Why is an electric force conservative?

1 Answer
Mar 17, 2018

Answer:

#\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.