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Imagine a world in which there are three kinds of electric charges instead of two. In this world, like charges repel and unlike charges attract. Let these types of charges be $A, B$ & $C$. Thus $A$ & $A$ repel, $A$ & $B$ attract, $A$ & $C$ attract and so on. Coulombs inverse square law is valid and superposition law is also valid like in our world. In this world like charges repel with force two times that of attraction between unlike charges.

A configuration of charged particles is neutral if each particle experiences no net force. Which of the following configurations is neutral?enter image description here

Assumptions: In option 2, it is an equilateral triangle. In option 4, it is regular hexagon. In option 5, it is a square. In option 6, it is regular pentagon.

One or more than one options can be correct.

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    $\begingroup$ What do you mean by neutral (net charge)? We have two kinds of charges, so we just find the net charge as the difference between + and - charges. If your world has 3 charges, how do we define neutral? $\endgroup$ – ghosts_in_the_code Feb 8 '16 at 14:32
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    $\begingroup$ @ghosts_in_the_code A neutral body is one which experiences no net force. $\endgroup$ – manshu Feb 8 '16 at 14:34
  • $\begingroup$ Is a neutral configuration a stable one? $\endgroup$ – T. Verron Feb 8 '16 at 14:36
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    $\begingroup$ @JulianRosen we dont need the definition of electric field to answer the question $\endgroup$ – manshu Feb 8 '16 at 17:10
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    $\begingroup$ "Equilibrium" makes more sense. You might also find Earnshaw's Theorem to be interesting reading. As written, it's only applicable to "real" charges, but I think you could apply the argument to your problem — first use it to prove that there can't be a stable equilbrium for the A charges by treating them as positive and the B's and C's as negative, then treat the B's as positive and the A's and C's as negative, etc. $\endgroup$ – Michael Seifert Feb 8 '16 at 18:26
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the force between electric charges are propotional to $\frac{1}{d^2}$ (distance between two charges.) So if charges are close to each other, the force is much higher. As a result (assuming the definition of neutral in the equation);

1&3

1 and 3 is neutral since both electric charges attracts each other if there is no friction.

2

If the angle is assumed to be 120 degrees between AAC, BAC and AAB, then the charge A in the middle will move towards between B and C with an initial accelaration then slow down since the distance between A-A getting lower. But the route cannot be foreseen since we do not know the distance and the weight of the charges. It can move back towards to A and go back again or just move outside of the range of B, C & A. NO Neutral.

4

This is the tricky but easy after knowing $\frac{1}{d^2}$ rule. It is symmetric for every charge so I take A as an example; A - A force is so little since the distance between A-A. So A will tend to go to A with a force from two Bs and two Cs. It will happen all of the charges and they will touch each other and the end and becomes stable since there are more attracting force than pushing. No Neutral.

5

Same logic as 4, but this time the distance may be important. Let's call A-A pushin force as F and $F=\frac{1}{2d^2}$ if we call the distance between A-B as d. A-B attracting forces are 2F from both sides. so $2\sqrt(2)F-F$ will be the force that attracts A to the middle until they all touch to each other and stick together just like 4. No Neutral

6

You can easily say that A will move towards the middle of the hexagon and then move like a come and go forever. B and C will tend to move just like A but slower. still No Neutral.

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Oh man, I did this scenario in college! Without the actual diagrams, of course.

1 is neutral, but only if there is no C charge within finite distance of it.

2 is not neutral; the two A's fly apart.

3 is neutral and probably the most likely outcome.

4 is not neutral (I think) - can't do the math off the top of my head - but my suspicion is that it's too attractive and collapses inwards due to the attraction being based off distance squared rather than distance.

5 is not neutral; although the attractive forces balance out the repulsive forces in strength, they don't in direction. All the charges fly away from each other.

6 is not neutral - there's nothing to stop A from collapsing towards the center.

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  • $\begingroup$ None of these are neutral if there is another charged body nearby, are they? And for the record, unless I did a mistake in my trigonometry, you're right for 4, the inwards pull is approx. 2.31 times stronger than the outwards push. $\endgroup$ – T. Verron Feb 8 '16 at 14:50
  • $\begingroup$ Actually, 3 is roughly neutral with respect to outside charges as well, thus the distinction. $\endgroup$ – Zerris Feb 8 '16 at 15:23
  • $\begingroup$ "roughly", yes. $\endgroup$ – T. Verron Feb 8 '16 at 16:11
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1 and 3 are neutral because they consist of unlike charges next to each other. In 2 and 6, all the other charges move respectively the A in the middle and at the top downwards. In 4, the force attracting the A at the top downwards is greater than one that would have repelled it upwards if the bottom A were in the middle, so there's no neutrality. In 5, a square, the force attracting the top left A to the bottom right is definitely greater than the force coming from farther and repelling it in the opposite direction, so none other than 1 and 3 is neutral.

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