Effect of an uncharged sphere touched to charged sphere

[5.98e24]

Homework Statement

Two identical metal spheres, each with a positive charge q, are separated by a centre-to-centre distance r. What effect will the following change have on the magnitude of the electric force exerted on each sphere by the other:

-one sphere is touched by an identical neutral sphere, which is then taken far away

Homework Equations

F₂ / F= q₁q₂ / q₁q₂

where F is the initial electrical force, and F₂ is the final electrical force.

The Attempt at a Solution

I looked at the solutions manual and it says this:

I just don't understand why one of the charges is halved. What does this have to do with the neutral sphere?

Thank you!

 

[gneill]

Think about what happens when the neutral metal sphere is touched to a charged metal sphere. Do the charges stay put?

 

[5.98e24]

Think about what happens when the neutral metal sphere is touched to a charged metal sphere. Do the charges stay put?

Since the stationary sphere is positive, that means that when a neutral sphere approaches it, the electrons in the neutral sphere will move towards the stationary sphere.

I don't know how that explains the halving of the charge, though.

 

[gneill]

Since the stationary sphere is positive, that means that when a neutral sphere approaches it, the electrons in the neutral sphere will move towards the stationary sphere.

I don't know how that explains the halving of the charge, though.

The sphere's touch. Are the electrons going to be satisfied moving only to the surface of the neutral sphere?

Also, you need not think about what the particular charge carriers are. Just think about the net charges on each sphere. The positive sphere has q positive charges. The neutral sphere has no charge. The positive charges on the charged sphere would like to get away from each other...

 

[5.98e24]

The electrons will move to the side with a positive charge. So the neutral sphere will become positively charged, and the previously positive sphere becomes negatively charged?

 

[gneill]

The electrons will move to the side with a positive charge. So the neutral sphere will become positively charged, and the previously positive sphere becomes negatively charged?

Almost right. The previously positively charged sphere will become LESS positively charged. Charges stop moving once there is no potential difference to drive them. If the spheres have identical dimensions, then the charge will be split equally between them.

 

[5.98e24]

Almost right. The previously positively charged sphere will become LESS positively charged. Charges stop moving once there is no potential difference to drive them. If the spheres have identical dimensions, then the charge will be split equally between them.

So you mean that q1 = q2 now? And since the charge is split equally, q1 = q2 = q/2?

 

[gneill]

So you mean that q1 = q2 now? And since the charge is split equally, q1 = q2 = q/2?

That would appear to be the logical conclusion

 

[5.98e24]

That would appear to be the logical conclusion

Great! Thanks a lot!

So this relationship holds true involving any two spheres of equal dimensions, as long as one is neutral and the other is either positive or negative to any magnitude, right?

 

[SammyS]

Since the stationary sphere is positive, that means that when a neutral sphere approaches it, the electrons in the neutral sphere will move towards the stationary sphere.

I don't know how that explains the halving of the charge, though.

I agree with you that to say the excess charge is equally shared by the two spheres coming in contact is questionable. In my opinion, it's a HUGE assumption.

I would only be confident in the halving of the charge if when the two spheres are in contact, then each of them is the same distance from the third sphere. Otherwise, the third sphere will tend to push more than half of the excess charge onto the more distant of the two spheres.

 

[5.98e24]

I agree with you that to say the excess charge is equally shared by the two spheres coming in contact is questionable. In my opinion, it's a HUGE assumption.

I would only be confident in the halving of the charge if when the two spheres are in contact, then each of them is the same distance from the third sphere. Otherwise, the third sphere will tend to push more than half of the excess charge onto the more distant of the two spheres.

What if there is no third sphere? Just two spheres?

 

[SammyS]

What if there is no third sphere? Just two spheres?

Which two spheres?

 

[5.98e24]

Which two spheres?

The ones in my original question. There were only 2 to start with.

 

[gneill]

I agree with you that to say the excess charge is equally shared by the two spheres coming in contact is questionable. In my opinion, it's a HUGE assumption.

I would only be confident in the halving of the charge if when the two spheres are in contact, then each of them is the same distance from the third sphere. Otherwise, the third sphere will tend to push more than half of the excess charge onto the more distant of the two spheres.

We can be sneaky and specify that the two contacting spheres are equidistant from the third sphere.

It's not clear that poster 5.98e24 has covered induction charges in his course as yet.

 

[SammyS]

If your question is: "What if two identical conducting spheres touch, will any excess charge be equally shared by the two spheres?" , then the answer is: "Yes, provided that they are isolated from any other charges object or externally produced electric field."

 

[5.98e24]

We can be sneaky and specify that the two contacting spheres are equidistant from the third sphere.

It's not clear that poster 5.98e24 has covered induction charges in his course as yet.

I'm a her .. :)

I did learn about induction, but that was a while ago. I haven't touched physics in a few years.

Thanks for all the help! Greatly appreciated!

 

[gneill]

I'm a her .. :)

My apologies

 

[SammyS]

We can be sneaky and specify that the two contacting spheres are equidistant from the third sphere.

It's not clear that poster 5.98e24 has covered induction charges in his course as yet.

It's still not clear whether 5.98e24 has covered induction charges, even after her response.

We could also move the single charged sphere to a very distant location while the other two touch, then bring it back to the original position after the other two separate.

5.98e24 is a puzzling name. It could represent a bit less than 10 moles.

 

[gneill]

5.98e24 is a puzzling name. It could represent a bit less than 10 moles.

Or the mass of the Earth in kg.

 

[SammyS]

That makes much more sense. Gives me an idea for 5.98e24's name, but I'll only post the guess if she's game.

 

[Quinzio]

The ones in my original question. There were only 2 to start with.

In the original problem, 3 shperes are involved.
Sphere A and B has qA, qB charge, C has zero charge.

A and C touches, so A shares half of the charge with C.
Then C is taken far away.

 

[SammyS]

...
A and C touches, so A shares half of the charge with C.
Then C is taken far away.

Only under specific conditions will they share exactly half the charge.

 

[Quinzio]

In the 1st message it is clearly said that the spheres are identical. What other conditions to take account of ?
The closeness of the other charged sphere ?
mmm ok, but I guess this is far from the problem aim.

 

[5.98e24]

That makes much more sense. Gives me an idea for 5.98e24's name, but I'll only post the guess if she's game.

Be my guest!

In the original problem, 3 shperes are involved.
Sphere A and B has qA, qB charge, C has zero charge.

A and C touches, so A shares half of the charge with C.
Then C is taken far away.

Ah right! I forgot about that.

In the 1st message it is clearly said that the spheres are identical. What other conditions to take account of ?
The closeness of the other charged sphere ?
mmm ok, but I guess this is far from the problem aim.

They're 10cm apart and attract each other with a magnitude of 3.0 x 10^-6 N. I don't think it matters though.

 

[SammyS]

My guess is Terra.

 

[5.98e24]

My guess is Terra.

Creative, but no :) Nice try though !