Electric flux through a surface

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Homework Help Overview

The problem involves calculating the electric flux through a square surface due to a point charge positioned above it. The context is rooted in electrostatics and the application of Gauss's law.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the nature of electric flux and whether it can be zero based on the entry and exit of electric field lines. There is mention of using integration to calculate the flux and considerations of symmetry in the setup.

Discussion Status

Some participants have provided guidance on integrating the electric field over the surface and have referenced Gauss's law to approach the problem. There is an acknowledgment of different methods being considered, including a simpler approach without integration.

Contextual Notes

Participants are navigating assumptions about the configuration of the charge and the surface, as well as the implications of symmetry in the electric field distribution.

Plasmosis1
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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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