Two large flat parallel sheets have opposite uniform surface charge densities to and are separated a distance d. A large, uncharged conducting slab of thickness d/3 is parallel to the charged sheets, centered between them. First, using Gauss' law, prove that charge must be induced on the surfaces of the slab, and find the surface density of this charge. Then find the electrostatic potential as a function of distance y perpendicular to the sheets. Take the reference potential Vo = = 0 and the origin y = 0 to be at the negative sheet. -

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Two large flat parallel sheets have opposite uniform surface charge densities to and are
separated a distance d. A large, uncharged conducting slab of thickness d/3 is parallel to
the charged sheets, centered between them. First, using Gauss' law, prove that charge must
be induced on the surfaces of the slab, and find the surface density of this charge. Then
find the electrostatic potential as a function of distance y perpendicular to the sheets. Take
the reference potential Vo = 0 and the origin y = 0 to be at the negative sheet.
Transcribed Image Text:Two large flat parallel sheets have opposite uniform surface charge densities to and are separated a distance d. A large, uncharged conducting slab of thickness d/3 is parallel to the charged sheets, centered between them. First, using Gauss' law, prove that charge must be induced on the surfaces of the slab, and find the surface density of this charge. Then find the electrostatic potential as a function of distance y perpendicular to the sheets. Take the reference potential Vo = 0 and the origin y = 0 to be at the negative sheet.
Expert Solution
Step 1

Given Data:
The flat sheet has charge density ±σ.
Difference between the two flat sheets d
The thickness of slab t=d3

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