Solve Gauss's Law: Cube w/E Field & Calculate Flux

[gunster]

Homework Statement

Consider a cube with one corner at the origin
and with sides of length 10 cm positioned
along the xyz axes. There is an electric field

E= (18, 235y, 0)

through the region that has a constant x component and a y component that increases linearly with y. Start by drawing a diagram showing ˆn and E~ for each side of the cube.
How much change is inside the cube?

Homework Equations

flux = integral (E * dA)
flux = Q/epsilon nought

The Attempt at a Solution

Tried integrating from 0 to 10 for the E to find change in the electric field

then set Q/A = integral (E * dA)

However, I don't know how to integrate it cause I don't know what the direction of dA is in this case. Is dA same direction as E? Also, what AREA do I use? I know its the surface area but idk what the proper area would be for flux going through a cube :/

 

[anathema]

Thank you for your post. In order to find the change in the electric field inside the cube, we first need to understand the direction of the electric field and the direction of the surface area vector, dA. In this case, the electric field is given by E= (18, 235y, 0), which means that it has a constant x component and a linearly increasing y component. The surface area vector, dA, is always perpendicular to the surface it is associated with. Therefore, for each side of the cube, the dA vector will be parallel to the surface of the cube and perpendicular to the electric field.

To find the change in the electric field inside the cube, we can use the equation flux = integral (E * dA), where the integral is taken over the surface area of the cube. In this case, the surface area of each side of the cube is 10 cm x 10 cm = 100 cm^2. Therefore, the total surface area of the cube is 6 x 100 cm^2 = 600 cm^2.

Now, let's consider each side of the cube separately. For the side with the normal vector ˆn = (1,0,0), the electric field and the surface area vector are both in the same direction, so the dot product E*dA will be positive. This means that the electric flux through this side of the cube will be positive. Similarly, for the side with the normal vector ˆn = (0,1,0), the electric flux will also be positive.

For the side with the normal vector ˆn = (0,0,1), the electric field and the surface area vector are perpendicular to each other, so the dot product E*dA will be zero. This means that there is no electric flux through this side of the cube.

For the remaining three sides of the cube, the electric field and the surface area vector will be in opposite directions, so the dot product E*dA will be negative. This means that the electric flux through these sides of the cube will be negative.

Therefore, the total electric flux through the cube will be the sum of the electric flux through each side of the cube. Since the electric flux is positive for two sides and negative for three sides, the total electric flux will be negative. This indicates that there is a net flow of electric field out of the cube