Finding Current Flux Through Closed Surface

[Gee Wiz]

Homework Statement

Let J=xy^2($\hat{x}$+$\hat{y}$+$\hat{z}$) A/m^2 denote the electrical current density filed -i.e, current flux per unit area - in a region of space represented in Cartesian coordinates. A current density of J=xy^2($\hat{x}$+$\hat{y}$+$\hat{z}$) A/m^2 implies the flow of electrical current in the direction J/abs(J)= ($\hat{x}$+$\hat{y}$+$\hat{z}$)/$\sqrt{3}$ with a magnitude of abs(J) = xy^2$\sqrt{3}$ Amperes (A) per unit area.

Calculate the total current flux J*Ds through a closed surface S enclosing a cubic volume V = 1m^3 with vertices at (x,y,z,) = (0,0,0) and (1,1,1) m.

Homework Equations

The Attempt at a Solution

I believe that I can use a fair amount of symmetry for this problem because this flux should be the sum of six surface integrals. I have done a fair amount of things with electric flux in the past..and I feel that this should be very similar however my mind seems to believe drawing a blank is the best solution. I tried to start with the integral of abs(J)*($\hat{z}$)dxdy=total current enclosed/eo. Is that kind of the right idea? Or should I be trying to set it up differently.

 

[BvU]

Postpone the symmetry idea and start with one of the six surface integrals. Direction of J is a constant (unit vector) and so is $d\vec S$. Only thing that varies is the magnitude. Write out the integral.

 

[Gee Wiz]

∫xy^2($\hat{z}$)dxdy

I kind of had this before. I then was thinking along the lines well okay for when I did electric flux if I had a electric field E and I needed to find the charge enclosed and I was given a charge density I would start:

E*2*π*r^2=ρ*π*r^2/εo
then things would cancel and I would get E=ρ/(2*εo)

However, I am unsure how to progress to that point for the current flux. I'm sure its some simple thing I'm not putting together

 

[BvU]

OK, you picked the one where symmetry does come in useful: z=0. Check with the opposite one (z=1) and bingo, two down, four to go.

Note that two of those four have magnitude of J = 0 ( the ones with x or y equal to zero ).

Leaves some work to do, still, but you'll manage !

 

[HalfLostAndSearching]

I would approach this problem by first identifying the relevant equations and variables involved. In this case, the relevant equation is the definition of current flux, which is given by J*Ds, where J is the current density and Ds is the surface area. The variables involved are the current density J and the surface area Ds.

Next, I would visualize the problem and identify any symmetries that could simplify the calculation. In this case, the cubic volume and the current density being constant throughout the volume suggest a high degree of symmetry. This means that we can use the symmetry to simplify the calculation.

To calculate the total current flux, we need to integrate the current density over the surface area. Since the current density is constant, we can pull it out of the integral. Then, we can use the symmetry to simplify the integral by breaking it up into six surface integrals, each corresponding to one face of the cube. The surface area for each face can be calculated using basic geometry.

Finally, we can use the given vertices of the cube to set up the limits of integration for each surface integral. Once we have all six surface integrals, we can add them together to get the total current flux through the closed surface.

In summary, to solve this problem, we need to use the definition of current flux, take advantage of the symmetry of the problem, and set up and evaluate six surface integrals to calculate the total current flux through the closed surface.