Simple application of Coulomb's Law/Equation

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1. Mar 10, 2015

kostoglotov

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

Two pieces of copper weighing 10 grams each are 10 cm apart and 1/1000 electrons are transferred from one to the other. Find the force of electrostatic attraction between them.

2. Relevant equations

Molar mass Cu: 63.5 g/mol

$$\frac{1}{4\pi \epsilon_0} = 9 \times 10^9$$

$e = 1.6 \times 10^{-19} \ C$

Avagadro: $6 \times 10^{23}$

3. The attempt at a solution

Actually, I only have one minor thing I don't understand.

In my solution I have

$$F = 9 \times 10^9 \frac{-15.12 \times +15.12}{0.1^2}$$

Whereas the solution outlined in the book only uses one charge in the numerator

$$F = 9 \times 10^9 \frac{15.12}{0.1^2}$$

So my answer is out by a factor of 15.12.

I don't get it. Wouldn't the charge on one of the copper pieces be +15.12 C and the other -15.12 C and therefore q1 x q2 would equal $- 15.12^2$...?

2. Mar 10, 2015

Staff: Mentor

Can you show how you arrived at 15.12 C for the charge? It seems a bit small to me for 1/1000 of the electrons in a 10g chunk of copper (29 electrons per atom).

3. Mar 10, 2015

haruspex

I believe each copper atom is only considered to provide one free electron.

4. Mar 10, 2015

Staff: Mentor

5. Mar 10, 2015

kostoglotov

My reasoning is exactly the same as the solution as outlined in the text. No different. Number of moles of copper in a 10g piece multiplied by avagadro multiplied by 1/1000.

I get 15.12 C the book says 15.12 C. If we're both wrong, that's interesting, but what I'd like to know is: Should it be 15.12 or -15.12^2 in the numerator of Coulomb's Equation?

6. Mar 10, 2015

Staff: Mentor

If we're questioning the book's solution then I believe that it behooves us to examine the details. But perhaps that's just my picky nature

One could answer the question by considering the units of the equation... kQ/r2 yields a field strength (V/m or N/C) rather than a force.

7. Mar 10, 2015

kostoglotov

Sorry, I sounded a bit pushy there, I went back and edited my comment pretty much immediately after, realizing.

Interesting! Thanks for the reply. I should have thought of doing dimensional analysis on it. I really should have. Thanks again.

8. Mar 10, 2015

kostoglotov

The text shows a worked solution, not just an answer, so I can see where my work differs from the texts, and it differs only in that it has only 15.12 in the numerator of Coulomb's equation, not -15.12^2. :)

9. Mar 11, 2015

BvU

I don't understand. The book is wrong by a factor of about 15 x 109 and all we care about is a factor of 15 ? I understand kosto, but I feel with gneill a whole lot more.
What book is this and what does teacher say ?

10. Mar 11, 2015

kostoglotov

How is the book wrong by a factor of 15 x 10^9? I don't think it is. It gives an answer of about 2 x 10^14 N for the force between the charges.

It is Physics for Competitions by C. G Agrawal Vol II A Master Book for IIT, PMT, AIEEE, UPSEAT, CET, PET, JEE, BCECE

11. Mar 11, 2015

BvU

Ah, thanks for clarifying. The book has the right numerical value but the 2 was missing.

I was forgetting the 1 in a 1000 , and ended up comparing that wrong value to 2 x 10^14/ 15000 which is also the wrong idea of what the book might have come up with. Two wrongs never make a right and there you are talking nonsense.

Time to slow down !
Once again. Oh boy.

12. Mar 11, 2015

kostoglotov

LOL, I think I need to adopt the sorry folks signature. Though at the moment I can pass it off as being a noob.

I think next time I should just write out everything, especially since folks here seem to want to get more deeply into explanations than in other parts of the internet. Which is excellent. I didn't want to write a longer post originally because I didn't want to waste anybody's time, just cut to chase sort of thing.

Next time I need to include more of the working.

13. Mar 11, 2015

kostoglotov

No teacher, teaching myself right now. I imagine I'll be posting around here a lot

14. Mar 11, 2015

BvU

[/QUOTE]kudos and good luck. look forward to the next post