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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: field due to a conducting plate
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esha89 (0)

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the elctric field due to infinte conducting sheet isq/epsilon and that due to plane sheet of charge is q/2epsilon.but if we had used the cylindrical guassian surface through the conducting sheet we wud hav got E=q/2epsilon....plz explain why it isn't so.....

    
esha89 (0)

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plz tell me why we hav to consider the portion inside?
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mukundmadhav (460)

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You have to consider a closed surface, na.. So you have to take into account flux through all the surfaces of the gaussian surface you've taken
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prateek2292 (134)

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since it is a closed surface then we have to consider both the surfaces.
for a plane sheet E=Q/2epsilon.
But, for a conducting sheet u have to consider both the surfaces because it acts as a plate and plate have some thickness.
therefore,bcauz of both the surfaces..
E=Q/2epsilon+q/2epsilon
=Q/epsilon.





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esha89 (0)

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thnx...i got confused coz in brilliants they have given directly..
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