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 $\infty$ $$ \Delta V = \frac{-KQ}{r} \Big|_\infty^r $$ $$ \frac{-KQ}{r} - \frac{-KQ}{\infty} $$