Flux calculation using Gauss’s Theorem

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SUMMARY

The discussion focuses on calculating electric flux using Gauss's Theorem in a uniform electric field of strength 300 N/C at a 30º angle. The cube has sides measuring 5 cm. The participants confirm that while Gauss's law can be applied to find the net flux, the individual flux through each face requires integration. The net flux through the entire surface is determined to be zero due to the absence of divergence within the cube.

PREREQUISITES
  • Understanding of Gauss's Law in electrostatics
  • Knowledge of electric field strength and direction
  • Familiarity with surface integrals in vector calculus
  • Basic concepts of electric flux
NEXT STEPS
  • Study the application of Gauss's Law in different geometries
  • Learn how to perform surface integrals for electric fields
  • Explore the concept of divergence in vector fields
  • Investigate the relationship between electric flux and charge distribution
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Students and professionals in physics, electrical engineering, and anyone interested in understanding electric fields and their applications in electrostatics.

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Qn: A uniform electric field of strength 300 N/C at an angle of
30º with respect to the x-axis goes through a cube of sides
5 cm. (a) Calculate the flux through each cube face:
Front, Back, Left, Right, Top, and Bottom. (b) Calculate
the net flux through the entire surface. (c) An electron is
placed centered 10 cm from the left surface. What is the
net flux through the entire surface? Explain your answer
Ans:
Applying Gauss’s law the net flux can be calculated.
And for option (B), i guess the flux will be 0. But not sure. can anyone explain all the 3 options?
 
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And for option (B), i guess the flux will be 0. But not sure.
Option B?
That is right, and Gauss gives you the reason (is there divergence of the field inside the cube?).

(a) cannot be solved with Gauss alone, but the integrals are easy to do.
 

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