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

• timn
In summary: 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}
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.)

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.

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

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

lanedance said:
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

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

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?

flyingpig said:
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.

Pengwuino said:
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.

flyingpig said:
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

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

Pengwuino said:
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

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

flyingpig said:
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 $d\vec x = - dr \hat{r}$

Last edited:

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

The letter C represents the capacitance in a circuit, which is a measure of how much charge can be stored per unit of voltage. In this case, the charge is being moved by an external force, such as a battery, and not by the capacitor itself. Therefore, the capacitance of the circuit does not affect the work being done to move the charge.

## 2. Is C irrelevant in the movement of this charge?

No, C is not irrelevant in this scenario. While the capacitance does not directly affect the work being done to move the charge, it does play a role in storing and releasing the charge once it has been moved. Without the capacitor, the charge would not be able to be moved as efficiently.

## 3. Can the value of C impact the amount of work needed to move the charge?

Yes, the value of C can indirectly impact the amount of work needed to move the charge. A higher capacitance means that more charge can be stored per unit of voltage, so a larger capacitor may require more work to move the same amount of charge compared to a smaller capacitor with a lower capacitance.

## 4. How does the presence of a capacitor affect the overall work done in a circuit?

The presence of a capacitor can affect the overall work done in a circuit by storing and releasing electrical energy as the charge is moved. This can impact the efficiency and effectiveness of the circuit, but the work being done by an external force to move the charge is not directly affected by the capacitor.

## 5. Can the work done to move the charge be completely eliminated by removing the capacitor?

No, even without a capacitor, there will still be work required to move the charge in a circuit. The capacitor may be able to store some of the charge and release it later, but the initial work to move the charge will still need to be done by an external force, such as a battery. Removing the capacitor may impact the efficiency of the circuit, but it will not eliminate the work required to move the charge.

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