Calculate net charge with nonuniform electric field

[ooohffff]

Homework Statement

In a cubical volume, 1.05 m on a side, the electric field is given by the formula below, where E₀ = 1.25 N/C and a = 1.05 m.

= E₀(1 + z/a) i + E₀(z/a) j

The cube has its sides parallel to the coordinate axes, see the figure. Determine the net charge within the cube.

Homework Equations

φ_e = ∫E⋅dA = q_enc/ε₀

The Attempt at a Solution

[/B]
So I know that I need to calculate the net flux through all the 6 faces of the cube in order to solve for q_enc. I know that φ_+z and φ_-z are equal to 0.

I think I am doing something wrong because it seems like they would cancel out?

φ_+x = E₀ a² ∫ (1+z/a) dz
φ_-x = -E₀ a² ∫ (1+z/a) dz
φ_+y = E₀ a² ∫ (z/a) dz
φ_-y = -E₀ a² ∫ (z/a) dz

Also, I would evaluate the integrals from 0 to a right?

 

[SammyS]

Homework Statement

In a cubical volume, 1.05 m on a side, the electric field is given by the formula below, where E0 = 1.25 N/C and a = 1.05 m.

= E₀(1 + z/a) i + E₀(z/a) j

The cube has its sides parallel to the coordinate axes, see the figure. Determine the net charge within the cube.

Homework Equations

φ_e = ∫E⋅dA = q_enc/ε₀

The Attempt at a Solution

[/B]
So I know that I need to calculate the net flux through all the 6 faces of the cube in order to solve for q_enc. I know that φ_+z and φ_-z are equal to 0.

I think I am doing something wrong because it seems like they would cancel out?

φ_+x = E₀ a² ∫ (1+z/a) dz
φ_-x = -E₀ a² ∫ (1+z/a) dz
φ_+y = E₀ a² ∫ (z/a) dz
φ_-y = -E₀ a² ∫ (z/a) dz

Also, I would evaluate the integrals from 0 to a right?

You have done a lot correct.

The units will be wrong.

Where does a² come from ?

 

[Qwertywerty]

Hello!

I would evaluate the integrals from 0 to a right?

Yes.

I think I am doing something wrong because it seems like they would cancel out?

You don't think this is possible? Notice that for a particular $z$, $E$ remains constant throughout the object. Also, what is the definition of flux?

 

[ooohffff]

You have done a lot correct.

The units will be wrong.

Where does a² come from ?

I think I was getting confused between ∫ E⋅dA and E⋅A, but since I'm doing the integral one then I don't need a². So,

φ_+x = E₀ ∫ (1+z/a) dz
φ_-x = -E₀ ∫ (1+z/a) dz
φ_+y = E₀ ∫ (z/a) dz
φ_-y = -E₀ ∫ (z/a) dz

from 0 to a.

 

[SammyS]

I think I was getting confused between ∫ E⋅dA and E⋅A, but since I'm doing the integral one then I don't need a². So,

φ_+x = E₀ ∫ (1+z/a) dz
φ_-x = -E₀ ∫ (1+z/a) dz
φ_+y = E₀ ∫ (z/a) dz
φ_-y = -E₀ ∫ (z/a) dz

from 0 to a.

The units are also incorrect this time.

You need to integrate over y for some & over x for others. Since E is independent of x & y the result of those is easy.