# Griffiths 7.3

1. Apr 23, 2008

### ehrenfest

[SOLVED] Griffiths 7.3

1. The problem statement, all variables and given/known data
Two metal objects are embedded in weakly conducting meterial of conductivity of $\sigma$ Show that the resistance between them is related to the capacitance of the arrangement by
$$R=\frac{\epsilon_0}{\sigma C}$$

2. Relevant equations

3. The attempt at a solution
I am not sure how sigma or epsilon_0 comes in. The only way I know how to do this problem is by getting the E-field and that is obviously not possible since they tell you nothing about the geometry.

2. Apr 23, 2008

### G01

You have to get the E-Field involved, you are correct about that.

HINT:

$$I=\int \vec{J}\cdot d\vec{a}$$

Start here. Can you get the E-Field involved in this equation from what you know from chapter 7? If so, then what about getting a charge term in the equation?

3. Apr 26, 2008

### ehrenfest

What is the surface that you are integrating over? If it is a closed surface containing the metal object 1, then we just apply Gauss's law and we get the answer. But can you explain why this is equal to I? Actually I am not even sure what the definition of I is here even.

Last edited: Apr 26, 2008
4. Apr 26, 2008

### G01

I here is a general current moving between the two metal objects (if I remember the problem correctly).

Last edited: Apr 26, 2008
5. Apr 26, 2008

### ehrenfest

But how do you defined the "general current moving between the two metal objects". It is obvious when they are connected by a wire but I do not think it is obvious here.

You still never said what you are integrating over which is a related question.

6. Apr 27, 2008

### G01

It is a current that is moving between the two metal objects. There really is nothing more you can say about it. It moves from one object to the other, through the material in between.

Pick any closed surface through which the current either enters or leaves and you'll be able to apply Gauss's Law. Thus, it could contain either object one, or object two.

You also haven't shown ANY work on this problem since I gave you an initial hint. I'm sorry if I'm being vague, but I will not be able to tell you any more about this problem until you show some work. My first hint should have been enough to get you started. For instance, you said:

How about showing me what you will do to solve the problem, once you are sure you can apply Gauss's Law? How do you manipulate the equation i gave you involving J to find the answer using Gauss's Law? Show me some work.

Last edited by a moderator: Apr 29, 2008
7. Apr 29, 2008

### ehrenfest

Trust me I can do the manipulations. I am trying to figure out what the problem is asking. There is no point in showing work when I don't understand the problem.

Can you please give me a non-ambiguous MATHEMATICAL DEFINITION of the current that I am supposed to calculate?

That is all I am asking for. I can do the rest by myself.

8. Apr 29, 2008

### G01

There are several problems with your post here, ehrenfest.

First off, I do not have to trust you to do the manipulations. You, on the other hand, DO have to show me your work. You should refresh yourself on the forum rules.

In all honesty, I should not have offered help at all since you did not show ANY work, and still haven't done so. I have seen you around on these forums and, from your other threads I knew you were one to put time into your problems, so I decided to help you.

Secondly, I have offered you a mathematical definition of the current:

$$I = \int J \cdot da$$

The problem is so general that you can't apply any more specific definition. This is the best I can do. If this is not specific enough, then there are two options for you.

1)You can assume I'm wrong about there being no other, more specific definition and wait for another homework helper to find the thread. This is very possible. I am not perfect and have already asked some other helpers if there is something I'm missing.

2)You can find a problem with more specific parameters.

Sorry, ehrenfest, but I cannot give you any more specific definition of the current. I just can't. The problem is too general. The definition I gave is the only one that can be applied, since it is the most general expression for any current.

Last edited: Apr 29, 2008
9. Apr 29, 2008

### ehrenfest

Sorry GO1. The definition you gave me in your first post

$$I=\int \vec{J}\cdot d\vec{a}$$

was perfectly fine. I kind of forgot about that. Sorry. I didn't mean to be rude in my last post. I will post my work shortly.

10. Apr 30, 2008

### ehrenfest

OK. We defined I as
$$I=\int \vec{J}\cdot d\vec{a}$$
where the integral is over any surface in the material containing one and only one of the conductors. By Ohms law that equals,

$$I=\sigma \int \vec{E}\cdot d\vec{a}$$

and by Gauss that equals

$$I=\sigma Q/\epsilon_0$$

and by the relation V=IR we get

$$V/R = \sigma Q/\epsilon_0$$

and the result is immediate. Sorry to be rude. I hate just ill-defined concepts. I wish everything were defined as nicely as it is in Rudin's Principles of Mathematical Analysis. I still like Griffiths a lot though. Some of his problems cannot really be solved from first principles though since you need experience to understand the problem.

11. Apr 30, 2008

### Ben Niehoff

This concept is actually not ill-defined at all. The question is specifically asking you to prove something very general, but what you've shown is that you can prove it; therefore it must have been well-defined! There are problems in Jackson that ask you to prove things even more generic than this. But it's always possible to get at them, starting from Maxwell's equations and using vector calculus.

Perhaps what you're confused about is exactly what surface to integrate over. Since you are finding the current "between two objects", you can choose any surface which divides all of space into exactly two regions, one containing Object A and one containing Object B. Because of the conservation of charge, it doesn't make a difference which surface you choose; whatever current flows out of A must flow into B.

12. Apr 30, 2008

### ehrenfest

That makes sense!

13. Apr 30, 2008

You got it!