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Electric Flux through the Face of a Cube

CDL

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1. Homework Statement
Griffiths' Introduction to Electrodynamics problem 2.10,
upload_2018-6-25_0-12-33.png


2. Homework Equations

Gauss' Law, ##\int_{S} \textbf{E}\cdot \textbf{dS} = \frac{Q_{\text{enc}}}{\epsilon_0}##



3. The Attempt at a Solution

It seems reasonable that the flux through the shaded surface and the front face of the cube are equal. I don't see any reason for them to be different. Assuming that they are equal, we should have that the flux through the front face added with the flux through the shaded face is ##\frac{q}{\epsilon_0}##, as the flux through the other faces is zero, since the electric field is tangential to those faces. Hence, we have ##\Phi_{\text{shaded}} = \frac{q}{2\epsilon_0}##. This is not the correct answer though! Why? Is it because the charge is on the boundary of the solid?

Let the charge as pictured be the origin, and say we solved this problem exactly for the charge at ##(\epsilon, \epsilon, \epsilon)## where ##\epsilon < \sqrt{3}##, then took the limit ##\epsilon \to 0##. Would we not get the flux to be ##\frac{q}{2\epsilon_0}##?

I understand the argument that we can consider a larger cube with q at the centre, and the final result comes to be ##\frac{q}{24 \epsilon_0}##, I would like to know why my method fails.
 

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TSny

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There are more than two faces which have non-zero flux.

Think of the charge as having a small, but finite radius. If it is exactly centered on a corner, how much of the charge is inside the cube?
 
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We should have that the flux through the front face added with the flux through the shaded face is ##\frac{q}{\epsilon_0}##, as the flux through the other faces is zero, since the electric field is tangential to those faces.
Nope.
The electric field through other faces is not purely tangential to them. The charge is a point charge, so its electric field has spherical symmetry, so the E. field lines won't be straight and parallel, they will be radially outward.
 

CDL

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There are more than two faces which have non-zero flux.

Think of the charge as having a small, but finite radius. If it is exactly centered on a corner, how much of the charge is inside the cube?
Oh yes, sorry, the top face would have non-zero flux too, but this gets it down to ##\frac{q}{3\epsilon_0}## only..

Nope.
The electric field through other faces is not purely tangential to them. The charge is a point charge, so its electric field has spherical symmetry, so the E. field lines won't be straight and parallel, they will be radially outward.
The top, front and right faces have non-zero flux but isn't the ##\textbf{E}## field produced by the charge tangential to the bottom, back and left faces?
 
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CDL

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Think of the charge as having a small, but finite radius. If it is exactly centered on a corner, how much of the charge is inside the cube?
An eighth? That would give ##\frac{q}{24 \epsilon_0}## :eek: Why can we think of the charge like this (shell theorem)? Why not some other shape? When applying Gauss' law, and a charge is on the boundary of a surface, do we usually think about charges this way?
 

TSny

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An eighth? That would give ##\frac{q}{24 \epsilon_0}## :eek: Why can we think of the charge like this (shell theorem)? [Why not some other shape?
The field of a point charge is radial and spherically symmetric. That's really all you need in order to see that when the point charge is at a corner of the cube, the total number of field lines that enter the cube is one-eighth of the total number of lines that leave the point charge. Also, the fact that the field is radial implies that none of the field lines will pierce through any of the three faces of the cube that touch the point charge. The spherical symmetry of the field of the point charge implies that the flux will be equal for the three other faces. Thus, each of these three latter faces will be pierced by one-third of one-eighth of the total number of field lines emanating from the charge.

To get this result, we only had to use the fact that the field of a point charge is radial and spherically symmetric. It is not necessary to consider the point charge as a limiting case of a uniformly-charged, finite-sized sphere in order to get the answer. Nevertheless, this limit approach can be helpful as another way to look at it. For a true point charge, all the charge is at one mathematical point. This can make you feel a little uncertain when the point charge lies on the gaussian surface and you want to know how much of the point charge is enclosed by the gaussian surface. The point charge also has the pathology that the magnitude of the field blows up to infinity as you approach the charge.

If you replace the point charge by a uniformly-charged sphere of radius R less than the length of an edge of the cube, you will get that there is no flux through three of the faces of the cube and the flux through each of the other three faces is ##\frac{q}{24 \epsilon_0}##. This is because the finite sphere also produces a field that is radial and spherically symmetric; and, outside the charged sphere, the field is the same as if the charge of the sphere is concentrated as a point charge (shell theorem). Thus, the flux through any one of the six faces does not depend on R as long as R is less than an edge-length. So, the flux through each face will remain constant as you take the limit of R going to zero.
 
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Orodruin

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This can make you feel a little uncertain when the point charge lies on the gaussian surface and you want to know how much of the point charge is enclosed by the gaussian surface. The point charge also has the pathology that the magnitude of the field blows up to infinity as you approach the charge.
I would solve this by using a different surface. Just because there is a cube in the problem does not mean we have to use that cube as our Gaussian surface. The square in question is 1/4 of one side of a cube with double the side length in which the charge is in the centre. By symmetry, 1/6 of the flux goes through each side of the bigger cube and so our original square gets 1/(4*6) = 1/24 of the total flux. This removes the need to consider your charge as a limit of a spread out distribution or having to deal with a delta on the surface. (For which the appropriate rule is that its contribution to the integral is the solid angle subtended by the area as seen from the charge divided by 4pi.)
 
