Relationship between Resistance and Capacitance in Weakly Conducting Material

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So in summary, the resistance between two metal objects embedded in a weakly conducting material of conductivity sigma is related to the capacitance of the arrangement by the equation R = (epsilon_0)/(sigma*C). To solve this problem, you can use the equation I=\int \vec{J}\cdot d\vec{a} and apply Gauss's Law by picking a closed surface through which the current either enters or leaves, which could contain either object one or object two. The current in this problem is a general current that moves between the two metal objects through the material in between.
  • #1
ehrenfest
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[SOLVED] Griffiths 7.3

Homework Statement


Two metal objects are embedded in weakly conducting meterial of conductivity of [itex]\sigma[/itex] Show that the resistance between them is related to the capacitance of the arrangement by
[tex]R=\frac{\epsilon_0}{\sigma C}[/tex]

Homework Equations


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.
 
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  • #2
You have to get the E-Field involved, you are correct about that.

HINT:

[tex]I=\int \vec{J}\cdot d\vec{a}[/tex]

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
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.
 
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  • #4
ehrenfest said:
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.

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

HINT: Your on the right track with your Gauss's law idea.
 
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  • #5
G01 said:
I here is a general current moving between the two metal objects (if I remember the problem correctly).

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
ehrenfest said:
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.

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.

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

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:
...then we just apply Gauss's law and we get the answer.

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.
 
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  • #7
G01 said:
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.

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
ehrenfest said:
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.

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:

[tex]I = \int J \cdot da[/tex]

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.
 
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  • #9
Sorry GO1. The definition you gave me in your first post

[tex]
I=\int \vec{J}\cdot d\vec{a}
[/tex]

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

[tex] I=\sigma \int \vec{E}\cdot d\vec{a}[/tex]

and by Gauss that equals

[tex] I=\sigma Q/\epsilon_0[/tex]

and by the relation V=IR we get

[tex]V/R = \sigma Q/\epsilon_0 [/tex]

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
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
That makes sense!
 
  • #13
ehrenfest said:
OK. We defined I as
[tex] I=\int \vec{J}\cdot d\vec{a}[/tex]
where the integral is over any surface in the material containing one and only one of the conductors. By Ohms law that equals,

[tex] I=\sigma \int \vec{E}\cdot d\vec{a}[/tex]

and by Gauss that equals

[tex] I=\sigma Q/\epsilon_0[/tex]

and by the relation V=IR we get

[tex]V/R = \sigma Q/\epsilon_0 [/tex]

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.


You got it!
 

What is Griffiths 7.3?

Griffiths 7.3 refers to a specific chapter in the textbook "Introduction to Electrodynamics" written by David J. Griffiths. This chapter focuses on the concept of multipole expansions and their applications in electromagnetism.

What topics are covered in Griffiths 7.3?

In this chapter, Griffiths covers the basics of multipole expansions, including the definition of multipole moments, the calculation of electric and magnetic fields from multipole moments, and the applications of multipole expansions in real-life situations.

What is the importance of understanding Griffiths 7.3?

Griffiths 7.3 is an essential chapter for any student or researcher in the field of electromagnetism. Understanding multipole expansions and their applications is crucial for solving complex problems in electromagnetism and for understanding the behavior of electric and magnetic fields in various situations.

Are there any real-life applications of Griffiths 7.3?

Yes, there are many real-life applications of multipole expansions discussed in Griffiths 7.3. Some examples include the analysis of electric and magnetic fields in atoms and molecules, the calculation of gravitational forces between celestial bodies, and the design of antenna systems for communication.

Is Griffiths 7.3 suitable for beginners in electromagnetism?

While some background knowledge in electromagnetism is recommended, Griffiths 7.3 can be understood by beginners with some effort. The chapter includes clear explanations and examples, making it a valuable resource for students learning about multipole expansions for the first time.

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