# Why electrostatic field is conservative?

1. I know, curl of static electric field (i.e. electric field generated by a time independent charge distribution) is zero, I know it can be derived from a scalar potential etc. But I want some intuitive arguments.

2. Actually, I have a point, consider a charged particle having charge $Q$ and mass $m$ is placed in an electrostatic field, so it will experience some force on it exerted by the field, so it will accelerate in some direction, now according to Maxwell's E-M theory (though I hardly understand this theory as it was not included in our 2nd year course) accelerating charged particle always radiates some energy. So, between any two points of space, how one can say that sum of potential and kinetic energy will be same as it also loses some energy as E-M wave?

• Quite simply, we neglect the energy loss due to radiation because it's usually small and generally quite complicated. In more advanced treatments of the subject, we take this energy loss into consideration. Aug 18, 2017 at 22:33
• "Why" questions are a funny business in physics. Physics describes the law of nature, but not "why" nature works like this. If you haven't seen it, have a look at Feynman's answer why magnets interact with each another. Aug 19, 2017 at 15:59

consider a charged particle having charge Q and mass m is placed in an electrostatic field, so it will experience some force on it exerted by the field, so it will accelerate in some direction

It could be kept from accelerating, for example by being attached to some more massive object.

now according to Maxwell's E-M theory(though I hardly understand this theory as it was not included in our 2nd year course) accelerating charged particle always radiates some energy.

When we talk about the electrostatic field, and probing the field with a charged particle, we always talk about the quasi-static limit. That is we take the limit of moving the particle very slowly. So slowly that its acceleration doesn't produce any radiation (or even a magnetic field).

Indeed if a charged particle is moving (or accelerating) quickly, then the quasi-static limit no longer applies, and we can no longer rely on the results obtained by considering the electrostatic potential.

The reason the curl of a static field is zero is actually simply the Maxwell-Faraday law :

$\nabla \times \mathbf E = \partial_t \mathbf H = \mathbf 0$

if the fields are considered static.

So if you have the intuition of this law, you have the intuition of your question

• "accelerating charged particle always radiates some energy. So, between any two points of space, how one can say that sum of potential and kinetic energy will be same as it also loses some energy as E-M wave?" I found no logic(not via. some mathematics) to discard this question, may be I am asking silly question or there may be something I am misinterpreting. It will be very helpful if you point out the mistake I've done in thinking
– sid
Aug 18, 2017 at 18:02