Point Charge above an Infinite Plate

[josh777]

Merged thread deleted first post next post is the first post.

 

[josh777]

Thanks for reading. I originally asked this question on Stack-Exchange, but I haven't received any satisfying answers yet.

The only answer I've received (as of posting this) uses the "method of images" concept...but, in the book where I got this example from (Purcell) the author hints that the "image - charge" wasn't necessary to explain it.

So, I thought I'd try my luck with this community.

You can see the original post here:
https://physics.stackexchange.com/q...late-and-point-charge-above-an-infinite-plate

------------------------------------Thanks for reading.

Say we have a point charge +q above an infinite plate. How can we show that the charge that accumulates on the upper surface of the plate is −q?

There are two ways I've usually seen this explained.

The first is with image charges...I accept that explanation. However, according to the books I've been learning E&M from, there's a different way...

The other argument I've heard is with a Gaussian cube whose bottom part goes through the plate and top and sides are infinitely far away...but, I don't understand this argument.

I understand that if we make a Gaussian surface that has the bottom inside the plate,then the flux through that bottom part must be zero.

However, I don't buy the argument that if we put the top and sides infinitely far, the flux through them would be zero.

After all, as we grow the Gaussian cube, the areas of the top and the sides would increase proportionally to r2r2, just as the strength of the electric field would decrease inversely proportionally to r2r2, where rr is the side-length of the cube.

Why is it that the flux through the sides and the top must be zero as well? I don't understand the infinity argument, because if we have an infinitely large sphere with a point charge inside, its not like we can say that the flux through the sphere is zero just because the sphere's walls are infinitely far away?

Thanks!

Edit:

I know that we can show it with the method of images. However, both in Purcell and in Griffiths, the authors hint that the answer could've been arrived at some different way:

Griffiths:

Purcell:

What is this different way, and why does it work? Thank you!

 

[kuruman]

You should not have not started a separate thread, but waited for the moderator(s) to move it, which is what they usually do in such cases. My answer to your question was

Here is a hand waving argument about how to visualize this. The key is that no matter how big a Gaussian surface you draw, the infinite plane will cut through it. Remember that electric field lines start at positive charges and end at negative charges. In this case the lines starting at +q above the plane will end at negative induced charges on the plane. The more of the plane that is included inside the surface, the more lines will end on the plane. When the Gaussian box extends to infinity all the lines generated by +q will be intercepted by the plane. This means that the total induced charge on the surface is -q.

 

[josh777]

You should not have not started a separate thread, but waited for the moderator(s) to move it, which is what they usually do in such cases. My answer to your question was

Here is a hand waving argument about how to visualize this. The key is that no matter how big a Gaussian surface you draw, the infinite plane will cut through it. Remember that electric field lines start at positive charges and end at negative charges. In this case the lines starting at +q above the plane will end at negative induced charges on the plane. The more of the plane that is included inside the surface, the more lines will end on the plane. When the Gaussian box extends to infinity all the lines generated by +q will be intercepted by the plane. This means that the total induced charge on the surface is -q.

Whoops! I'll just delete the old thread...next time I'll wait for the moderators. I'll copy my response from there to here and message a moderator to delete it (it doesn't let me delete it for some reason...sorry, I'm not very familiar with PhysicsForums).

Thank you kuruman...I'm a little confused as to why all the field lines must end on the plane though...what about the electric field lines facing overall upwards?

It would seem sensible to me that no matter how far up we go, the force will still be in the upwards direction so long as we stay relatively close to the point charge in the horizontal directions (why would the field lines curve down?)

Also, do field lines all end at negative charges? Why? I feel like if this is true...I don't understand field lines as well as I thought I did. What if we just have an isolated point charge in space?

Thanks!

 

[kuruman]

It would seem sensible to me that no matter how far up we go, the force will still be in the upwards direction ...

What force?

Also, do field lines all end at negative charges? Why?

Because it is a law of nature also known as Gauss's law. Positive charges are sources of electric field lines and negative charges are sinks of electric field lines. If you believe that positive charges start electric field lines you must accept that negative charges end them.

