Gauss Law Problem direction of Area

In summary, the conversation discusses the calculation of electric flux (∅) using the formula ∅ = ∫E.dA, where E represents the electric field and A represents the area. The direction of E is going out of the net towards the positive i axis, but there is confusion about the direction of the area, which could be either positive or negative i axis. The conversation also mentions the calculation of ∅ = ∫3.dA, where ∫dA is the area of a circle, which is then used to find the total flux through the surface. The answer is negative, and it is explained by considering the dot product of the area vectors with the electric field on the netting.
  • #1
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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  • #2
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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1. What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface.

2. How is Gauss's Law applied to solve problems?

In order to apply Gauss's Law, we must first choose a closed surface, known as a Gaussian surface, and determine the direction of the area vector for that surface. Then, we can use the formula Φ = ∫E⃗ · dA⃗ = Qenc/ε0 to solve for the electric flux Φ, where E⃗ is the electric field, dA⃗ is the differential area element, Qenc is the enclosed charge, and ε0 is the permittivity of free space.

3. How do we determine the direction of the area vector for a Gaussian surface?

The direction of the area vector for a Gaussian surface is always outward, perpendicular to the surface. This means that the area vector points away from the enclosed charge for a positive charge, and towards the enclosed charge for a negative charge.

4. Can Gauss's Law be applied to any closed surface?

No, Gauss's Law can only be applied to closed surfaces that are symmetrical and have a constant electric field. This is because the electric field must be able to be pulled out of the integral in the formula Φ = ∫E⃗ · dA⃗, which is only possible for symmetrical surfaces with constant electric fields.

5. How does Gauss's Law relate to Coulomb's Law?

Gauss's Law is essentially a more general and powerful version of Coulomb's Law. While Coulomb's Law only applies to point charges, Gauss's Law can be used to solve problems for any symmetrical charge distribution. Additionally, Gauss's Law allows us to solve for the electric field without having to know the exact location of the charges, which can be difficult to determine in some situations.

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