Electric potential derivation The definition of electric potential is the work done per unit charge in moving the charge from infinity to that distance.
Now from Coulomb's law $f=\frac{KQ_1Q_2}{r^2}$
So we can now rearrange for the electric field strength.
$\frac{F}{Q_1}=\frac{KQ_2}{r^2}$
The next bt is where my confusion lies. To get the electric potential equation we clearly have to multiply by r to get F x R=W. But is this not the distance away from the charge and so the work done in moving the charge through that distance as opposed to the distance from infinity?
 A: For a constant force, we can obtain the work by multiplying the force by the distance it is applied over.
If we however have a non-constant force, then the work is $W = \int \vec{F} d\vec{s}$.
In your case, this translates to $W = - \int_{\infty}^{r} \frac{KQ_2}{r'^2} dr'={\frac{KQ_2}{r'}}|^r_{\infty}=\frac{KQ_2}{r}$.
So the fact that you obtain the right answer needs to be called chance because you are not calculating your work correctly.
A: In general, the work done by a force ${\bf F}$ is
$$
W_{ab} = \int_a^b {\rm d}{\bf l}\cdot {\bf F}
$$
The expression $W = RF$ is only true if the force is constant, which is not the case here. As a matter of fact, the work in bringing the particle from infinity to a distance $r$ is
$$
W = \int {\rm d}{\bf l}\cdot \frac{kQq}{r^2}\hat{\bf r} = \int_{r}^{+\infty}{\rm d}{r}\frac{kQq}{r^2}= \frac{kQq}{r}
$$
A: lectric potential is the work done in moving a unit charge from infinity to a point in an electric field. Electric potential due to point charge: V = − ∫ ⃗ E ⋅ d ⃗ s V=−∫E→⋅ds→ V = − ∫ E ⋅ d s c o s ϑ V=−∫E⋅dscosϑ if the stationary charge is positive and if the test charge is is moved from infinity to point P then V = − ∫ E ⋅ d s c o s 180 V=−∫E⋅dscos180 V = − K Q ∫ 1 r 2 d s c o s 180 V=−KQ∫1r2dscos180
Reference https://www.physicsforums.com/threads/derivation-of-electric-potential-due-to-point-charge.528528/
