LianDAlane
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An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity σ1 = 0.35 μC/m2. Another infinite sheet of charge with uniform charge density σ2 = -0.54 μC/m2 is located at x = c = 31 cm.. An uncharged infinite conducting slab is placed halfway in between these sheets ( i.e., between x = 13.5 cm and x = 17.5 cm).. 1)What is Ex(P), the x-component of the electric field at point P, located at (x,y) = (6.75 cm, 0)?. 2). What is σa, the charge density on the surface of the conducting slab at x = 13.5 cm?

Verified Answer :

meerkat18

using a cylinder for a gaussian surface we find the flux through each end means that the surface integral will be double (the field has to be split between both directions, if you like)∮E⃗ ⋅d⃗ A=E(2A)=qϵ0=σAϵ0→E=σ2ϵ0which does not depend on distance from the slab.
The conducting slab in the middle picks up an induced charge from each charged slab, so it does not contribute to or hinder the field strength. Hence the total of the E-fields at any point should be∑E=E1+E2=σ12ϵ0+σ22ϵ0

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