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Assuming we have a charge existing in an electric field, then we have a work done on this charge called the internal work, if we apply an external electric field why does it have to be in the opposite direction of the internal one? (did we just assume it this way or what?), and why does the external electric field equal to the internal one (in magnitude)? in other words, i'm having trouble with the following laws :

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Why does the electric field in the case of internal work and external work has the same magnitude?

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  • $\begingroup$ > "then we have a work done on this charge called the internal work, if we apply an external electric field why does it have to be in the opposite direction of the internal one?" This is odd, where did your read it? What is the "internal electric field"? $\endgroup$ – Ján Lalinský Nov 21 '17 at 15:10
  • $\begingroup$ The internal electric field is the one causing repulsion or attraction between charges and produces internal work, the external electric field is caused by any source outside the system of charges and produces external work on the charge in the system, my question is why are the internal and external fields are equal in magnitude and why are they in opposite directions ? $\endgroup$ – Mohamed A. Nov 21 '17 at 17:17
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Assuming I understand the question being asked correctly, I think the external work in this case is the reduction in work potential, but there is some context I am not quite getting here.

Imagine two charged objects in the box, one positive, and one negative. In the initial state, there is potential work that can be extracted there, because there is a force differential. Imagine one of the charged objects is attached to a pump lever - as the objects moves, it operates the pump. So that is our work. (This isn't necessary, the energy will be emitted either way, but it may help visualize the process).

I think the second, negative equation might represent potential work, the work we could have gotten out of the system - we use up the potential work, so that value is negative. As the objects move together, the potential work decreases, in proportion to the work extracted (exact proportion, in an ideal frictionless problem).

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  • $\begingroup$ Thanks for your reply, but i don't think this answered my question $\endgroup$ – Mohamed A. Nov 22 '17 at 20:29

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