Question

Why is an electric force conservative?

Answer 1

`\oint_lk(q_1q_2)/r^2 \hatr* 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(\vec r) = k(q_1q_2)/(r^2)\hatr`

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

Work is given as:

`\oint_l F(\vec r) * dl = \oint_l k(q_1q_2)/r^2 \hatr* dl`
where `dl \equiv dr\hatr + rd\theta \hat\theta + r sin\theta d\theta \hat\phi`

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

`= \int_a^a k(q_1q_2)/r^2 \hatr * (dr\hatr + rd\theta \hat\theta + r sin\theta d\theta \hat\phi)`

`= \int_a^a k(q_1q_2)/r^2 dr`

`= kq_1q_2\int_a^a 1/r^2 dr`

`= -2kq_1q_2 [1/r^3]_(a)^(a)`

`= -2kq_1q_2 (1/a^3 - 1/a^3)`

`= -2kq_1q_2 (0) = 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:

`\grad xx \vec F = \vec 0`

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

`W = \oint_l \vec F*d\vecl = 0`

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

`\vec F = -\grad \Phi`

I demonstrated (2) in the answer given.