# Gauss' Law

I know i'm reposting this... but no one seems interested in my thread.. maybe becuase it already has replies (which doesn't mean its been solved!) but whatever

cube is placed with its edge at (0,0,0) as in the diagram. The sides of the cube are1 .4m. Find the electric flux and the charge inside the cube if the electric field is given by
a)3yN/C j
b) -4N/C i + [6N/C + (3Nm/C)y]j

for a) $$\Phi = \int E \bullet dA \hat{j}$$
$$\Phi = \int (3y)(dA) \hat{j} \bullet \hat{j}$$
$$\Phi = 3 \int ydA$$
but y is constant and A is constant soo
$$\Phi = 3 (1.4)^3$$
taht gives it for ONE face. But what about hte other face of the cube? I dont want to proceed with the other question since it is similar to this. ANYWAY., if you could help me it would be really awesome

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No worries, you're "fluxation" worries will be taken care of. Just as an inital though, if a field vector is constant for the entire domain, chances are pretty good the the net flux is zero throughout the gaussian surface. IN order to investigate though, you need to examine the flux at each face. since you have the dimensions of the cube and the orientation of the electric vector field, then evaluate the flux at each surface. If you're numbers show a NET FLUX of zero, then your in good shape with the general analysis. But if not and/or you're not sure about your numbers, post up what you have done and either myself or some other brave sould will take a look at it.

but in ths case the electric field is not consatnt its 3y, poiting in the Y direction... Are you sure that the electrifc field would be zero then?

good call, I was wondering that as well. PLease clarifyand leave nothing out. That value of Y (constant or not) could affect the whole problem.

i posted the question as it appeared in teh book.

Is the flux really zero through this cube? The field is non uniform...

soc na nayone help me with the qeustion?

the field is not zero.. since the field is not uniform, or doesi t work out to be zero? Please advise!

Doc Al
Mentor
stunner5000pt said:
for a) $$\Phi = \int E \bullet dA \hat{j}$$
$$\Phi = \int (3y)(dA) \hat{j} \bullet \hat{j}$$
$$\Phi = 3 \int ydA$$
but y is constant and A is constant soo
$$\Phi = 3 (1.4)^3$$
taht gives it for ONE face. But what about hte other face of the cube? I dont want to proceed with the other question since it is similar to this. ANYWAY., if you could help me it would be really awesome
You'd find the flux through the other relevant side in exactly the same way. (Of course, it's trivial since y = 0 for the other side.) Since the field is only in the $\hat{j}$ direction, only the two sides perpendicular to the y-axis could have any flux (the sides at y = 1.4m, and y = 0).

What makes you think the total flux is zero?

so would the electric flux of the other side be the negative value since the angle between E and dA on that side is pi?
THIS is whats making me think the flux is zero... but i cant believe that since the field is non uniform...

for the other side
$$\Phi = \int E \bullet dA = \int 3\hat{j} dA\hat{j} cos 180$$
i dont think this is right...

Doc Al
Mentor
Before you worry about the angle that the field makes with dA, first tell me the value of the field on the side y = 0.

the value is the same as the other side because the same calculation from #1 follows, does it not? that is 3 (1.4)^3?

Doc Al
Mentor
No. The field is given as 3yN/C j; it depends on y! (The same method is used to calculate the flux for each side, but you have to use the appropriate value of y--which is different for each side.)

on the left the y value is zero
on the right side the value is 1.4
so what happens to the dA however...?

Doc Al
Mentor
Who cares what happens to dA? If the y value is zero, what is the value of the electric field?

Doc Al said:
Who cares what happens to dA? If the y value is zero, what is the value of the electric field?
the value of the elctric field is then zero if y is zero
right?

Doc Al
Mentor
Right! So the flux through that side is zero.

awesome

now for the other side... is what i have correct?

Doc Al
Mentor
If you are asking about the flux you caculated through the side with y = 1.4m (that you posted in #1): yes, that looks good.

so the total flux for the first question that is part a is 3(1.4)^3 nm^2/C and thats that? To find the induced charge... i simply use this $$\Phi = \frac{q}{epsilon_{0}}$$

THank you for the help so far!

I will solve part b and post my solution up in a little while.

Once again.. thank you for the help!

Doc Al
Mentor
That's all there is to it.

ok for the second part where $E = (-4N/C)\hat{i} + [6N/C + (3N/Cm)y]\hat{j}$
ok since the field is uniform for the -4N/C i part and the 6N/C j part... the flux is zero through the x=0 and x=1.4 sides and the y sides which are affected by the 6N/C field

however for the 3N/cm y j field - for the y=0 the flux is zero, but the y=1.4m the flux is
$$\Phi = \int 3y dA = 3 \int y dA$$ since y is constnat and A is also constnat and y = 1.4 and A = 1.4^2
$$\Phi = 3(1.4)^3$$
Am i good now? Did i get it right?
So the only flux is 3(1.4)^3 and the other components of the field have no effect on the flux in the cube?

Doc Al
Mentor
You got it. The field in part b is just the field from part a with a constant term added everywhere. But that constant term will not add to the the net flux through the closed cube. Thus the flux is identical in both parts.