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 confirmyour answer to the electric field given above.

Given that the electric field in slab 1 is -(sigma)/ 4e_0 z-hat direction and the electric field in slab 2 is - (sigma)/ 3e_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 confirmyour answer to the electric field given above.

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Question

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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