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TSny

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I would solve this by using a different surface. Just because there is a cube in the problem does not mean we have to use that cube as our Gaussian surface. The square in question is 1/4 of one side of a cube with double the side length in which the charge is in the centre. By symmetry, 1/6 of the flux goes through each side of the bigger cube and so our original square gets 1/(4*6) = 1/24 of the total flux. This removes the need to consider your charge as a limit of a spread out distribution or having to deal with a delta on the surface. (For which the appropriate rule is that its contribution to the integral is the solid angle subtended by the area as seen from the charge divided by 4pi.)
This is a very nice way to get the answer. The OP indicated he/she is familiar with this method but nevertheless wanted to solve the problem by sticking with the original cube.
 
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Orodruin

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This is a very nice way to get the answer. The OP indicated he/she is familiar with this method but nevertheless wanted to solve the problem by sticking with the original cube.
I always tell my students never to put delta functions on the integration boundaries unless they are very certain of what they are doing. The problem is that there are many different one-parameter functions that converge to the delta function in the distributional sense, but that give different results for the enclosed charge (of course they all give the same result for the flux through the square). A more appropriate way to handle the point charge is to cut out a corner of the smaller cube using a spherical surface of radius ##\varepsilon## instead. The flux into the volume in this corner must then be equal to the flux out of the three sides away from the centre and can be easily computed to be the solid angle covered by the corner (##\pi/2##) divided by ##4\pi##.
 

TSny

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A more appropriate way to handle the point charge is to cut out a corner of the smaller cube using a spherical surface of radius ##\varepsilon## instead. The flux into the volume in this corner must then be equal to the flux out of the three sides away from the centre and can be easily computed to be the solid angle covered by the corner (##\pi/2##) divided by ##4\pi##.
Beautiful. I never thought of doing it that way. This sort of trick should work any time you have a point charge sitting on a surface.
 

Orodruin

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Beautiful. I never thought of doing it that way. This sort of trick should work any time you have a point charge sitting on a surface.
Indeed it does. That is why I mentioned the solid angle earlier.
 

TSny

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Indeed it does. That is why I mentioned the solid angle earlier.
Right. I still remember a problem from Purcell's text from my student days 50 years ago. It required setting up a Gaussian surface that passed through a point charge. I recall feeling uncomfortable at the time with dealing with this. Your approach would have helped! If you're interested, here is the problem:
upload_2018-6-25_14-0-4.png

Purcell had already worked out the induced charge density, σ, on the plane earlier in the text.
 

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CDL

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The flux into the volume in this corner must then be equal to the flux out of the three sides away from the centre
Is this because we have ##\nabla \cdot \textbf{E} = \textbf{0}## everywhere but the point charge? When is this principle applicable?

and can be easily computed to be the solid angle covered by the corner (π/2) divided by 4π.
By direct computation using a surface integral? Cool!
 

CDL

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Right. I still remember a problem from Purcell's text from my student days 50 years ago. It required setting up a Gaussian surface that passed through a point charge. I recall feeling uncomfortable at the time with dealing with this. Your approach would have helped! If you're interested, here is the problem:
View attachment 227299
Purcell had already worked out the induced charge density, σ, on the plane earlier in the text.
Interesting problem, I shall give it a go when I get up to conductors : ~ )
 

Orodruin

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Is this because we have ##\nabla \cdot \textbf{E} = \textbf{0}## everywhere but the point charge? When is this principle applicable?
Yes, if the field divergence inside the volume is zero, then the net flux into the volume must be zero. What goes in through one part of the surface must come out through another.

By direct computation using a surface integral? Cool!
For example. You can also argue by symmetry here as well.
 

CDL

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Yes, if the field divergence inside the volume is zero, then the net flux into the volume must be zero. What goes in through one part of the surface must come out through another.
So here we are considering the volume bounded by the sphere and the cube, what comes in goes out, so the sphere and the three cube surfaces share the same flux?

For example. You can also argue by symmetry here as well.
For example, could we also say that we know the flux through the whole sphere, and since the electric field due to the charge is constant on our sphere of radius ##\epsilon##, the flux through this part of the sphere is the fraction of the total flux ##\frac{\Omega}{4\pi}##, and as you said in this case we have ##\Omega = \frac{\pi}{2}##? Sorry, I am not very familiar with this stuff and want to make sure :)
 

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So here we are considering the volume bounded by the sphere and the cube, what comes in goes out, so the sphere and the three cube surfaces share the same flux?
They must have the same magnitude and opposite signs.

For example, could we also say that we know the flux through the whole sphere, and since the electric field due to the charge is constant on our sphere of radius ##\epsilon##, the flux through this part of the sphere is the fraction of the total flux ##\frac{\Omega}{4\pi}##, and as you said in this case we have ##\Omega = \frac{\pi}{2}##? Sorry, I am not very familiar with this stuff and want to make sure :)
Right.
 
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CDL

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They must have the same magnitude and opposite signs.
Yep, if we consistently orient the surface. Thanks! you have been a great help.
 

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