So I was introduced to Gauss’ Law last year and now I’m taking an electromag course. One of the standard problems is to use GL to show that inside a conducting sphere there exists no electrical field. To this day, I absolutely cannot understand this result. Its great that the math works, but I don’t understand the physics behind it. None of the profs I’ve asked have ever given me a good answer - they just point to the board and say “well, if you se GL…” :rolleyes:
Lets say the shell is positively charged. Visualizing the shell, I can see a whole bunch of electric field vectors pointing into the centre. In this case, I can the see the zero result occuring due to a sum of vectors. Ie, the fields all cancel. BUT there is still a field there.
So really the question is, does the zero occur because of a canceling of fields or some other principle?
PS. No math is permitted in answering this question.
IF there is a non-zero electric field, then an electron in the field would be accelerated in one direction. Inside a metal box, which way do you propose that the electron would move toward?
No math? Right, and then I’ll explain why orbits are ellipses without using math. Physics is math.
There’s nothing wrong with thinking of it as the cancelling of fields. You know that a nearby small charge can produce the same force as a faraway large charge, and that’s what’s happening. Draw a line from the North pole to the South pole of the sphere, and imagine that you (a test charge) are on this line, closer to the North pole than the South pole. You know that there’s a lot less charge North of you than South of you, but the stuff North of you is a lot closer than the stuff South of you. So you’ll get pushed in the South direction by some amount, and the North direction by some amount. Maybe you can see why it’s reasonable that the two forces are equal, but I don’t see how you can see that they are equal without doing some math.
CurtC, I don’t think what you’re saying works in general. This doesn’t hold for a cube, for instance.
Can you think of an experiment you could do to distinguish between these two situations: 1) No electric field 2) Two identical but opposite electric fields that complete/y cancel each other at all points?
You can’t, because there is no physical difference between the two.
It’s impossible to prove without maths, but perhaps it’s possible to understand.
Imagine the situation where there is a charged objet in the centre of the hollow sphere, it’s clear that this object won’t move. If you move slightly it along some axis (and what’s true for one axis within a sphere by symmetry is true for all axes and therefore all points nside the sphere. It is now nearer some charge than it was in the centre position, but it’s further away than most of the charge. If the attraction between the sphere and the object obeyed a linear law the object would simply return to the centre, but as you know it doesn’t obey a linear law and it just so happens that the forces exerted by all points of the sphere exactly cancel out (this is the fudge as it’s impossible to prove without some for of mathematics (or perhaps ‘crypto-mathematics’)). This is true along all points of the axis (another fudge).
I can do it if you allow me a big whopping dose of mathematics at the beginning, although this form might be a little more palatable: the voltage inside the conductor obeys Laplace’s equation, and solutions don’t allow local maxima or minima except on the boundary of the region you’re solving in. These are two highly non-trivial statements, but once you’ve accepted them the rest is easy: since we’re assuming our box is perfectly conducting, there can’t be any voltage minima or maxima on the box or the electrons would have piled into/ran out of such a region, thereby changing the voltage. So there are no local voltage minima or maxima anywhere in the box or on the surface, i.e.\ the voltage is constant everywhere and therefore there’s no electric field.
I’ll see if I can come up with a better explanation, but as has been said it’s not an easy thing to explain without at least a small amount of mathematics.
You place your proton in the middle. Then you push it a little north. It feels more “above” charge then “below” charge and is pushed back into the middle. Thus we see an electric field is present inside the sphere.
But that isn’t what happens. You’re assuming that the field is only zero in the centre. That’s not the case. It’s zero everywhere inside the sphere. You push the proton a little north, and it stays there. It doesn’t get pushed back to the centre.
Aha! So the proton doesn’t move back to the centre!
Of course, that doesn’t really resolve the issue. Imagine drawing the field lines around the protons in the shell. What is it about the metal/air boundary that somehow stops 'em from getting in there?
Dammit I had a whole explanation and then somehow I deleted it or something.
Anyway I’m taking E&M this semester too.
In a conductor, any net charge will reside on the surface of the sphere because Coulomb’s law says that like charges will repel eachother. So these like charges will distribute themselves equally around the surface of the sphere to get as far away from one another as possible. If we then put a test charge into the conductor, it will be either equally pulled or equally pushed from all angles by all the charges on the sphere, so the net force it experiences will be 0 because it is being pulled/pushed equally in all directions. Electric field, by definition, describes the force a test charge experiences due to another charge or charges. Therefore, if the net force on the test charge is zero, the electric field is 0.
Here’s my question, in a similar vein to the OP: Gauss’ law predicts the electric field for an infinitely long, thin metal sheet with a uniform charge density will be = charge density / 2(permativity of free space.) But the electric field is only perpendicular to the sheet, so shouldn’t the test charge just experience the field due to the one differentially small piece of charge whose field line is perpendicular to the plane and intersects the point at which the test charge results, thus yielding the same force as a point charge interacting with a test charge (and thus being distance dependent)? In other words, why does the fact that the a charge is part of a sheet of charge if the rest of the sheet’s field doesn’t go through the test charge’s location?
