Question

Calculate the total electric flux leaving the cubical surface formed by the
six planes x, y, z = ±5 if the charge distribution is: (a) twopoint charges, 0.1 μC
at (1,−2, 3) and 17μC at (−1, 2,−2); (b) a uniform line charge of π μC/m at
x = −2, y = 3; (c) a uniform surface charge of 0.1 μC/m2 on the plane y = 3x.

Answer 1

use gauss' law and symmetry

Flux through closed surface = Charge Enclosed inside surface / epsilon_zero

Answer 2

@IrishBoy123 I don't think that that will work here. Guass law only applies well to surfaces where the electric field is uniform. The symmetries exist in a point charge, an infinitely long line or cylinder, and an infinitely expansive plane, as I remember. That is so \(\int \vec E\dot\ \vec da=E\int da\) over the surface.

Since we don't have anything but a point charge, which is simple, I think Coulomb's law is most appropriate. Unfortunately, it's time consuming.

We do have a point, line, and plane. But I don't know if we can use the line or plane as if it were contained in a shell because then part of it is still out of the cube and the flux in is then going to be considered, too. I think that the exact approach is necessary.

Answer 3

Coulomb's law, or one form of it (for continuous charge like the line or plane), is
\[E=\frac1{4\pi\epsilon}\int{\frac \rho{\mathscr r^2}\text dV}\]
However, \(\rho\) is the volume charge density, and \(\text dV\) is for the volume.
For the plane, we should use \(\sigma\) (surface charge density) and \(\text da\).
For the line, we should use \(\lambda\) (line charge density) and \(\text dl\).

Otherwise, we can use
\[E=\frac1{4\pi\epsilon}\sum_i{\frac {q_i}{\mathrm r_i^2}}\]

Answer 4

hi @theEric, you point is well made but i think you are getting 2 ideas mixed up.

you seem to be describing situations where Gauss is used, effectively in reverse, to establish the **electric field**. in that case, you do need symmetry. so if we wanted the **field** due to a point charge, we would place it at the centre of a sphere and know that the **flux** is evenly distributed over all the area allowing is to describe E in terms of r.

however, the question here is much simpler and asks for the "total electric **flux** leaving the cubical surface". for the 2 point charges:
-- 0.1 μC at (1,−2, 3) and
-- 17μC at (−1, 2,−2)
that will be 17.1 micro C.

@sohiub
some information is missing for part (b).
plus you have to assume that the plane ends at y = +/- 5 .

Answer 5

There, \(\rm r\) is the distance from source point to field point. That is, the distance from where the charge is to where the electric field is being found.

One way to write this is \({\rm \vec r }= \vec r-\overrightarrow r'\), where \(\vec r\) is the vector from origin to the point where we want to know the electric field and \(\overrightarrow r'\) is the vector from origin to the charge.

Answer 6

@IrishBoy123 I didn't do a whole lot with electric flux in class, but flux, I think, is generally the sum over a surface of the scalar (dot) product of the vector field and the surface normal at each point on the surface.

But I think I see what you're saying. Are you saying that the fluxes through all surfaces that contain the charges are equivalent?

Answer 7

"Are you saying that the fluxes through all surfaces that contain the charges are equivalent?"

No, they're definitely not, because, as you remarked, there is no symmetry.

however, the question doesn't care how the flux is getting out, it just wants to know how much is getting out.

Answer 8

Well, would it make a difference if I changed "surfaces" to "closed surfaces"?

Answer 9

@theEric
indeed, yes it would.

the flux through that box should be the flux through a concentric 100x100x100 box. provided there are no other charges getting in the way.

the field strength at the new surface E would of course change as the flux is spread over a bigger area and E loosely is flux/area

Answer 10

I was thinking that \[\int{\vec E_1\dot\ d\vec a_1} = \frac1{4\pi\epsilon}\int{\frac\rho{\rm r^2}\text dV}= \int{\vec E_2\dot\ d\vec a_2}\]

So then any closed surface that contains the charges would have the same flux. So Gauss's law is still useful?

But, for the line,So Gauss's law won't apply. I guess all that matters is the charge enclosed though. \[\int{\vec E\dot\ d\vec a} = \frac{Q_{enclosed}}{\epsilon}\]

So finding the charge is sufficient to find the flux, since we know \(\epsilon_0\).

Good luck, I have to go!

Answer 11

My \[\int{\vec E\dot\ d\vec a}\] are for closed surfaces.

Answer 12

Hi! I was being foolish before! This equation that I posted earlier is nonsense!
\[\int{\vec E_1\dot\ d\vec a_1} = \frac1{4\pi\epsilon}\int{\frac\rho{\rm r^2}\text dV}= \int{\vec E_2\dot\ d\vec a_2}\]
Really,
\[E = \frac1{4\pi\epsilon}\int{\frac\rho{\rm r^2}\text dV}\]
That's Coulomb's law.

And\[\int{\vec E\dot\ \text d\vec a}=\frac{Q_{enclosed}}\epsilon\]is Gauss's law.

The flux is as simple as @IrishBoy123 was getting at.

Answer 13

For \(\sf any\) closed surface, the flux is the enclosed charge divided by the permittivity of the material. Here, we assume an ideal vacuum, where the permittivity of free space can be used (\(\epsilon_0\)).

So, if you find the charge enclosed, you can find the flux pretty easily.

IrishBoy123's first post was
"use gauss' law and symmetry
Flux through closed surface = Charge Enclosed inside surface / epsilon_zero "

I'm not sure what symmetry applies, but integration will definitely find the charge!

For point charges, you can add them simply. For the line and plane of charges, you have a continuous charge distribution (given) and you use integration to "sum up" all of the charges.

For a line you have\[Q=\int{\text{line charge density}\dot\ \text dl}=\int{\lambda\dot\ \text dl}\]

For a plane you have\[Q=\int\int{\text{surface charge density}\dot\ \text da}=\int\int{\sigma\dot\ \text da}\]


Setting up the limits of integration will be crucial. For the line, you have a constant \(x\) and \(y\). So the only variable is \(z\). Since we only to consider the total charge inside the cube, we only integrate over the line segment inside the cube. The bounds for \(z\) are given in the problem description.