Flux for an Infinite charge line through a cylinder

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SUMMARY

The discussion focuses on calculating the total electric flux (Φ) through a cylindrical surface due to two infinite lines of charge with charge densities λ1 = 2.9 μC/cm and λ2 = -8.7 μC/cm. The challenge arises from the non-constant electric field (E) around the cylinder when both lines are present. The key equation used is ∫E*dA = qencl/ε, which allows for calculating flux by considering the enclosed charge (qencl) and the permittivity of free space (ε). The recommended approach is to utilize the right-hand side of the equation for flux calculation, simplifying the process when E is not uniform.

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  • Understanding of electric flux and Gauss's Law
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  • Knowledge of cylindrical coordinates in electrostatics
  • Basic calculus for evaluating integrals
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mrshappy0
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Homework Statement


An infinite line of charge with charge density λ1 = 2.9 μC/cm is aligned with the y-axis as shown. a = 7.9 cm. h = 8.8 cm. λ2 = -8.7 μC/cm. Another infinite line of charge with charge density λ2 = -8.7 μC/cm parallel to the y-axis is now added at x = 3.95 cm as shown
Physics_3.jpg


What is the total flux Φ that now passes through the cylindrical surface? Enter a positive number if the net flux leaves the cylinder and a negative number if the net flux enters the cylnder. (This question is what got me confused).

Homework Equations



∫E*dA=qencl


The Attempt at a Solution



I was able to correctly find the net E along x for point P and I was also able to find the flux for the centered line for the cylinder but once the second line was added I wasn't sure where to go from there because E is no longer constant around the cyclinder. I am sure I am missing something very fundamental but I can't wrap my mind around what that is.
 
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mrshappy0 said:

Homework Equations


∫E*dA=qencl
The left-hand-side is the basic definition of flux, and provides one way of calculating it (where it is the component of E perpendicular to A that is used.

However, it turns out the flux is also equivalent to the right-hand-side of the above equation, so that provides an alternative way to calculate the same quantity.

As you said, it's difficult to calculate the integral since E is not constant everywhere around the cylinder -- also E is not everywhere perpendicular to A. Therefore, I suggest using qencl/ε to calculate the flux instead.
 

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