How Does Distance Affect Electric Flux Through a Square?

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SUMMARY

The discussion focuses on calculating the electric flux through a square when a point charge of -3.2μC is positioned at a distance of d/2 above its center. The side length of the square is specified as d = 0.34 cm. To determine the electric flux, participants emphasize the importance of the electric field, defined by the equation E = kq/(r^2), where r represents the distance from the charge to the square. The relationship between distance and electric flux is crucial, as it directly influences the electric field strength impacting the square.

PREREQUISITES
  • Understanding of electric flux and its relation to electric fields.
  • Familiarity with Coulomb's Law and the equation E = kq/(r^2).
  • Basic knowledge of geometry, specifically regarding squares and cubes.
  • Concept of dot products in vector calculus, particularly in the context of E dot dA.
NEXT STEPS
  • Study the concept of electric flux in detail, focusing on Gauss's Law.
  • Learn how to apply Coulomb's Law to different geometrical configurations.
  • Explore the implications of distance on electric field strength and flux calculations.
  • Investigate the use of vector calculus in physics, particularly in electromagnetism.
USEFUL FOR

Students in physics, particularly those studying electromagnetism, educators teaching electric fields and flux, and anyone interested in the mathematical relationships between charge, distance, and electric field strength.

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Homework Statement


In the figure, a point charge -3.2μ C is a distance d/2 directly above the center of a square of side d = 0.34 cm . What is the magnitude of the electric flux through the square? (Hint: Think of the square as on face of a cube with edge d.)
(in N*m^2/C)


Homework Equations





The Attempt at a Solution


Ok so I am not asking specifically for the solution, but when the charge is located a specific distance from a finite plane such as this, how does the distance from the square(d/2) play into the solution?
 
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Well since Flux = E dot dA, you need to find the Electric field going through the side of the square. So the distance from the square is r in the equation for an electric field due to a point charge ( E = kq/(r^2) ).
 

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