# Homework Help: Mechanical force on unit area of charged conductor.

1. Nov 20, 2012

### AlchemistK

1. The problem statement, all variables and given/known data

OK first, I didn't know where to post this since it is and is not homework,anyways, posting here would be safest.

Now, the attached scan is a proof from my (not so trustworthy) textbook and I have some doubts in it.

The first doubt is in statement S1 : Every element of charged conductor experiences a normal outwards force.
This is sort of a general doubt, why normal? That should be a special case when the body is symmetrical, right? Why is it necessary for all the electric field vectors to cancel out in way to produce a normal resultant?
Something vague to back me up : One known fact about electric field lines in conductors is that they are always perpendicular to the surface; for a curved object to have perpendicular lines, the lines themselves should be curved. Thus the lines passing through dS (in figure) must be curved and wont give a normal resultant unless the magnitude of the fields from opposite directions is equal.
Now, all of this I thought was based on one assumption I unknowingly made, that the charge density is uniform. Having non uniform charge density will change everything. So in the end the question is whether S1 is always true. Is it?

The second doubt is S2. I have no problem with them taking the field like that, just that when the proof is done, will we be able to use this for conductors of any magnitude and dimensions and use it to calculate force over a finite area rather than an area element?

Now, relating to this is a question which I think is meant to be solved using this result. (Hardworking people could try double integration but the question was ideally meant to be solved in 3 minutes) An alternative method will be Highly appreciated.

Q: A conducting spherical shell of radius R is given a charge Q. Find the force exerted by one half on the other half.

The attempt is done below but my question is once again regarding S2, we derived the result using the electric field as σ2/2ε, can it be used here too? Conversely, does this imply that the hemisphere has a field of σ2/2ε at the point where the charge of the other hemisphere can be thought of to be concentrated?

2. The attempt at a solution

df = σ2/2ε * dS

Taking the integral of dS as the projected area of the hemisphere, the center circular plane
pi*R2 (R being radius of the the hemisphere)
σ= Q/(2*2*pi*R2)

Thus, f = Q2/(32ε*pi*R2)

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2. Nov 20, 2012

### SammyS

Staff Emeritus
For a conductor in electrostatic equilibrium: It is true that the electric field lines are perpendicular to the surface at their point of intersection with the surface. The reason is much the same as the reason for the electric field to be zero within the conducting material itself. It comes about because in a conductor, charges may move relatively freely.

If you were to produce a situation in which there was a non-zero component of the electric field parallel to the surface at the surface of a conductor, that would induce charges to move in such a way that eventually they would achieve a configuration which would cancel the parallel component.
For a conductor in electrostatic equilibrium: Yes, statement S1 true, because the electric field lines are perpendicular to the surface. Therefore, any excess charge on the surface will "feel" a force perpendicular to the surface.
I've got to go now.

More, later, unless the issues are resolved in the meantime.

3. Nov 20, 2012

### haruspex

The statement is correct. If there were any tangential component to the field at some point on the surface then it would give rise to a current along the surface. So the field is normal to the surface everywhere. The only quibble I have about the statement is the reference to a force. The element experiences a field. Only charges will experience forces.
Exactly. That's what this statement is saying.
No, a spherical surface is curved, but the field lines are straight. More generally, curvature is a second order effect (like, second derivative) whereas normality is first order. I.e. sufficiently close to the surface you can see the normality but not the curvature.
You appear to be questioning the validity of differential and integral calculus. To make the argument rigorous would require rather more maths. It turns out that although the result is inexact for any given dS size, the limit as dS vanishes is exactly correct.
k is missing?
Note that force is a vector. In the above equation, dS is a vector normal to the surface. So when you integrate there will be some cancellation.

4. Nov 20, 2012

### AlchemistK

OK, my confusion about S1 is from the fact that I thought that dS is actually not a real element but a hypothetical one, from when the complete smooth conductor was cut and gave rise to the surface of dS .Now that I think about it, its a terribly stupid mistake, since σ would change too if it was cut, along with other stuff. I think S1 is clear.

The element is charged. (Again something that I missed, dS is actually a part of the conductor)

ε=εoK , but leave that, trivial stuff.

Alright then, I think everything is done. Thank you!!

An alternative way to solve the question? I hadn't heard of this result till after my test so it was a bit impossible for this to strike me while solving the question. Maybe something more natural, even though a bit lengthy?(Not double integration)
Is there a technique to find the position of the point where the charge of the hemisphere can be said to be located?

5. Nov 20, 2012

### haruspex

No. It is not in general possible to substitute a point charge to represent a charged conductor. That you can do it for outside a spherical shell (or uniformly charged spherical shell insulator) appears to be just a lucky fact of inverse square laws in a three dimensional world. (And likewise, the absence of a field inside a uniformly charged spherical shell.) For other charge distributions, where the 'equivalent' point charge would sit depends on what part of the field you look at.