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The principle of conservation of electric charge (or law of conservation of electric charge) affirms that:

in a closed (isolated) system, which does not exchange any matter with the outside, the algebraic sum of all the positive and negative charges and negative charges in it remains unchanged in the time.

It represents a postulate of the physics.

We suppose that I have an isolated system with this drawing where there are three charges $q_1, q_2$ and $q_3$ with $q_2$ have positive sign where I know the value. In addendum I know $r_{12}$ and $r_{23}$; hence the $r_{12}+r_{23}=r_{13}$:

enter image description here

I know that the system is isolated and therefore named the $F$ the generic coulombian forces:

$$\sum_{\forall \, \text{coulombian force}} F =0.$$

Equivalentely:

\begin{cases} \dfrac{k_0|q_1q_2|}{r_{12}^2} =\dfrac{k_0|q_2q_3|}{r_{23}^2} \\ \dfrac{k_0|q_1q_2|}{r_{12}^2} =\dfrac{k_0|q_1q_3|}{r_{13}^2}\, \text{ or }\, \dfrac{k_0|q_2q_3|}{r_{23}^2} =\dfrac{k_0|q_1q_3|}{r_{13}^2}\\ \end{cases}

Now I will find the values of $q_1$ and $q_3$ knowing $q_2$.

My question is: If I have many charges in an isolated system is it possible to have this condition? $$\sum_{i=1}^N q_i=0 \tag 1$$ i.e. exist this condition?

Is it possible to apply in general the $(1)$ in the case of my drawing?

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  • $\begingroup$ You claim you want to know $q_1$ and $q_3$ given $q_2$, but you also claim you know $q_2$ and $q_3$. Can you clarify? $\endgroup$ Commented Oct 6, 2021 at 21:22
  • $\begingroup$ @ZeroTheHero I would to know if the relation $(1)$ (in general) is it true or false and why? After if it is true what is the reason? For your comment I want to know $q_1$ and $q_3$ given $q_2$. $\endgroup$
    – Sebastiano
    Commented Oct 6, 2021 at 21:30
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    $\begingroup$ (1) is false. The constant on the right hand side (it is a constant which depends on the initial conditions of each problem) not be 0. $\endgroup$ Commented Oct 6, 2021 at 23:38
  • $\begingroup$ @ZeroTheHero Please, can you add an exhaustive and detailed example for you last comment, please? THANKS. $\endgroup$
    – Sebastiano
    Commented Oct 7, 2021 at 18:57

2 Answers 2

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There is no law $$\sum q_i=0$$ if the system is closed its true that $$\sum q_i=Q=\text{ const.} $$

In your task somebody just put the three charges at his will.

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  • $\begingroup$ If $q_s$ is the charge of a scarf and a pen is also present; since the system is isolated it is obvious that the total charge if there is no rubbing with a pen $Q_{\text{tot.}}=0$. But if I rub the pen $q_p$ with the scarf it must not be $Q_{\text{tot.}}=q_p+q_s=0$? $\endgroup$
    – Sebastiano
    Commented Oct 5, 2021 at 21:38
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    $\begingroup$ if the charge in the beginning is qs, it will be qs in the end, since no charges are created, just exchanged $\endgroup$
    – trula
    Commented Oct 5, 2021 at 23:01
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    $\begingroup$ @Sebastiano there is no reason to believe the system had 0 net charge to start with. An isolated electron will conserve its charge but the system of the electron alone does not have 0 charge. $\endgroup$ Commented Oct 5, 2021 at 23:06
  • $\begingroup$ In the meantime thank you very much (+1) and @ZeroTheHero can you put a possible answer? Thanks also to you. $\endgroup$
    – Sebastiano
    Commented Oct 6, 2021 at 20:24
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To answer the specific questions:

Suppose $q_1=q_2=q_3=e$, the absolute value of the charge of the electron. Then clearly $q_1+q_2+q_3\ne 0$.

It still holds that $\sum \vec F=0$ since

  1. The net force on charge $q_1$ is $k_0q_1q_2\frac{\vec r_1-\vec r_2}{r_{12}^3}+ k_0q_1q_3\frac{\vec r_1-\vec r_3}{r_{13}^3}$,
  2. The net force on charge $q_2$ is $k_0q_1q_2\frac{\vec r_2-\vec r_1}{r_{12}^3}+ k_0q_2q_3\frac{\vec r_2-\vec r_3}{r_{23}^3}$,
  3. The net force on charge $q_3$ is $k_0q_1q_3\frac{\vec r_3-\vec r_1}{r_{13}^3}+ k_0q_2q_3\frac{\vec r_3-\vec r_2}{r_{23}^3}$.

Summing all the forces we get \begin{align} k_0\left( q_1q_2\frac{\vec r_1-\vec r_2}{r_{12}^3}+ q_1q_3\frac{\vec r_1-\vec r_3}{r_{13}^3} + q_2q_1\frac{\vec r_2-\vec r_1}{r_{12}^3}+ q_2q_3\frac{\vec r_2-\vec r_3}{r_{13}^3} +q_1q_3\frac{\vec r_3-\vec r_1}{r_{13}^3}+ q_2q_3\frac{\vec r_3-\vec r_2}{r_{23}^3} \right)=0 \end{align} since the first and third terms cancel, the second and fifth terms cancel, and the fourth and sixth terms cancel.

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  • $\begingroup$ Thank you very very much. I have appreciated. $\endgroup$
    – Sebastiano
    Commented Oct 7, 2021 at 22:06

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