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Given that the electric field in slab 1 is -(sigma)/ 4*e_0 z-hat direction and the electric field in slab 2 is - (sigma)/ 3*e_0 z-hat direction.
The bound surface charges were found to be -(sigma)/4 for slab 1, -(sigma)/6 for slab 2, and the boundary in between slab 1 and slab 2 is - 5(sigma)/12.
The bound volume charge is 0.
Question: Now that you know all the charge (free and bound), recalculate the field in each slab, and confirm
your answer to the electric field given above.
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