I hope this post is as clear as I felt my original was.
I’m not entirely sure which field lines you’re talking about ‘getting in’ (do you mean getting into the conductor?). I realized a few points may need clarification.
There is no electric field inside a conductor. CurtC gave the most succinct statement of why - the electrons are free to move, and if you had an electric field inside a conductor, there’d be a Coulomb force, causing acceleration, and they’d move until equilibrium was reached.
If you’re talking about a hollow shell, there can be a field inside it - but it can only be generated by charges inside the sphere. Nothing can get through the ‘wall’ of the conductor. So if you have a proton inside the sphere, it still has its field. The field stops at the conductor, and can’t go ‘through’ it.
If I’m saying what you already know, but you don’t understand why it stops at the conductor, consider the case of just a conducting wall.
You have a proton (point charge) next to an infinite conducting plane. Looking at just before you touch the wall, the field is strongest closest to the proton. It’s weaker if you go further along the wall away from the proton. If this field went into the conductor, there’d be a net potential from the point closest to the proton to all points further away. So the electrons redistribute so there’s no net field.
It may be clear that the field shouldn’t just reappear on the other side of the conductor if it doesn’t go inside it. You could say the induced redistribution of charge is canceling the effect of the proton from being ‘seen’ on the other side. (Even if the point charge is an electron, the effect is the same.)
Qwertyasdfg, I only saw your post on preview, so this may be wrong, but I think you’re using the result – that the E-field is perpendicular – to assume that there’s no force from the rest of the plane. The E-field is perpendicular and uniform only because that’s the way all the charge in the plane affects the test charge. (This is remarkably similar to my above example, but for those playing at home, note that a test charge produces no inductive effects.)
My original point was much broader than this particular quesiton about the sphere. You seemed to be struggling with the idea that there is a difference between “no field” and “a net zero field”. There is no difference, and you cannot design an experiement to distinguish the two.
Arrgh, no, I understand that you can’t distinguish between no field vs. cancelling fields. That wasn’t my point, and I’m sorry if that’s what my OP seemed to be asking.
Firstly, I was unsure as to whether or not a test charge, inside a hollow conducting sphere, would be forced into equillibrium at the centre. If this were true, then there would be an electric field present, except at the very centre. Desmostylus stated that this is NOT correct and that the particle will not be “forced” back to the centre. However, now we see Qwertyasdfg once again claiming that a test charge will be forced to the centre.
So which is it? Will the proton be forced to the centre or not? Let us answer only this question first, then I will proceed.
(BTW, Qwertyasdfg, I asked the same question regarding the infinite charged plate. And like this scenario, the answer I got seemed very muddled.)
I’m not sure that Qwertyasdfg was saying that. But anyway, a test charge inside a conductive box won’t move, because there is no field to push it in any direction. Even if it’s not at the center.
Exactly. And this is the reason, for example, that the collection sphere of a Van de Graff generator is fed fromthe inside. No matter how much charge accumulates on the outer surface, there always exists some potential difference between the inner and outer surfaces.
Panama Jack, I think you’re right. The forces from throughout the plane act on the test charge, but all non-perpendicular components cancel, so the E-field is perpendicular to the plane but has a force greater than that of the dQ perpendicular to it.
I’d like to add, John Mace is right. If you place the charge q at the center, than move it north, it will be closer to an amount of charge on the surface of the sphere that we can call Q[sub]1[/sub]. However, it will get only slightly further away from a larger amount of surface charge we’ll call Q[sub]2[/sub]. So, the fact that Q[sub]1[/sub] < Q[sub]2[/sub] will compensate for the fact that r[sub]1[/sub] < r[sub]2[/sub]. Then with Coulomb’s Law we can say
(kqQ[sub]1[/sub]/r[sub]1[/sub][sup]2[/sup]) = (kqQ[sub]2[/sub]/r[sub]2[/sub][sup]2[/sup]), but that the forces are pushing/pulling in opposite directions. So the test charge will not move, which means the electric field is 0.
I should state that I was never explcitly taught this line of reasoning, but I’m 99 percent sure its right, and if you’d like I’ll run it by my teacher next week.
Not a brilliant analogy, but it’s kind of like the way you can’t have a sloped surface on a pool of water. If you introduce a “step” by emptying a bucket into one end of the pool, all the water shifts around until it’s level again.
Introduce a conductor into an electric field, and all the free charge redistributes itself as a “skin” on the conductor’s surface, until there’s no field within the conductor. This must happen because free charges move under the influence of a field, so they’ll only stop moving when the field ceases to exist.