# The force applied by two charges on each other when there is another charge nearby them

QUESTION 1

Let me give two simple scene of charged particles to make you understand my question. There is a fixed positive charged particle in the center.

Scenerio 1 : A fixed negative charged particle is introduced there at distance $x$ from the positive charged particle. Force of attraction between them is $F_1$.

Scenerio 2 : Two fixed negative charged particles (of the same charge as that in the last scenario) are introduced such that the positive charged particle is at same distance from them (i.e. $x$, same as last scenario). And the force of attraction now between the positive charged particle and a single negative charged particle is $F_2$. And the force of attraction between the positive charged particle and another negative charged particle is $F_3$

Now, is $F_1 = F_2 = F_3$ or $F_1 = F_2 + F_3$ or is it something else? I think it's answer is $F_1 = F_2 = F_3$ because the distances and the product of charges is same in the calculation of all three forces. I have taken all the charges as fixed in these scenarios so that there is no change of magnitude of force with time.

QUESTION 2

If the answer to the last question is $F_1 = F_2 = F_3$, then

We know that in an atom, for example, in a neutral oxygen atom there are 8 protons and 8 electrons, i.e., 8 positive charged particles and 8 negative charged particles. We know it's nucleus can only carry 8 electrons around it. Now my question is why can't it carry so many electrons, like why don't it carry 13 electrons or 14 or even 16?

$F_1 = F_2 = F_3$.

This is essentially the superposition principle.

We know that in an atom, for example, in a neutral oxygen atom there are 8 protons and 8 electrons, i.e., 8 positive charged particles and 8 negative charged particles. We know it's nucleus can only carry 8 electrons around it. Now my question is why can't it carry so many electrons, like why don't it carry 13 electrons or 14 or even 16?

1. We know of lots of cases where the number of electrons in an atom is not exactly equal to the number of protons in its nucleus. These are called ions, and they're very common.

2. Say you have some electrons near the oxygen nucleus with its 8 protons. All of those electrons would feel the same attractive force from the nucleus. But each of them would also feel a repulsive force from the other electrons.

If you had an oxygen ion with, say, 10 electrons, and another electron approached from a distance, it would feel a larger repulsive force from those 10 electrons than the attractive force from the 8 protons. So it would not be likely to join up to that -2 ion and form a -3 ion.

Or if you had an oxygen ion with only 6 electrons, and another electron approached from a distance, it would feel a stronger attractive force than repulsive force and it would be very likely to be captured, changing the +2 ion into a +1 ion. So a highly positively charged ion is not likely to remain in that state for very long if there are any free electrons in the vicinity.

Obviously this is a very simplified picture of the situation, ignoring quantum mechanical considerations, orbital theory, etc. It's basically based on the "plum-pudding" atomic model which we've made many improvements on over the years. But I think even this very simple model is sufficient to capture the answer to your question at the most basic level.