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I'm trying to derive the point charge equation for voltage by integrating the point charge equation for an electric field over distance ($dr$) traversed:$ \int (KQ/r^2)\cdot dr$

This is my reasoning:

1) Take a source charge at $r = 0$ and a point charge at $r=-\infty$

2) Assume that both charges are positive. Therefore, as the point charge moves from $-\infty$ to $r$, an electric field acts against the charge, decreasing its kinetic energy.

3) Account for the force and distance being in opposite directions (hence the dot product $(KQ/r^2)\cdot dr$) and set $cos\theta$ to $cos(180)$

4) Setup the energy integral: $$ \Delta KE = \int_{-\infty}^{r} (KQ/r^2) dr * cos(180) $$

4 Multiply by $-1$ to find delta potential energy. My answer is $-KQ/r$ and not the actual $KQ/r$

Why?

$$ \Delta V = -1 \int_{-\infty}^{r} (KQ/r^2) dr * cos(180) $$

The math:

$$ \Delta V = -1 * cos(180)\int_{-\infty}^{r} (KQ/r^2) dr $$ $$ \Delta V = \int_{-\infty}^{r} (KQ/r^2) dr $$ $-\infty$ might as well be $\infty$ $$ \Delta V = \frac{-KQ}{r} \Big|_\infty^r $$ $$ \frac{-KQ}{r} - \frac{-KQ}{\infty} $$

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Thanks laying out your work so neatly in the question.

I think the solution is the following

$$\Delta KE= \int_{r_a}^{r_b}{ KQq \over r^2} dr$$

where $r_a$ is the initial position and $r_b$ is the final position (and I have added $q$ as the charge of the point charge).

so, for example, if the point charge goes from $r$ to $2r$ we have two positive charges and the change in kinetic energy will be positive as the point charge is pushed away from the source chare. - lets try that in the equation above....

$$\Delta KE= \int_{r}^{2r}{ KQq \over r^2} dr$$

$$ = -\left[{KQq \over r}\right]_r^{2r} = -\left({KQq \over 2r}-{KQq \over r}\right)$$

$$ = -{KQq \over r}\left({1 \over 2} - 1\right) = -{KQq \over r}\left({-1 \over 2}\right) = + {KQq \over 2r}$$

so we expected a positive $\Delta KE$ and we got one.

Now this equation for kinetic energy works from any $r_a$ to any other $r_b$. If we go in the positive direction where $r_b > r_a$ it works fine as we saw for $r$ to $2r$. If we go in the other direction towards the origin (so $r_b < r_a$) we do not need to put in the $cos 180$ term because that is looked after in the integral because

$$\int_{r_a}^{r_b}{ KQq \over r^2} dr = -\int_{r_b}^{r_a}{ KQq \over r^2} dr$$

[more generally of course $\int_a^b f(x)dx=-\int_b^a f(x)dx$ ]

Thus the solution to your question is that you do not need the $cos(180)$ term which is $-1$ - take this out and your solution is perfect.

apologies that my first answer missed this - thanks for showing more working.

so to put it all together

$$\Delta PE = - \Delta KE = -\int_{\infty}^{r}{ KQq \over r^2} dr$$ $$ = +\left[{KQq \over r}\right]_\infty^r = {KQq \over r}-0$$

to get the potential we divide by $q$ the charge of the point charge.

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  • $\begingroup$ Could you show your steps -- the actual integration? $\endgroup$ – Muno Dec 14 '14 at 2:21
  • $\begingroup$ Ah ok, sorry I see from your last edit... will edit answer $\endgroup$ – tom Dec 14 '14 at 2:33
  • $\begingroup$ @user21945 - sorry I missed the solution in my first post - I hope this answer makes sense. - One other comment is that in your question we would normally go from $+\infty$ to $r$, which saves having to go through the origin on the way.... - but this does not change the logic of your working in the question at all. $\endgroup$ – tom Dec 14 '14 at 2:50
  • $\begingroup$ many thanks for your help. I still can't understand why the identity takes cos 180 -- the dot product -- into account? By your logic, the right-hand portion of the identity also gives kinetic energy. So we have multiply it by -1 before we can get potential energy. That gives us the integral on your first LATEX line, and that integral doesn't evaluate to a positive value. Thanks $\endgroup$ – Muno Dec 14 '14 at 5:36
  • $\begingroup$ @user21945 more edits just gone in at the top - see what you think - comment if it does not make sense. $\endgroup$ – tom Dec 14 '14 at 8:22

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