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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 bit is where my confusion lies. To get the electric potential equation, we must multiply by $r$ to get $F r = W$.

But is this not the distance away from the charge and the work done in moving the charge through that distance, as opposed to the distance from infinity?

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For a constant force, we can obtain the work by multiplying the force by the distance over which it is applied.

If, however, we have a non-constant force, then the work is $$W = \int \vec{F} \cdot \mathrm{d}\vec{s}.$$
In your case, this translates to $$W = - \int_{\infty}^{r} \frac{KQ_2}{r'^2} \mathrm{d}r'={\frac{KQ_2}{r'}}\bigg|^r_{\infty}=\frac{KQ_2}{r}.$$

So the fact that you obtain the right answer is by chance because you are not calculating your work correctly.

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  • $\begingroup$ Ah I see so. The easiness of what appeared to be the correct derivation drew me in too much! $\endgroup$
    – Jake
    Commented May 22, 2017 at 21:13
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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

$$ W = \int {\rm d}{\bf l}\cdot \frac{kQq}{r^2}\hat{\bf r} = \int_{r}^{+\infty}{\rm d}{r}\frac{kQq}{r^2}= \frac{kQq}{r} $$

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  • $\begingroup$ so is it just a coincidence that it looks like my derivation? $\endgroup$
    – Jake
    Commented May 22, 2017 at 21:07
  • $\begingroup$ @JakeRose You can think of this as moving the charge by small amounts $dr$, and them adding them all up. You're missing that part $\endgroup$
    – caverac
    Commented May 22, 2017 at 21:09
  • $\begingroup$ I have not reached the level of these type of integrals so it is confusing me slightly $\endgroup$
    – Jake
    Commented May 22, 2017 at 21:12
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Electric 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: $$V = − \int\mathbf{E}\cdot\mathrm{d}\mathbf{s} =−\int E\cos\theta \mathrm{d}s.$$ If the stationary charge is positive and if the test charge is moved from infinity to point P, then $$V = − \int E \cos180° \mathrm{d}s = − KQ_1 \int \cos180°\frac{1}{r^2} \mathrm{d}s.$$

Reference: https://www.physicsforums.com/threads/derivation-of-electric-potential-due-to-point-charge.528528/

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