Resistance and Ohm's Law Problem

Click For Summary

Homework Help Overview

The problem involves calculating the resistance of a solid cube of silver and determining the average drift speed of electrons under a potential difference. The context is centered around concepts of electrical resistance and properties of materials, specifically focusing on silver's resistivity and electron behavior.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to look up the resistivity of silver and how to use the given formulas for resistance. There are questions about how to derive necessary dimensions from mass and density, and uncertainty regarding the appropriate formula for calculating drift speed.

Discussion Status

Some participants have suggested looking up resistivity values and have confirmed the relevance of certain formulas. There is ongoing exploration of how to apply the equations to the problem, particularly regarding the drift speed calculation.

Contextual Notes

Participants note the presence of multiple unknown variables and the need for specific values, such as resistivity, to proceed with the calculations. There is also mention of the formulas available in the relevant section of the textbook, which may not explicitly cover drift speed.

clairez93
Messages
113
Reaction score
0

Homework Statement



A solid cube of silver (density = 10.5 g/cm^3) has a mass of 90.0 g. (a) What is the resistance between the opposite faces of the cube? (b) If there is one conduction electron for each silver atom, what is the average drift speed of electrons when a potential difference of 1.00 x 10^-5 V is applied to opposite faces? (The atomic number of silver is 47, and its molar mass is 107.87 g/mol)



Homework Equations



(listing all the equations given in the section this problem is relevant to)

J = \sigmaE
J = nqv_{d}
R = \ell / \sigmaA = \DeltaV / I
\rho = 1 / \sigma
R = \rho\ell / A

The Attempt at a Solution



I am completely clueless, at how to start. I think I'll have to use one of the R formulas somehow, but there seems to be too many unknown variables if I try to use any of those. Could someone help me get started?

The book's answer is (a) 777 n\Omega (b) 3.28 \mum/S
 
Physics news on Phys.org
Hi clairez93,


About there seeming to be too many unknowns, I would guess that they want you to look up the resistivity of silver; you should have a table of those values in your book. Once you have that, the other values in the problem will be all you need to work with.
 
R = \rho\ell / A
I am thinking this is the one I will need to use. For \rho I think I will need to look up resistivity. For ell and A, I will need to use the density and mass to get a volume number, I believe, correct?

I'm not sure how to figure out prat b for a drift speed though. I don't see a formula in the section the problem is supposed to belong to for a drift speed. Is there a certain formula I should be using?
 
clairez93 said:
R = \rho\ell / A
I am thinking this is the one I will need to use. For \rho I think I will need to look up resistivity. For ell and A, I will need to use the density and mass to get a volume number, I believe, correct?

That's sounds right.

I'm not sure how to figure out prat b for a drift speed though. I don't see a formula in the section the problem is supposed to belong to for a drift speed. Is there a certain formula I should be using?


One of the formulas you listed in the relevant equations has the drift speed in it. There is a very similar formula that has the current instead of the current density that might be a bit more straightforward.
 
Part A:

R = \rhor / A

Part B:

J = I / A = nqv_{d}

Hope this is still useful.
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 15 ·
Replies
15
Views
10K
  • · Replies 2 ·
Replies
2
Views
31K
  • · Replies 38 ·
2
Replies
38
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
6K
Replies
2
Views
7K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K