# Find the enclosed charge

1. Jul 28, 2013

### chipotleaway

1. The problem statement, all variables and given/known data
I'm to find the enclosed charge of the closed surface in the image (not sure how to describe the shape).

3. 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 $ab$...

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2. Jul 28, 2013

### Staff: Mentor

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.