Gauss Law Problem direction of Area

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SUMMARY

The discussion centers on applying Gauss's Law to determine the electric flux through a surface. The user calculates the flux as ∅ = ∫E.dA, with E directed outward along the positive i-axis. The area of the circular face is computed as A = π * (0.11)^2 = 0.038 m², leading to a flux of ∅ = 3 * 0.038 = 0.114 NC/m². However, the user questions the sign of the net flux (Φ_net) given that the total charge (Q) is zero, indicating that Φ_net must be negative, necessitating a deeper understanding of the dot product between area vectors and the electric field.

PREREQUISITES
  • Understanding of Gauss's Law and its mathematical formulation.
  • Familiarity with electric field concepts and vector calculus.
  • Knowledge of surface integrals and area vector orientation.
  • Basic principles of electrostatics, particularly charge distribution effects.
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  • Study the implications of charge neutrality on electric flux in electrostatics.
  • Learn about the dot product in vector calculus and its application in physics.
  • Explore advanced applications of Gauss's Law in different geometries.
  • Investigate the relationship between electric field lines and surface area orientation.
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Physics students, electrical engineers, and educators seeking to deepen their understanding of electric flux and Gauss's Law applications.

Physicslearner500039
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Homework Statement
In Fig. 23-28, a butterfly net is in a uniform electric field of magnitude E = 3.0 mN/C. The rim, a circle of radius a = 11 cm, is aligned perpendicular to the field. The net contains no net charge. Find the electric flux through the netting.
Relevant Equations
NA
p4.PNG


My attempt is
∅ = ∫E.dA.
The direction of E is going out of the net towards +ve i axis.
I am not clear on the direction of the Area, it can be either +ve i-axis or -ve i-axis. Which direction should i consider?
∅ = ∫3.dA = 3*∫dA ---->1
∫dA is the area of the circle.
A = π * (0.11)^2 = 0.038 m2

Hence
∅ = 3*0.038 = 0.114 NC/m2. Please advise, the answer is negative of this.
 
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The total flux through the surface consisting of the net and the circular face, taking the area vectors to point out of the surface (i.e. positive flux ##\implies## field lines leaving the surface) $$\Phi = \oint \vec{E} \cdot {d\vec{A}} = \Phi_{net} + \Phi_{circle} = \Phi_{net} + 0.003\pi a^2$$Since ##Q=0##, what must be the sign of ##\Phi_{net}##? Can you then rationalise this by considering the dot product of the area vectors with the electric field on the netting?
 
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