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Two solid conductors are attached together orthogonally and there's an infinite plane with charge density σ at π/4 Rad with the conductors. Calculate the field dist. In the region with the plane

http://dl.dropbox.com/u/11897904/image_2.png

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So first take away the conductors and draw what you would see if the conductors were mirrors. – FrankH Nov 9 '11 at 13:40
Thanks for your reply. I tried that and i got two planes intersecting orthogonally with opposite charges on them -σ and +σ does this mean that there is no E field? this doesn't make sense. – AVenkat Nov 9 '11 at 13:51
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OK. So you have two infinite intersecting planes. For each infinite plane draw what the vector E field would be for each plane and then just add those E vectors together. – FrankH Nov 9 '11 at 13:55
E(for infinite plane) = σ/(2ε0) therefore using superposition we have -σ/(2ε0)+σ/(2ε0) = 0 is this logic valid? is this what you were talking about? – AVenkat Nov 9 '11 at 13:59
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Almost. But E is a vector so it has a direction so you have to add them as vectors. Which way do the vectors point for the two infinite planes? Then add the vectors. It should not be 0. I'm signing off so good luck or maybe someone else will help you. – FrankH Nov 9 '11 at 14:19
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closed as too localized by David Zaslavsky♦ Nov 9 '11 at 17:50

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