Why is C not involved in the work to move this charge?

[timn]

Homework Statement

Calculate the work required to bring a positive ion of charge e from infinity to a distance s from another ion of the same charge.

e<--s--->e

Homework Equations

$$E = C \frac{q_1 q_2}{r^2}$$

The Attempt at a Solution

Integrate the energy from infinity to s besides the fixed ion, which resides on origin.

$$W = \int_s^\infty E dr = C\frac{e^2}{s}$$

The answer should not have Coloumb's constant in there. What did I miss?

(Note that this only is a part of problem 1.8 from Purcell's Electricity and Magnetism. I might have missed something in my simplification.)

 

[Pengwuino]

Why would it not have Coulomb's constant? It basically tells you the strength of the electrical interaction. A stronger (weaker) force would require more (less) energy to bring the particle in from infinity, so there must be a dependence on Coulomb's constant.

 

[timn]

Pengwuino: Perhaps my simplification was wrong then? The whole question is:

Calculate the potential energy, per ion, for an infinite one-dimensional ionic crystal, that is, a row of equally spaced charges of magnitude e and alternating sign.

The suggested solution does not seem to care about Coloumb's constant. See "1-8" on the first page here: http://www.ssl.berkeley.edu/H7B/H7B_files/h7b_hw4.pdf

 

[lanedance]

i think your derivation is fine, some units systems or simplifications for calcs just assume C=1 and put it back in when numbers are required

 

[Pengwuino]

i think your derivation is fine, some units systems or simplifications for calcs just assume C=1 and put it back in when numbers are required

This is probably the case.

http://en.wikipedia.org/wiki/Gaussian_units

 

[timn]

I see. I just assumed that SI units were used. Thank you!

 

[flyingpig]

I got $$-C\frac{e^2}{s}$$, negative work, which make sense because you are pushing a like charge to antoher like charge. Notice how you said you were going from infinity to s?

 

[Pengwuino]

I got $$-C\frac{e^2}{s}$$, negative work, which make sense because you are pushing a like charge to antoher like charge. Notice how you said you were going from infinity to s?

If you do negative work, you'd be putting it into a bound state which is not what you have if you have similar charges. If you have to push the particle inward toward the ion and subsequently release it, it's going to fly off.

 

[flyingpig]

If you do negative work, you'd be putting it into a bound state which is not what you have if you have similar charges. If you have to push the particle inward toward the ion and subsequently release it, it's going to fly off.

What's bound state? He has two + charges and he is pushing them to each other, that's minus work.

 

[Pengwuino]

What's bound state? He has two + charges and he is pushing them to each other, that's minus work.

I believe that would be negative work done on whatever was actually doing the pushing. Positive work is done to the actual electron

 

[timn]

The ion doesn't move there by itself, so the work required to push the ion is positive.

 

[flyingpig]

I believe that would be negative work done on whatever was actually doing the pushing. Positive work is done to the actual electron

But it says it is positive charged.

I guess I am learning this too then.

Here is what I did to get a minus work

$$\sum W = \int_{\infty}^{s} \vec{E}\cdot d\vec{r} = Ce^2\lim_{t\to \infty} \int_{t}^{s}\frac{1}{r^2}dr= -Ce^2\lim_{t\to \infty} \frac{1}{s}-\frac{1}{t} = \frac{-Ce^2}{s}$$

 

[Pengwuino]

But it says it is positive charged.

I guess I am learning this too then.

Here is what I did to get a minus work

$$\sum W = \int_{\infty}^{s} \vec{E}\cdot d\vec{r} = Ce^2\lim_{t\to \infty} \int_{t}^{s}\frac{1}{r^2}dr= -Ce^2\lim_{t\to \infty} \frac{1}{s}-\frac{1}{t} = \frac{-Ce^2}{s}$$

The problem is your 'dr' points opposite of the field. The relevant dot-product goes like $\vec{E} \cdot -d\vec{r}$ because your $d\vec x = - dr \hat{r}$