# Why is an electric force conservative?

Mar 17, 2018

$\setminus {\oint}_{l} k \frac{{q}_{1} {q}_{2}}{r} ^ 2 \setminus \hat{r} \cdot \mathrm{dl} = 0$ ; from point $a$ to point $a$

The work done by the electric force along a path starting and ending at the same point $a$ is 0. Hence, the electric force is conservative.

#### Explanation:

We are given:

$F \left(\setminus \vec{r}\right) = k \frac{{q}_{1} {q}_{2}}{{r}^{2}} \setminus \hat{r}$

Force is conservative if work done along a path that starts and ends at the same point is 0.

Work is given as:

$\setminus {\oint}_{l} F \left(\setminus \vec{r}\right) \cdot \mathrm{dl} = \setminus {\oint}_{l} k \frac{{q}_{1} {q}_{2}}{r} ^ 2 \setminus \hat{r} \cdot \mathrm{dl}$
where $\mathrm{dl} \setminus \equiv \mathrm{dr} \setminus \hat{r} + r d \setminus \theta \setminus \hat{\setminus} \theta + r \sin \setminus \theta d \setminus \theta \setminus \hat{\setminus} \phi$

Let's make our path start and end at the same point $a$ to create a closed path:

$= \setminus {\int}_{a}^{a} k \frac{{q}_{1} {q}_{2}}{r} ^ 2 \setminus \hat{r} \cdot \left(\mathrm{dr} \setminus \hat{r} + r d \setminus \theta \setminus \hat{\setminus} \theta + r \sin \setminus \theta d \setminus \theta \setminus \hat{\setminus} \phi\right)$

$= \setminus {\int}_{a}^{a} k \frac{{q}_{1} {q}_{2}}{r} ^ 2 \mathrm{dr}$

$= k {q}_{1} {q}_{2} \setminus {\int}_{a}^{a} \frac{1}{r} ^ 2 \mathrm{dr}$

$= - 2 k {q}_{1} {q}_{2} {\left[\frac{1}{r} ^ 3\right]}_{a}^{a}$

$= - 2 k {q}_{1} {q}_{2} \left(\frac{1}{a} ^ 3 - \frac{1}{a} ^ 3\right)$

$= - 2 k {q}_{1} {q}_{2} \left(0\right) = 0$

We have shown that the work done by the electric force along a path starting and ending at the same point is 0. Hence, the electric force is conservative.

=======================EDIT=======================
There are three equivalent statements for showing a force is conservative:

• (1) The cross product of the force vector is 0:

$\setminus \nabla \times \setminus \vec{F} = \setminus \vec{0}$

• (2) The work done by the force vector along a path that starts and ends at the same point is 0:

$W = \setminus {\oint}_{l} \setminus \vec{F} \cdot d \setminus \vec{l} = 0$

• (3) A force vector is the negative gradient of the potential:

$\setminus \vec{F} = - \setminus \nabla \setminus \Phi$

I demonstrated (2) in the answer given.