0
$\begingroup$

Feynman claims is his lectures the following statement:

Suppose we have two charges $q_1$ and $q_2$ separated by the distance $r_{12}$. There is some energy in the system, because a certain amount of work was required to bring the charges together. We have already calculated the work done in bringing two charges together from a large distance. It is $$\frac{q_1q_2}{4\pi\epsilon _0 r_{12}}.$$

I see some contradiction in the statement above with the definition of work. If we have some object with a force $\mathbf{F}$ applied to it, then the work $W$ is obtained by multiplying $\mathbf{F}$ with $\mathbf{s}$ where $\mathbf{s}$ is the displacement of an object in the direction of the force. $$W=\mathbf{F}\cdot\mathbf{s}$$

Clearly the distance $s$ or as per Feynman $r_{12}$ is the displacement, i.e. the distance the charge has travelled through space to reach the current position (the position is characterized by $r_{12}$). This displacement is obviously not the distance between two charges.

I see an important clause in the Feynman's declaration––he says the charges were brought together from a large distance, which means we can assume the force of, say, $q_1$ on $q_2$ was reasonably small in the beginning and can be neglected, yielding the work done for large distances between charges to be zero.

However, this assumption doesn't explain what happens, when two charges get reasonably close to each other, so that the force can no longer be neglected. And of course it doesn't explain to me, why Feynman claims the net displacement of the traveling charge is simply the distance between it and another charge he's experiencing force from.

I am very likely missing some important detail, but at the moment I can't see it.

$\endgroup$
1
$\begingroup$

$r_{12}$ is the distance between the charges.

$r_{12}$ is not the distance that the charge has travelled.

To calculate the work done to move the two charges together we would use

$\int_\infty^a F.ds$

which is like the formula you have in your question - your formula works if $F$ is constant - here we need to use an integral as the force changes as $r$ changes.

in this case

$\int_\infty ^{r_{12}} {1 \over 4 \pi \epsilon _0} {q_1q_2\over r^2} .dr$

$= {1 \over 4 \pi \epsilon _0}[{-q_1q_2 \over r}]_\infty ^{r_{12}} = {- 1 \over 4 \pi \epsilon _0} {q_1q_2 \over r_{12}}$

does this help?

$\endgroup$
  • 1
    $\begingroup$ Isn't it 1/4πε ? $\endgroup$ – George Smyridis Nov 17 '14 at 17:58
  • $\begingroup$ @GeorgeSmyridis yes sorry - thanks for spotting that ..... editted and fixed $\endgroup$ – tom Nov 17 '14 at 17:59
  • $\begingroup$ Indeed, I was just thinking that I had omitted the fact the force is not constant, when you threw the answer. Thanks, that helps. $\endgroup$ – Dmitry Kazakov Nov 17 '14 at 18:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.