Here is another way of thinking about your problem that avoids mathematical infinities in favor of physical infinities, i.e. distances that are very very small relative to some other distance. Imagine point charge $+q$ at the center of a conducting shell the size of the Earth. If you construct a concentric Gaussian surface entirely between the iner and outer radius of the shell, the flux through the shell will be zero because the electric field on the shell is zero. This means that the charge enclosed by the shell is zero. You know already that the only free charge is $+q$ at the center so you must have total charge $-q$ on the inside surface of the shell.

This charge will be uniformly distributed because $+q$ is at the center. What would happen if you started moving this charge along a radius towards the wall? The total charge on the inner surface will not change but the distribution on it will. The closer you move $+q$ to the wall, the more negative charge will be moving from the farther parts of the inner surface as near it as possible. If it is 1 cm away from the wall and the shell is the size of the Earth, the vicinity around the charge will look like a flat plane and most of the induced charge will be in the immediate area. The approximation gets even better when the $+q$ is 1 nm away from the wall and the shell is the size of the solar system. That's what I mean by physical infinity as opposed to mathematical infinity.

 

[haruspex]

DO field lines all end at negative charges? What if we just have an isolated point charge in space?

Yes, they can all be taken as ending at negative charges.
Is it possible to have a universe with a net nonzero charge? Probably not.

 

[josh777]

What force?

Because it is a law of nature also known as Gauss's law. Positive charges are sources of electric field lines and negative charges are sinks of electric field lines. If you believe that positive charges start electric field lines you must accept that negative charges end them.

Here is another way of thinking about your problem that avoids mathematical infinities in favor of physical infinities, i.e. distances that are very very small relative to some other distance. Imagine point charge $+q$ at the center of a conducting shell the size of the Earth. If you construct a concentric Gaussian surface entirely between the iner and outer radius of the shell, the flux through the shell will be zero because the electric field on the shell is zero. This means that the charge enclosed by the shell is zero. You know already that the only free charge is $+q$ at the center so you must have total charge $-q$ on the inside surface of the shell.

This charge will be uniformly distributed because $+q$ is at the center. What would happen if you started moving this charge along a radius towards the wall? The total charge on the inner surface will not change but the distribution on it will. The closer you move $+q$ to the wall, the more negative charge will be moving from the farther parts of the inner surface as near it as possible. If it is 1 cm away from the wall and the shell is the size of the Earth, the vicinity around the charge will look like a flat plane and most of the induced charge will be in the immediate area. The approximation gets even better when the $+q$ is 1 nm away from the wall and the shell is the size of the solar system. That's what I mean by physical infinity as opposed to mathematical infinity.

Thank you! I like this explanation, and I accept Gauss law always holds, but...how can we do the same with a similar Gaussian surface for the infinite plane? I understand that it may be more complicated because of "mathematical infinities"...but there's got to be a way!

Also, by "force" I meant field-lines. As in, so long as we moved up from the point-charge and not horizontally parallel to the plane very far, the electric-field would still point upwards, regardless of how far up we move.

 

[kuruman]

Do you mean a mathematically infinite plane? See post #2. Imagine a Gaussian sphere of infinite radius with the infinite plane cutting in half. The electric field is zero on the Gaussian surface so the flux through the surface is zero. You know you have $+q$ somewhere inside. Because the flux is zero, there must be total $-q$ on the plane. Now if you are wondering where did this charge come from if the plane is uncharged, the answer is "from infinity". The plane must be an equipotential all the way out to infinity where the potential is zero. If you bring a positive charge near it, you will raise its potential unless negative charges are drawn in from infinity.

A finite Gaussian surface enclosing an initially neutral conducting plate will enclose net charge $+q$ unless the plate is grounded. Infinities in electrostatics do funny things when you take them literally as mathematical infinities.

 

[rude man]

$\nabla \cdot \mathbf D = \rho$ and $\rho = 0$.
There would have to be divergence somewhere in the E field which you can't have if there is no charge density anywhere in the field. All field lines must begin and end on charges in an electrostatic field.