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Let take two positive charges $q_1$ and $q_2$. There are four points. $q_1$ was initially located at (a), and $q_2$ was at (d). Distance between (a) to (d) is taken 1 lightsecond. Distance between (b) and (c) is also about 1 lightsecond (but a little less)

(a)...(b)................(c)...(d)

$q_1$.................................$q_2$

$q_1 $is moved from (a) to (b) toward $q_2$, and in a similar way, $q_2$ is moved from (d) to (c) toward $q_1$. Let say they both were pushed for half a second. Due to speed of electric field, during the push, $q_1$ and $q_2$ will not feel any change in position of each other. For the force applied during the time of the push, we assumed that $q_2$ was at (d) for $q_1$, and for $q_2$, $q_1$ was at (a). The confusion here is that the energy transferred that $q_2$ got (as a signal) would be greater than the energy needed to push $q_1$, because for $q_2$, $q_1$ is moved, and the distance between them is less than the distance that of $q_1$ felt during push.

(For Energy given) $\text{Force on } q_1 = \dfrac{Kq_1q_2}{r^2 }$

(For energy output) $\text{Force on } q_2 = \dfrac{Kq_1q_2}{r'^2 }$

When we think about $r$, we found that it would be somewhere between distance of (a) - (d) and (b) - (d), as the electric field of $q_2$ will be there the same for 1 second. But for $r'$, it is between (a)-(d) and (b) - (c), so $r>r'$. Therefore, the force on $q_2$ is greater.

Assume that you are $q_1$ and are pushed for 1/2 second over a very short distance. You will find that field of $q_2$ were the same there for that time period as a result of the velocity of the electric field. So, the $q_2$ for $q_1$ was at (d) during push. Thus, $q_1$'s average distance from $q_2$ was somewhere between (a) - (d) and (b) - (d).

However, for $r'$ taken for force on $q_2$ for receiving signal was different from $r$. Here, I am $q_2$ and was moved at the same time, but when your changed electric field reached me after 1 second, I was also moved closer. So, for me, you moved closer, and I also moved a little closer to you, which makes my average distance from $q_1$ less than $r$.

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    $\begingroup$ Hello! It is preferable to use MathJax (LaTeX) to display formulas. You can find a tutorial at MathJax basic tutorial and quick reference. Please edit your question accordingly. Thanks! $\endgroup$ – Jonas Apr 8 at 21:31
  • $\begingroup$ Yes i repeated a thing few times like showing that r is greater than r' so that one can get why i said that. Describing the whole situation. Ya i didn't know more about equations that's why i tried to use statements $\endgroup$ – Predaking Askboss Apr 9 at 16:09
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This does not cause any problems with the conservation of energy. The conservation of energy in EM holds locally and is described by Poynting’s theorem. As the theorem says, if the matter gains energy then the fields lose it and vice versa.

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  • $\begingroup$ You told me about field to be energetic kind of things but what about Coulombs law. The force acted in this law is due to product of charges distance between them. And force for a time gives a an amount of energy as displacement ocuurs on object. The above things i discussed above were based on this law. And if this is wrong, are you saying that culombs law was wrong. Please confirm that also $\endgroup$ – Predaking Askboss Apr 9 at 16:04
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    $\begingroup$ Absolutely. Coulomb’s law is only valid for electrostatic situations, which this is not $\endgroup$ – Dale Apr 9 at 16:39
  • $\begingroup$ Even you take the Coulombs law valid or not for moving charges Electric fux present there and charge's magnitude will show the amount of electric force (am i right). So it is the cause for force on q1 and q2 in this case. And as i moved q2 closer to q1 it got more average flux for t time compared to that flux q1 got when it was pushed.( what is wrong here) $\endgroup$ – Predaking Askboss Apr 9 at 20:21
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    $\begingroup$ Nothing is wrong. The mechanical energy is indeed not conserved. The difference in mechanical energy is equal to the change in the electromagnetic field energy, as described by Poynting’s theorem. Just like I said in the answer. $\endgroup$ – Dale Apr 10 at 0:31
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    $\begingroup$ @Predaking Askboss first, electric flux does not determine the force on a charge. That is the electric field, which is not the same as the electric flux. Second, you still seem to be ignoring the energy in the field as described by Poynting’s theorem, despite me mentioning it several times. Anyway, since there has been no upvotes it is not worth spending any more time on it so I am done commenting here. $\endgroup$ – Dale Apr 11 at 17:27

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