Calculating Volume Charge Density for Electric Flux in a Box

[FatoonsBaby71]

Homework Statement

The electric flux density in space is given as D = 2xy(ax)+x^2(ay) C/m^2. A box is given as: 0<=x<=1; 0<=y<=2;0<=z<=3 (m). It is found that the total electric flux out of the box is (Wo). The charges creating the electric flux density D are now removed and replaced by a sphere of radius a =0.2m centered at (0.2,0.2,0.2) m and charged with a uniform volume charge density d. It is found that the total electric flux out of the box again is (Wo). Find the volume charge density d.

Homework Equations

Double Intergral D * ds = Q (Gauss Law)
dflux = D* dS
Charge elements: Volume dQ = pv * dv

The Attempt at a Solution

Well i first found the total charge in side the box using gauss' law. Which resulted in 18 C. (Hope this is right!) Since we know that the flux is equal to the charge, I figured all i needed to do is use the equation dQ = pv * dv and solve for pv which is the volume charge density d. However how would I be able to do this, I can't divide dQ/dv = pv can I? Is this the logical way to go behind the problem??

Thanks for your help

 

[Doc Al]

Not sure why you are using differentials, since you are told that the density is uniform. Thus ρ = Q/V. What's the volume of that sphere?

(I didn't check your answer for the charge.)

 

[FatoonsBaby71]

Oh ok that makes sense. However, I wanted to check if I computed the charge correctly.
I had two double integrals...

The first double integral went from 0 to 3 then 0 to 2 of (2xy) * dy dz which resulted in 17 C
The second double integral went from 0 to 3 then 0 to 1 of x^2 * dx dz which resulted in 1 C for a total of 18 C... did I do that correctly?

Another quick questions now since there are no differentials involved is V just the volume of a sphere 4/3 pi r^3?

Thanks

 

[Doc Al]

The electric flux density in space is given as D = 2xy(ax)+x^2(ay) C/m^2.

I assume that's:

\vec{D} = 2xy\hat{i} + x^2\hat{j}

Right?

Oh ok that makes sense. However, I wanted to check if I computed the charge correctly.
I had two double integrals...

There are six sides to consider, but several are obviously zero flux.

The first double integral went from 0 to 3 then 0 to 2 of (2xy) * dy dz which resulted in 17 C

Looks like you are doing the side where x = 1. So how did you get 17?

The second double integral went from 0 to 3 then 0 to 1 of x^2 * dx dz which resulted in 1 C

Looks like you are doing the side where y = 2. But what about the y = 0 side?

Another quick questions now since there are no differentials involved is V just the volume of a sphere 4/3 pi r^3?

Right.

 

[FatoonsBaby71]

I assume that's:

\vec{D} = 2xy\hat{i} + x^2\hat{j}

Right?


There are six sides to consider, but several are obviously zero flux.


Looks like you are doing the side where x = 1. So how did you get 17?

Looks like you are doing the side where y = 2. But what about the y = 0 side?


Right.

\vec{D} = 2xy\hat{i} + x^2\hat{j} is correct.

When you say that several are zero flux, I assume your referring to the flux in the z direction because there is no z component. Another thing that points out is the fact that the y component is x^2. Due to symmetry, I am assuming that both fluxes will cancel out. So that leaves us with just the x component. When doing the side when x = 1, I see that you plug in x for 1 and my result is 12 C which sounds reasonable. My real question is why do you plug in the x = 1?? How come we don't intergrate the way I was doing leaving the x in there.

Thanks for your help

 

[Doc Al]

When you say that several are zero flux, I assume your referring to the flux in the z direction because there is no z component. Another thing that points out is the fact that the y component is x^2. Due to symmetry, I am assuming that both fluxes will cancel out. So that leaves us with just the x component.

All good.

When doing the side when x = 1, I see that you plug in x for 1 and my result is 12 C which sounds reasonable.

OK.

My real question is why do you plug in the x = 1?? How come we don't intergrate the way I was doing leaving the x in there.

I'm not sure what you mean. You can certainly leave the x there when you integrate, it's just a constant.

∫2xy dy dz = 2x ∫y dy dz