# Gauss Law and Flux

## Homework Statement

The electric field has been measured to be horizontal and to the right everywhere on the closed box shown in the figure. All over the left side of the box E1 = 90 V/m, and all over the right, slanting, side of the box E2 = 400 V/m. On the top the average field is E3 = 120 V/m, on the front and back the average field is E4 = 175 V/m, and on the bottom the average field is E5 = 245 V/m. How much charge is inside the box? Use the accurate value ε0 = 8.85e-12 C2/N·m2.

## The Attempt at a Solution

I have been working these homework problems no problem when it is just rectangles and cylinders and other shapes where Nhat and E both point in the same direction, however I thought that when it was not the same you had to use cosin of the angle made between Nhat and E. However it does not give me an angle. Am I going about this the wrong way?

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SammyS
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It doesn't give you the angle, but it gives enough information for you to find the cosine of the angle from the geometry of a right triangle.

I asked a friend after posting this. They said you use the area of the left side and multiply it by the electric field coming out of the slope as if it was just a rectangle and there was no slope at all. No need for any cosin or angles at all. I tried this and it gave me the right answer but I am still confused as to how it works?

SammyS
Staff Emeritus
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The cosine of the angle is 4/12, multiply (4/12) times the area of the slant rectangle, 12×5, gives the area of the vertical rectangle.

ahh clever

I don't understand, isn't it just 0? The net flux through a closed surface is not 0 here???

SammyS
Staff Emeritus