I'm to find the enclosed charge of the closed surface in the image (not sure how to describe the shape).
The Attempt at a Solution
Let [itex]a=4cm[/itex] and [itex]b=5cm[/itex]. The total flux in is then [itex]abE_1[/itex] and flux out is [itex]abE_2[/itex]. Applying Gauss' law then gives the total enclosed charge as [itex]Q=ab\epsilon_0(E_2-E_1)[/itex].
The step I'm a little unsure about is taking the total flux out to be [itex]E_2[/itex] times the area [itex]ab[/itex], since the electric field [itex]E_2[/itex] is through a slanted surface, not perpendicular. On the other hand, the total flux through that slope would have eventually to pass through an area of [itex]ab[/itex]...
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