Calculate net charge with nonuniform electric field

AI Thread Summary
To determine the net charge within a cube in a nonuniform electric field, the electric flux through all six faces must be calculated using the formula φe = ∫E⋅dA = qenc/ε0. The electric field is defined as E0(1 + z/a) i + E0(z/a) j, and it was noted that the flux through the +z and -z faces cancels out to zero. The integrals for the flux through the other faces were initially set up incorrectly, particularly with the inclusion of a², which is unnecessary for the integration. The integrals should be evaluated from 0 to a, and care must be taken to integrate over the correct dimensions for each face. Correcting these aspects will lead to the accurate calculation of the net charge within the cube.
ooohffff
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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.

boldE.gif
= 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.

22-43.gif

Homework Equations



φe = ∫E⋅dA = qenc/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 qenc. 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?
 
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ooohffff said:

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.

boldE.gif
= 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.

22-43.gif

Homework Equations



φe = ∫E⋅dA = qenc/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 qenc. 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 ?
 
Hello!

ooohffff said:
I would evaluate the integrals from 0 to a right?
Yes.
ooohffff said:
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?
 
SammyS said:
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.
 
ooohffff said:
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.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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