Why is an electric force conservative?

Question

Leo Abeyta · Accepted 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.

We receive: #F(\vec r) = k(q_1q_2)/(r^2)\hatr# If the work done along a path that begins and ends at the same point is zero, then force is conservative. Task is assigned as: #\oint_l F(\vec r) * dl = \oint_l k(q_1q_2)/r^2 \hatr* dl# where #dl \equiv dr\hatr + rd heta \hat heta + r sin heta d heta \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 heta \hat heta + r sin heta d heta \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 demonstrated that the electric force is conservative since it does no work along a path that begins and ends at the same place. There are three statements that are equivalent to demonstrating that a force is conservative: #\grad xx \vec F = \vec 0# #W = \oint_l \vec F*d\vecl = 0# #\vec F = -\grad \Phi# I provided an example of (2) in the response.

Charles Aeschliman · Answer

undefined An electric force is conservative because the work done by the force on a charged particle depends only on the initial and final positions of the particle, not on the specific path taken. This is a consequence of the electrostatic potential energy associated with the electric force, and it follows the principles of conservative forces in physics.