B Collision of a charged and an uncharged particle

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A charged and an uncharged particle collide. What will be their charges after the collision? The average of the total charge?
 

anorlunda

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If total charge is conserved, what do you think the answer is?
 
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If total charge is conserved, what do you think the answer is?
If the total charge is conserved, that only means that the total charge in the system is the same before and after the collision. But it doesn't say anything about its distribution.
 
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My guess would be that the two particles will have the same charge, therefore the distribution is determined by the arithmetic mean, but that's only a guess for now.
 

anorlunda

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But it doesn't say anything about its distribution.
Good. That's true. But when things collide they might share their charge or keep them separate. There is no universal answer to your question. It depends on what the objects are.
 
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Good. That's true. But when things collide they might share their charge or keep them separate. There is no universal answer to your question. It depends on what the objects are.
I was a bit careless. The objects are not particles in general, but they are metal balls.
 

ZapperZ

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I was a bit careless.
Yes you were, because this is now a different problem than just using "particles".

The objects are not particles in general, but they are metal balls.
You are careless once more, because the size of these "metal balls" should also be specified! Are they of the same size? Different?

Zz.
 
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A charged and an uncharged particle collide. What will be their charges after the collision? The average of the total charge?
The average cannot be because of charge conservation, already said. If you also intended to ask how the totale charge is shared into the final particles, theni t depends on a lot of different cases, for example: the first particle's charge is +1e (or -1e) ?
Then this (elementary) charge cannot divide in parts so the final particles cannot have fractionary charge: one of them must have +1e (-1e) charge and the other zero;
the first particle's charge is +ne (or -ne) n>1? Then the final particles can be both charged but the sum of their chsrges must be +ne(or - ne), etc;
the initial particles are elementary ones? Then their charge is fixed, they cannot have different charges (even if total charge conservation is respected).
Edit: I see now that the OP question has been refined.
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ZapperZ

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Same size. They are completely identical, the difference is just their charge.
I'm not sure why this isn't intuitively obvious to you. The "collision" part is irrelevant. This is high-school level static electricity problem.

When they touch, the charge will try to spread out as evenly as possible between the two metallic spheres. Because they are of identical size, no one sphere is more preferable than the other. This means that each sphere will have the same amount of charge. So when they separate, each sphere will carry half of the charge of original sphere. The two of them combined will carry the same amount of total charge as before.

Zz.
 
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I was a bit careless. The objects are not particles in general, but they are metal balls.
You made quite a mess: "particles" and then metal balls, "average charge": if you have two equal metal balls, why you didn't simply say "half of total charge on every ball"? :smile:

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You made quite a mess: "particles" and then metal balls, "average charge": if you have two equal metal balls, why you didn't simply say "half of total charge on every ball"? :smile:

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Yes, it made sense to me too, but this is part of a problem of a competition at my university and I was suspicious that it is too simple. Thank you for your help.
 

sophiecentaur

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You made quite a mess: "particles" and then metal balls, "average charge": if you have two equal metal balls, why you didn't simply say "half of total charge on every ball"? :smile:

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If the resulting two balls have different diameters then their relative capacitances would make the sharing unequal. The capacitance of a sphere is proportional to its Radius so the charges would be distributed proportionally to their radii. If one ball were twice the radius of the other, the original charge would be shared in the ratio 2:1
 
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If the resulting two balls have different diameters then their relative capacitances would make the sharing unequal. The capacitance of a sphere is proportional to its Radius so the charges would be distributed proportionally to their radii. If one ball were twice the radius of the other, the original charge would be shared in the ratio 2:1
Sure. But in post #8 the OP stated "Same size. They are completely identical, the difference is just their charge."
Bye!

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sophiecentaur

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Sure. But in post #8 the OP stated "Same size. They are completely identical, the difference is just their charge."
Bye!

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Yes, that's right but the more general case is of interest and the result is pretty simple.
 

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