Net force acting on a charged particle ##+Q##

AI Thread Summary
The net force acting on the central charge ##+Q## from twelve equally spaced charges around a circumference is zero due to symmetry, as the forces cancel each other out. If one charge is removed, the net force on ##+Q## is exerted solely by the charge diametrically opposite to the removed charge, resulting in a non-zero force. Removing two charges from the arrangement leads to a system that may still reach equilibrium, depending on the remaining charges' configuration. The discussion highlights the importance of vector addition and symmetry in determining net forces in electrostatic systems. Understanding these principles is crucial for solving similar problems in physics.
Mutatis
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Homework Statement



Twelve equal particles of charge ##+q## are equally spaced over a circumference (like the hours in a watch) of radius R. At the center of the circumference is a particle with charge ##+Q##.

a) Describe the net force acting over ##+Q##.

b) If the charge located at "3'o'clock" has been taken off, what's the net force acting over ##+Q##?

c) If both charges located at "3'o'clok" and "9'o'clock" has been taken off, describe the force acting over +Q this time.

Homework Equations



##\vec F= \frac {1} {4 \pi \varepsilon_0} \frac {Q_1 Q_2} {r^2} \vec r##

The Attempt at a Solution



a) As the charges are equally spaced over that circumference, the net force acting over ##+Q## is zero, because the forcs diametrically opposed to each other cancel each other out.

b) If you take the "3'o'clock" particle out the circumference, the only particle that exerts force over ##+Q## is the particle diametrically oposed to "3'o'clock" particle: $$ \vec F= \frac {1} {4 \pi \varepsilon_0} \frac {q Q} {R^2} \vec R $$

c) The system back to equilibrium state.

What do you think guys? I'm not sure about my answers...
 
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In (b) you say
Mutatis said:
... the only particle that exerts force over ##+Q## is the particle diametrically oposed to "3'o'clock" particle
Have the other particles stopped exerting forces on ##+Q##? Your conclusion is correct but your justification needs to be improved.
 
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kuruman said:
In (b) you say
Have the other particles stopped exerting forces on ##+Q##? Your conclusion is correct but your justification needs to be improved.

They're still exerting force over ##+Q##, but as they're are diametrically opposed to each other, so they cancel out. Right?
 
Right. As an enrichment to this problem, what would the net force on the central charge be if you had 11 equally spaced charges on the circumference instead of 12? Your cancellation in pairs argument will no longer work. What if you removed one of these 11 and were left with 10?
 
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kuruman said:
Right. As an enrichment to this problem, what would the net force on the central charge be if you had 11 equally spaced charges on the circumference instead of 12? Your cancellation in pairs argument will no longer work. What if you removed one of these 11 and were left with 10?

In the first case, the net force is going to be a sum of the individual contributions of each charge acting over ##+Q##, superposition principle. And then if I was left with 10 equally spaced charges the system is going to equilibrium state.
 
Mutatis said:
In the first case, the net force is going to be a sum of the individual contributions of each charge acting over ##+Q##, superposition principle. And then if I was left with 10 equally spaced charges the system is going to equilibrium state.
Sorry, you are incorrect on both counts, in fact you got it backwards. First draw the 11 force vectors acting on the central charge. Then add them using the "tail-to-tip" graphical method for adding vectors. What do you get? Also consider this. If you remove a charge from the distribution, the electric force at the center would be the same as the electric force from the old distribution plus a charge of opposite sign placed at the location of the charge that you removed. This is superposition. So if the old distribution gives zero force at the center, removing one ##+q## charge from the 12 o' clock position will provide at the center the force of a ##-q## charge placed at the 12 o' clock position.
 
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kuruman said:
First draw the 11 force vectors acting on the central charge. Then add them using the "tail-to-tip" graphical method for adding vectors. What do you get?

If the arrangement is symmetrical, then the central charge cannot move in any direction, as you can rotate the system by ##2 \pi/11## and get an equivalent system where it is now moving in a different direction.
 
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kuruman said:
Sorry, you are incorrect on both counts, in fact you got it backwards. First draw the 11 force vectors acting on the central charge. Then add them using the "tail-to-tip" graphical method for adding vectors. What do you get? Also consider this. If you remove a charge from the distribution, the electric force at the center would be the same as the electric force from the old distribution plus a charge of opposite sign placed at the location of the charge that you removed. This is superposition. So if the old distribution gives zero force at the center, removing one ##+q## charge from the 12 o' clock position will provide at the center the force of a ##-q## charge placed at the 12 o' clock position.

I'll try to write up this when I get home. This exercise have got my brain confused. At the first question that I've posted, would you, if you were my physics teacher, consider it right?
 
Mutatis said:
I'll try to write up this when I get home. This exercise have got my brain confused. At the first question that I've posted, would you, if you were my physics teacher, consider it right?

You need to say what that vector ##\vec{R}## means. Otherwise, it's fine.
 
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PeroK said:
If the arrangement is symmetrical, then the central charge cannot move in any direction, as you can rotate the system by ##2 \pi/11## and get an equivalent system where it is now moving in a different direction.
The symmetry argument is, of course, applicable and correct in this case. More generally though, if the sum of the forces forms a closed polygon (regular or not) when drawn graphically, then that sum is zero. That's the connection that I wanted @Mutatis to establish.
 
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