Electric flux through a surface

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SUMMARY

The discussion centers on calculating the electric flux through a square surface due to a point charge located above it. The correct expression for the electric flux is derived using Gauss' law, specifically \(\Phi = \frac{q_{enc}}{\epsilon_0}\). The participants clarify that the integral of the electric field \(E\) over the surface area \(dA\) must be considered, and due to symmetry, the flux through each face of an enclosing cube can be uniformly calculated. The consensus indicates that the answer is \( \frac{2q}{\epsilon_0 d} \), confirming the importance of integrating the electric field across the surface.

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  • Knowledge of surface integrals in vector calculus
  • Basic principles of symmetry in physics
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Homework Statement


A point charge q is situated a distance d/2 above a square surface of side d as shown (see attached).
a) It cannot be determined
b) 2q/ε0d
c) q/4\piε0
d) q/\piε0d2
e) q/6ε0

Homework Equations



\Phi=qenc0=\ointE*dA

The Attempt at a Solution


Shouldn't the answer be zero because whatever enters the area also leave the area? If not I know I should use the integral of E*dA but I don't know what E would be.
 

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If the area were a thin volume you'd be correct, the net flux into the volume = 0, but flux thru a surface A = the integral over the surface of the dot-product of the electric field E and an element of surface area dA: flux = ∫∫ E*dA.


So yes, you need to integrate E*dA over the surface. dA is normal to the surface at every point on the surface.
 
You can manage without integration.
Imagine the charge is in the middle of a cube with side d.
Due to symmetry, the flux through each face of the cube is the same.
And the total flux through the cubic, closed surface can be found from Gauss' law.
 
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Nasu has come up with the right and simple way to solve this.
 

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