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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.

Julian Rosen
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manshu
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    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? – ghosts_in_the_code Feb 08 '16 at 14:32
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    @ghosts_in_the_code A neutral body is one which experiences no net force. – manshu Feb 08 '16 at 14:34
  • Is a neutral configuration a stable one? – T. Verron Feb 08 '16 at 14:36
  • @T.Verron The only difference in our and the other world is stated in the question. Other than that...use the science of our world. – manshu Feb 08 '16 at 14:39
  • Your definition of "neutral" is given for a body. If we are to apply the same definition, all configurations are neutral because there's nothing else. So do you mean that all bodies in the configuration experience no net force? – T. Verron Feb 08 '16 at 14:41
  • @T.Verron "In this world like charges repel with force two times that of attraction between unlike charges." This sentence makes all the difference – manshu Feb 08 '16 at 14:46
  • Maybe it does, but I fail to see how: any body, charged or not, put in an environment without any charge, will not experience any net force, right? – T. Verron Feb 08 '16 at 14:50
  • You give a definition of electric field, but if I understand correctly the definition of neutrality doesn't reference electric field (so that the definition of electric field is not needed to answer the question). Is this correct? – Julian Rosen Feb 08 '16 at 17:05
  • @JulianRosen as much as i know about electrostatics...no – manshu Feb 08 '16 at 17:07
  • I don't understand. Is "no" the answer to my question "is this correct"? – Julian Rosen Feb 08 '16 at 17:08
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    @JulianRosen we dont need the definition of electric field to answer the question – manshu Feb 08 '16 at 17:10
  • Your use of "net force" is confusing. Assuming that Newton's Third Law holds in this Universe, none of these configurations will experience a "net force", since the force exerted by (say) Charge #1 on Charge #2 will be balanced out by an equal and opposite force on Charge #2 exerted by Charge #1. So the net force experienced by all six of these configurations is zero, and by your definition they're all neutral. Surely that's not what you meant to ask, though. – Michael Seifert Feb 08 '16 at 17:55
  • @MichaelSeifert thanks for pointing it out. Your comment surely helped me in finding the mistake in the question (and maybe in my understanding of forces). And now i think a more proper word could have been 'equilibrium'. – manshu Feb 08 '16 at 18:18
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    "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. – Michael Seifert Feb 08 '16 at 18:26

3 Answers3

2

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.

Oray
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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.

Zerris
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  • 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. – T. Verron Feb 08 '16 at 14:50
  • Actually, 3 is roughly neutral with respect to outside charges as well, thus the distinction. – Zerris Feb 08 '16 at 15:23
  • "roughly", yes. – T. Verron Feb 08 '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.

Nautilus
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