Calculation of electric flux through a face of a cube.

Aug 2019
88
23
India
A charge \(\displaystyle q \) sits at the back corner of a cube, as shown in the attachment. What is the flux of \(\displaystyle \mathbf{E}\) through the yellow shaded side?

The way I thought fro proceeding was that should take differential area element \(\displaystyle d\mathbf{A}\) in the yellow face, now the direction of this element will be pointing outside the cube i.e. in positive y-direction (see the second attachment). We know that the electric field will be the vector pointing from the origin to this area element and hence the position vector
\(\displaystyle \vec{r}= x\hat i + a \hat j + z \hat k \)
I have taken the y-component to be \(\displaystyle a \) because that's the distance our position vector will traverse in y-direction to get to \(\displaystyle d\mathbf{A}\) (I know I'm not making myself clear over here, but I'm trying). Now,
\(\displaystyle \hat r = \frac{1}{\sqrt{x^2 +a^2+z^2}} \times (x \hat i + a \hat j +z \hat k) \)

Now the problem comes, if we evaluate the first integral we will get
\(\displaystyle \frac{a^2}{(a^2+z^2)(\sqrt{z^2 + 2a^2})} \)


Flux = \(\displaystyle \int\limits_S \mathbf{E} \cdot d\mathbf{A}\)
Flux = \(\displaystyle \int\limits_S \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat r ~ dA \hat j\) (dA points in the positive y direction)

Flux = \(\displaystyle \frac{q}{4\pi\epsilon_0} \int\limits_S \frac{a}{(x^2+a^2+z^2)^{3/2}} dx~dz \)

Flux = \(\displaystyle \frac{q}{4\pi\epsilon_0} \int_{z=0}^{z=a}dz \int_{x=0}^{x=a} \frac{a}{(x^2+a^2+z^2)^{3/2}}dx \)

Now the problem comes, if we evaluate the first integral we will get
\(\displaystyle \frac{a^2} {(a^2 +z^2) (\sqrt{z^2 + 2a^2})} \)

and then
Flux = \(\displaystyle \frac{q}{4\pi\epsilon_0} \int_{0}^{a} \frac{a^2}{(z^2+a^2)(\sqrt{z^2+2a^2})} dz \)

This is integral is not solvable (by me) and hence I'm stuck. I want to know why my approach is wrong? Did I make any mistake? Where does my mistake lie?

I apologize if I'm not too clear in my question, actually I couldn't make a good picture of a vector pointing from the charge to the area element and therefore tried my best to compensate it through my explanatory text.

Thank you. Any help will be much appreciated.
 

Attachments

Jun 2019
493
262
USA
There are many things I don't recall about the relationship of the electric and magnetic fields and converting flux integrals to line integrals.

Luckily, we don't need them for this problem. The electric field will emanate from the charge, correct? So the flux through the three faces touching the origin will be zero. By symmetry, the flux going through the face shaded in yellow will be 1/3 of the flux through the entire cube. Also by symmetry, the flux through the cube will be 1/8 of the total flux from the charge. So find the total electric flux and divide by 24. (This was an old trick they made us learn for things like determining the net amount of radiative heat transfer from a heated cylinder onto a semi-hexagonal interior surface.)
 
  • Like
Reactions: 1 person
Aug 2019
88
23
India
I couldn't understand this " Also by symmetry, the flux through the cube will be 1/8 of the total flux from the charge " line of yours. I request you to please explain it just a little more.
Thank you.
 
Jun 2019
493
262
USA
You have a cube in the (+,+,+) octant. There will also be electric flux in the (+,+,-) octant, and the (+,-,+) octant, (+,-,-), etc..

If you imagine a $2a \times 2a \times 2a$ cube centred at the origin, your surface is 1/24th of the cube. The cube completely encloses the charge, so (if I remember my E&M correctly), the flux across the cube should equal the total flux from the charge. And each $a \times a$ square is symmetrically placed, so each one should have 1/24 of the total flux.
 
  • Like
Reactions: 1 person