How Does Side Length Influence Electric Flux Through a Cube?

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SUMMARY

The electric flux through a cube with a point charge of 6.50 nC at its center is calculated using Gauss's Law. The total electric flux through the surface of the cube is determined to be 734.5 N·m²/C. Consequently, the electric flux through one of the six faces of the cube is 122.4 N·m²/C, derived by dividing the total flux by six. The side length of the cube does not directly influence the calculations for electric flux in this scenario.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric flux concepts
  • Knowledge of Coulomb's Law
  • Basic proficiency in unit conversions (N·m²/C)
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  • Learn about electric field calculations for point charges
  • Explore the relationship between charge distribution and electric flux
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kirby2
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A point charge 6.50 nC is at the center of a cube with sides of length 0.260 m.

PART A: What is the electric flux through the surface of the cube?

PART B: What is the electric flux through one of the six faces of the cube?

ATTEMPT:

A)
Total flux enclosed by the cube = Q / eo
6.5E-9/8.85 x 10^-12= 734.5

B)
flux through one of the six faces is
= Q / 6 eo
so 734.5/6 = 122.4

with the exception of units, are these right? i think I'm doing it wrong because i didn't use the side length in any of the calculations.
 
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hi kirvy2! :wink:
kirby2 said:
… i think I'm doing it wrong because i didn't use the side length in any of the calculations.

no, your method is fine :smile:
 
Agreed. It is one of the interesting things about the Coulombic force :) Maybe they were throwing you a red herring by giving you the length of the sides.
 

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