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If we assume that energy was first absorbed by q1 and then released by it which then reaches q2 , the same as , electric field not being absorbable and somehow magically gives the energy to q2 and q1 with no reduction as seems by assuming that charge can absorb it. THE WHOLE SITUATION IS MADE TO BE STATIC

Energy gained by q2 in $t$ time $= (E/4(22/7)(2d)^2)+(E/4×(22/7)d^2)$

Which is different from assuming there is nothing like absorbtion, as it gives different value of energy gained by q2 in $t$ time $= (E/4(22/7)d^2)×1/4(22/7)d^2 + (E/4(22/7)d^2)$

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EDIT: i think a more close assumption for that is , Electric fields do contain energy and can also pass through charges body and can also get absorbed by it But the absorption of that energy only happens when the field actually accelerate any charge(work done by producing motion) OR charge can neither accumulate more or less energy density or electric field density. Take electric field strength(electric flux) proportional to Energy density

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    $\begingroup$ I see your question has a close vote due to lack of clarity. I think I understand what you are trying to ask, but try adding more words to make it more clear. To me it sounds like you're thinking that if energy is really in the field, then work done on Q1 by Q should remove energy from the field thereby reducing the energy available for Q to do work on Q2. But the big flaw in your question might be to start using your own made up model and assumptions as if they were correct and then asking a question based on it. $\endgroup$
    – DKNguyen
    Commented Jun 26, 2021 at 19:31
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    $\begingroup$ You might want to distinguish between when you are talking about the electric field produced by any single charge on its own and the net electric field of all charges in your question. $\endgroup$
    – DKNguyen
    Commented Jun 26, 2021 at 19:42
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    $\begingroup$ I agree with @dknguyen that you should rephrase your question for it is very obscure now but I think it is worth rewording using mathjax. $\endgroup$
    – hyportnex
    Commented Jun 26, 2021 at 19:44
  • $\begingroup$ I have added the electric field of both Q anf q1 to provide energy to q2 in both cases means that i have considered the net field or net energy in both cases $\endgroup$
    – Predaking Askboss
    Commented Jun 27, 2021 at 15:59

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Taken in the context of your previous question (Can electric field pass through a charge particle so that the charge particle behind the one can get energy?), it seems you're envisioning an electric charge as continuously generating a "flow" of outgoing electric field, and that this flow has associated to it an energy flux. The former point may be arguable and up to a choice of metaphysical perspective (particularly given that changes propagate at $c$), but the latter is definitively not in line with standard E&M. The flow/flux of energy associated to electromagnetic fields is exactly what's described by the Poynting vector:

$$\vec S = \frac{1}{\mu_0} \vec E \times \vec B $$

This encodes both the magnitude and direction of energy flow through electromagnetic fields. Notice that it is identically zero in any electrostatic setup (as in your scenario, with stationary charges), where $\vec B = 0$. This means that there is no energy flow in electrostatics, and your charges aren't accumulating energy over time simply by virtue of being in each other's presence. Once you allow the charges to start moving in response to each other's fields, however, this situation will change as the fields do work on the charges and the charges radiate.

Closely related to the above is energy we associate with the field itself through the energy density, $$ u = \frac{1}{2}\left( \varepsilon_0 E^2 + \frac{1}{\mu_0}B^2 \right) .$$ This quantity encodes both the interaction energies of the various charges sourcing the fields, as well as energy of propagating electromagnetic waves which can exist in the absence of sources. From this one infers that the electric field does indeed contain energy, but notice that in static scenarios with stationary charges, the energy density at each point in space is constant over time, so once again there is no flow of energy between our static charges. This association of energy to the field is problematic in the case of literal point charges (see What is the energy of a single charge system?)-- an indication we must accept that ideal point charges are an approximation and something changes at very small scales-- but your scenario can be considered just as well with finite spheres of surface charge, and the essential picture above does not change at all.

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  • $\begingroup$ You can always add an arbitrary solenoidal (divergence-free) vector field to the Poynting vector without any difference in measurable quantities; energy conservation cannot be measured/verified locally and that restriction includes energy density. Even if E and B are static fields the $E \times B$ is circulating while no charges are moving. $\endgroup$
    – hyportnex
    Commented Jun 26, 2021 at 19:53
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    $\begingroup$ Were Newtonian energy flow and conservation the only properties constraining the preferred form of $\vec S$, I'd agree, though two additional considerations indicate otherwise: relativistic considerations of the stress-energy tensor (see Jackson, section 12.10) and the experimental verification and utilization of radiation pressure as modeled by $\frac{1}{c^2} \vec S$ representing momentum density of the fields. Nonetheless, I've edited my answer to qualify "not in line with standard E&M". $\endgroup$
    – jawheele
    Commented Jun 26, 2021 at 20:27
  • $\begingroup$ @hyportnex "No charges are moving" in electrostatics means there is no magnetic field. You're right that I may have been using the term electrostatics loosely, however, to include this implicitly (I think this is fairly standard usage, distinguished from magnetostatics, but perhaps not universal). In any event, it's certainly the scenario relevant to OP's considerations. $\endgroup$
    – jawheele
    Commented Jun 26, 2021 at 20:34
  • $\begingroup$ you can have a stationary (nonmoving) electric charge sitting in the static magnetic field of a permanent ferromagnet, your $E \times B \ne 0$ circulating without any measurable effect on anything. $\endgroup$
    – hyportnex
    Commented Jun 26, 2021 at 21:46
  • $\begingroup$ @hyportnex Classically, a permanent ferromagnet would not fall under the purview of electrostatics (one could only model it via current loops, i.e. moving charges), but fine. It seems you're just taking issue with the notion that energy can be flowing through the fields without "doing" anything? A packet of light propagating in a vacuum doesn't have any measurable effect on anything, either (until you put charges in its way, and one could say the same about your setup)-- do you take issue with describing that as a flow of energy? $\endgroup$
    – jawheele
    Commented Jun 26, 2021 at 22:33

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