Gauss's Law and Parallel Line Charges

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SUMMARY

The discussion focuses on calculating the net electric field between two infinite line charges with line charge densities λ and -λ, positioned symmetrically around a point at distance R. Using Gauss's Law, the electric field for each line charge is derived as ##\vec E##1 and ##\vec E##2, leading to a net electric field expression of ##\vec E##net = (xλ / π∈0R²)##\hat i##. The method employed is confirmed as correct, utilizing the principle of superposition to combine the effects of both line charges.

PREREQUISITES
  • Understanding of Gauss's Law and its application to line charges
  • Familiarity with vector addition in electric fields
  • Knowledge of electric field equations for infinite line charges
  • Basic geometry involving triangles and distances in the xy-plane
NEXT STEPS
  • Study the principle of superposition in electrostatics
  • Learn about electric field calculations for different charge distributions
  • Explore advanced applications of Gauss's Law in complex geometries
  • Investigate the effects of varying line charge densities on electric fields
USEFUL FOR

Students and educators in physics, particularly those studying electromagnetism and electric fields, as well as anyone looking to deepen their understanding of Gauss's Law and its applications in electrostatics.

goober
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Homework Statement


Find an equation for the net electric field at a point, above and between, two infinite line charges, one with line charge density λ and the second with line charge density -λ. The point is a distance R from both line charges, a distance y above the midpoint between charges, and the distance between charges is 2x. For reference, if the lines are viewed as parallel to the z-axis in an xy-plane, the line charges and the point would form a triangle with base 2x and height y.

Homework Equations


##\vec E## = (λ / 2π∈0r)##\hat r## (line charge field from Gauss's Law)
##\hat r## = ##\vec r## / r
R = (x2 + y2)1/2

The Attempt at a Solution


##\vec E##1 = (λ / 2π∈0R) [ (x##\hat i## + y##\hat j##) / R ]
##\vec E##2 = (-λ / 2π∈0R) [ (-x##\hat i## + y##\hat j##) / R ]
##\vec E##net = ##\vec E##1 + ##\vec E##2 = 2xλ / 2π∈0R2##\hat i## = (xλ / π∈0R2)##\hat i##

I found the electric field for each line charge individually using the equation given in the textbook for Gauss's law on an infinite line charge. I then added the two field vectors and found a final result where the y-components canceled and the x-components doubled. I just need to know if this is the correct method for solving the problem, because I was led to believe it would be more complex.
 
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Hi goober and welcome to PF.

goober said:
I just need to know if this is the correct method for solving the problem, because I was led to believe it would be more complex.
That's it. It's known as superposition.
 

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