Expert: Daniel Mazur - 10/26/2009

Question
QUESTION: Hi Daniel,

Consider any imaginary closed surface and a charge q located outside it.
Now the gauss's law states that the net flux through this surface equals q_in/eps_0
And as q_in is 0, the net "flux" through the surface is 0.

But that doesn't necessarily mean that the electric field at a point on the surface is also 0 , does it?? the charge q outside does have a net field at different points of the surface,
It's just the net flux that's 0. right??
If that's the case, then how can we in so many problems, use this to find out the field at a point??
For ex, if we had an imaginary shell, the eq. would be as follows,
E(4pier^2) = 0
SO we would conclude that E at any point lying on the surface of this shell is 0.
But that shouldn't be the case....where am I going wrong??
And then does it work the same way for the amperes circuital law??

Also I came around a problem which considered a steady current carrying wire.
There were alternatives given out of which the correct ones were to be chosen.
One of the options was:
"the electric field at the axis of the wire is 0."

this is not one of the correct alternatives.
I don't understand why
if the wire is uniform and homogeneous, why isn't the field 0 at the axis?
has it got anything to do with a current in the wire

Thanks,
Shikhin

ANSWER: Hi Shikhin,

for your first problem, you need to understand, what "net flux" means. It does not mean "no flux through any part of the surface", it means "everything that comes in through the surface, comes out through the surface as well". Hence, there is no contradiction between zero net flux and a field produced by an external charge.
[Q]For ex, if we had an imaginary shell, the eq. would be as follows,
E(4pier^2) = 0[/Q]
This is vague writing again, against which I have warned you. Electrical flux is something different than electrical field. What is "4pier^2"? Where is the "r" measured from - the center of the shell or the external charge?
[Q]SO we would conclude that E at any point lying on the surface of this shell is 0.[/Q]
No we don't. No accepted physics textbook would say such a thing. Where do you have it from? See above, learn the difference between flux and field and if there is anything unclear afterwards, I'll be ready to answer a properly phrased question. I have no way of finding, what you mean otherwise.

[Q]And then does it work the same way for the amperes circuital law??[/Q]
Some things are analogous, some are different between electrostatics and magnetostatics. In Gauss' Law we integrate over a closed surface (or volume thereby enclosed), in Ampere's Law we integrate over a closed loop (or else over an area encircled by it). There is much more playing with vector calculus, where magnetism is involved, than in electrostatics. I recommend that you don't assume analogies. Learn the two areas separately and after you master them, fish out the analogies that emerge.

To your second question:
A current-carrying wire (unless it is explicitly said that it's superconducting) has a high el. potential at one end, low el. potential at the other end and in beween the potential is monotonically decreasing along the wire's length. At one point along its length the potential may be zero, but it just cannot be zero along any length of the wire (axis or not axis).

Cheers,
Daniel


---------- FOLLOW-UP ----------

QUESTION: Hi Daniel,

I'm sorry. The first question was absolutely sensless.
I soon realized the fault, which is as follows:
If the charge q was placed inside and at the very center of the sphere, instead of being placed outside, then the eq. would have been as follows:
integral(E.dS) = q/eps_0

Now in this integral on LHS, we can take the E out of the integration process since we can use the argument of symmetry and prove it to be constant throughout the surface and so in this case, it becomes
E(4pie r^2) = q/eps_0

In my last question, I incorrectly assumed E to be constant all throughout the shell when the source of E , i.e. the charge q is placed outside the shell.
It won't be constant and in fact if we integrate the LHS then, using the law of dot product of vectors, it will indeed turn out to be 0.

Apart from that, I had another problem.
Consider again the thin shperical conducting shell.
Let us place a charge q at its center and also let us give a charge Q to the shell itself which is distributed uniformily over its surface. Now if we draw an imaginary concentric sphere passing through the points b/w the inner and outer radii of the thin shell( which are aprox. equal as its a thin shell),
we can apply the gauss's law on this surface and then we get the charge distribution on the iner and outer surface of the shell as -q and Q+q respectively.
Now if we bring another charge q' near the sphere(assume q,Qand q' to be of same sign), the charge on the shell will redistribute.
Now the question asks, what will be the net force on the charge at the center of the shell after bringing the charge q' near the shell.
The answer is 0. and the reason is that the charges on the spherical shell redistribute themselves in such a way so that the field created due to this redistributed charge distribution on the charge q at the center is exactly equal and opposite to the field created by q' at the center.
I want to know the reasoning behind this redistribution which exactly opposes the external field.
I know that in conductors, charges redistribute until the field created by the redistributed charge distribution is exactly equal and opposite to the external field so that the interior of the conductor always has 0 net field.
And though, we have a conducting shell in this problem too, but it is not continous. i.e. it has an empty space in between.
Does this law of ideal conductors having a 0 net field in the interior even in the prescence of an external field, apply to hollow conductors as well, as in our case,??
And if it does, how can it be proved exactly.

There's still another thing that I wanted to know.
Both coulomb's law and Gauss's law basically show the inverse square dependence of the electric force and thus can be derived from each other.
And similarily it goes with biot-savart law and ampere's circuital law, i.e. they can be derived from each other.
Can you tell me how ampere's law can be derived from biot-savart's law....without using vector calculus...
And if the derivations are long enough and time consuming to be typed down, can you tell me anywhere I could find it, on the internet??

Thanks a lot,
Shikhin

Answer
Dear Shikhin,

I appreciate the advance you've made in your electrostatic study. Regarding your hollow shell as opposed to a continuous conductor I can say that yes, the same laws work in either of them. The one necessary condition for complete cancellation of the external fields is that the shell, as a surface, is completely enclosed. Only then has the free charge the necessary freedom of redistribution and can indeed cancel all external fields. If you drill a hole in your shell, you can calculate that in the immediate vicinity the external field will palpably penetrate inside the shell. In principle, even the tiniest hole causes penetration of the external field to all places inside the shell...

Observe that this requirement makes certain exercise problems from text books impossible to realize: Your (months ago) system of a three shells, where the middle one is grounded, cannot be realized "exactly", because in order to physically ground the middle shell, you need to drill a hole and pass a cable through the outer shell... In such cases we consider (and the limits can be found analytically) small holes just small perturbations (with something like a polynomial decay along the line into the drilled shell's center), which need not be included in the calculation.

About the Ampere's law derivation from Biot-Savart's law:
I don't know, how it can be generally done without vector calculus and I don't think it is really possible. You can do it for certain simple geometries, which make the vector operations (scalar and vector multiplication) quite simple. I think this is impossible to do generally - you have a vector operator inside the closed-curve integral you need to solve, how can we get past that without some vector calculus? Either you do the vector calculus "properly", i.e. invariant of any choice of coordinate system, or you choose a system of coordinates (Cartesian, spherical, cylindrical,...) and do the proof by vector components. Even in the latter case one must use the rules of calculation of vector and/or scalar product... and that's not what I'd call "proof without vector calculus". Let me know, anyway, if proof by components is what you had in mind.

I have an observation for you at this point regarding your previous questions about analogies between magnetostatic and electrostatic laws: In electrostatics a charged spherical shell acts so that the total field anywhere in its hole is zero. This can be proven easily as the sphere elements lying symmetrically on the opposite sides of any point of interest inside the shell exactly cancel each other's field. In magnetostatics, however, the opposite lying elements of one current-carrying loop ADD, not subtract, their magnetic field! This example shows, how different the two situations are, despite the outward similarity of the equations' shape.

I hope my answer was some help.

All Best,
Daniel