How do I Find the Enclosed Charge of a Closed Surface?

[chipotleaway]

Homework Statement

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 a=4cm and b=5cm. The total flux in is then abE_1 and flux out is abE_2. Applying Gauss' law then gives the total enclosed charge as Q=ab\epsilon_0(E_2-E_1).

The step I'm a little unsure about is taking the total flux out to be E_2 times the area ab, since the electric field E_2 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 ab...

 

[gneill]

Homework Statement

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 a=4cm and b=5cm. The total flux in is then abE_1 and flux out is abE_2. Applying Gauss' law then gives the total enclosed charge as Q=ab\epsilon_0(E_2-E_1).

The step I'm a little unsure about is taking the total flux out to be E_2 times the area ab, since the electric field E_2 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 ab...

The area ab is correct. You want to sum the component of the flux which is normal to the surface it passes through. If you do the geometry you'll find that for a flat surface at an angle to the flux, this amounts to the cross sectional area which is perpendicular to the field vectors.