 cm from the rod along its perpendicular bisector.

Homework Equations



dE= (kdQ)/r^2


The Attempt at a Solution



dQ = charge density (a) * dL --> dE = intergral from -L/2 to L/2 of (k*a*dL)/(R^2 + L^2/4)
E= k*a/ (R^2 + L^2/4) * L ---> charge density (a)= Q/L
E = kQ/(R^2 + L^2/4)

It might be a bit hard to understand all the broken notations. Sorry. I also tried to attached a diagram, please look at that to clarify. I don't think I set up the integral correctly. Help please! Thanks
 

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Assuming the line of charge is vertical, you only need to consider the x-components of the force; the y-components will cancel out. Why? Symmetry.

Can you find an expression for the x-component of the force?

If so, just integrate that over the integration region and you'll have your answer.
 

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