- #1
iScience
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hi all, gauss's law states that for any surface, the total electric flux coming through an open surface is always q/ε0. okay.. i understand this...
so for a sphere.. the simplest case... http://imgur.com/bwJzf7t,2KFO8Xw,O9Mgafg
[itex]\Phi[/itex]=[itex]\int[/itex][itex]\vec{E}[/itex][itex]\bullet[/itex][itex]\hat{n}[/itex]dA=EA= [itex]\frac{q}{4(pi)ε}[/itex] (4[itex]\pi[/itex]r2)=[itex]\frac{q}{ε}[/itex].okay... so this says... that ALL of the electric field produced by the charge q0 contributes to the value of [itex]\Phi[/itex]. However, let's now look at the case of two charged particles...
http://imgur.com/bwJzf7t,2KFO8Xw,O9Mgafg#1
forgive the ambiguousness if the image is not very clear at first.. i just have two charges and the lines/curves are the electric fields, more specifically.. Ʃ(E1+E2).. in otherwords superposition.
now, at every point that does NOT lie on the plane where the two electrons sit, there will exist a y-component of [itex]\vec{E}[/itex]. However at every point that lies on the plane where the two electrons sit, there will exist only an x-component [itex]\vec{E}[/itex].
okay.. so then let's consider a plane/sheet of charge carriers.
http://imgur.com/bwJzf7t,2KFO8Xw,O9Mgafg#2
everywhere i look i see that the two gaussian surfaces are supposed to be JUST the two top and bottom circular areas and that the flux produced/going through these two areas is supposed to be equal to [itex]\frac{q}{ε}[/itex]. i don't agree with this (obviously i am wrong so please show me where my reasoning is wrong) because..
if we go back to the point when i earlier said.. that ALL of the electric field lines went into contributing to the value of [itex]\Phi[/itex]. however if we now look at just the top and bottom two circular surfaces, i don't see how they incorporate ALL the field lines produced by the charges enclosed in the cylindrical volume, ie.. the 2-D sheet. the two top and bottom circular areas should ONLY account for those electric field lines that have an initial y-component of [itex]\vec{E}[/itex] (speaking with respect to the plane of the sheet). what about all the rest of the x-component [itex]\vec{E}[/itex]? their superpositioned [itex]\vec{E}[/itex] doesn't get pushed up into the y-direction at all. which means that these x-component [itex]\vec{E}[/itex] reside on the plane of the sheet. so why then do we only include the two top and bottom circles for gauss's law?
EVEN IF the x-component [itex]\vec{E}[/itex] were to all cancel out.. the y-component [itex]\vec{E}[/itex] alone should not contribute to the entire value of "[itex]\frac{q}{ε}[/itex]"... but it apparently does... why?
so for a sphere.. the simplest case... http://imgur.com/bwJzf7t,2KFO8Xw,O9Mgafg
[itex]\Phi[/itex]=[itex]\int[/itex][itex]\vec{E}[/itex][itex]\bullet[/itex][itex]\hat{n}[/itex]dA=EA= [itex]\frac{q}{4(pi)ε}[/itex] (4[itex]\pi[/itex]r2)=[itex]\frac{q}{ε}[/itex].okay... so this says... that ALL of the electric field produced by the charge q0 contributes to the value of [itex]\Phi[/itex]. However, let's now look at the case of two charged particles...
http://imgur.com/bwJzf7t,2KFO8Xw,O9Mgafg#1
forgive the ambiguousness if the image is not very clear at first.. i just have two charges and the lines/curves are the electric fields, more specifically.. Ʃ(E1+E2).. in otherwords superposition.
now, at every point that does NOT lie on the plane where the two electrons sit, there will exist a y-component of [itex]\vec{E}[/itex]. However at every point that lies on the plane where the two electrons sit, there will exist only an x-component [itex]\vec{E}[/itex].
okay.. so then let's consider a plane/sheet of charge carriers.
http://imgur.com/bwJzf7t,2KFO8Xw,O9Mgafg#2
everywhere i look i see that the two gaussian surfaces are supposed to be JUST the two top and bottom circular areas and that the flux produced/going through these two areas is supposed to be equal to [itex]\frac{q}{ε}[/itex]. i don't agree with this (obviously i am wrong so please show me where my reasoning is wrong) because..
if we go back to the point when i earlier said.. that ALL of the electric field lines went into contributing to the value of [itex]\Phi[/itex]. however if we now look at just the top and bottom two circular surfaces, i don't see how they incorporate ALL the field lines produced by the charges enclosed in the cylindrical volume, ie.. the 2-D sheet. the two top and bottom circular areas should ONLY account for those electric field lines that have an initial y-component of [itex]\vec{E}[/itex] (speaking with respect to the plane of the sheet). what about all the rest of the x-component [itex]\vec{E}[/itex]? their superpositioned [itex]\vec{E}[/itex] doesn't get pushed up into the y-direction at all. which means that these x-component [itex]\vec{E}[/itex] reside on the plane of the sheet. so why then do we only include the two top and bottom circles for gauss's law?
EVEN IF the x-component [itex]\vec{E}[/itex] were to all cancel out.. the y-component [itex]\vec{E}[/itex] alone should not contribute to the entire value of "[itex]\frac{q}{ε}[/itex]"... but it apparently does... why?
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