How to Rank Charge Density of Conductive Spheres?

[isukatphysics69]

Yes all these points are kind of known to me and they were lead me to think that C>A>B>E>D>0. Hmm your teacher says that the charge in sphere 1 is much smaller than that of sphere 3, maybe try C>E>D>A>B>0 as last try for me, though I still don't think that's the case. For me C>A>B>E>D>0 is the correct one.

Incorrect, I think there may be something wrong here with the webpages backend answer key or something

 

[isukatphysics69]

As Sphere 2 goes from touching Sphere 1 and then moving towards Sphere 3, Sphere 2 has much less net charge than Sphere 3. That's because the fact that Spheres 1 & 2 have any net charge is due to the polarization produced by the Electric field due to Sphere 3.

By the way: Did I see somewhere (maybe in your old thread) some statement about the distances relative to sphere size?​

You're right, as Sphere 2 approaches Sphere 3, a lot of polarization occurs on Sphere 2.but not much on Sphere 3, because of the very different amount net charge on these two spheres.

As for Sphere 1, when it and Sphere 2 just separate, points B and C have charge near zero, similar to when they were touching. As Sphere 2 moves away point B gains a little positive charge and point A looses some. Don't forget. These are conductors. The (like) charges on the surface are also repelling each other.
...

Just now I see the responses from your prof.. I generally agree with them.

@Delta² 's idea in Post #49 looks reasonable..

I believe you mean this

. Note also that the center of Sphere 1 is about 10 times farther from the center of Sphere 3 as the radius of Sphere 3.

I'm done for the day here, I just don't even know anymore. I sent another text to professor hopefully response in the am

 

[SammyS]

Incorrect, I think there may be something wrong here with the webpages backend answer key or something

Only other quick idea now is do the same order with E = D .

Personally I don't like it.

 

[SammyS]

I believe you mean this

. Note also that the center of Sphere 1 is about 10 times farther from the center of Sphere 3 as the radius of Sphere 3.

I'm done for the day here, I just don't even know anymore. I sent another text to professor hopefully response in the am

10 times the big radius... Yes, I thought I saw that somewhere.

Good night all.
(Good morning Δ² )

 

[isukatphysics69]

I have a new thought, A could potentially be the greatest, yes it has the least net charge in that sphere but the charges in the 2/3 system come to equilibrium spreading out the charge equally

 

[SammyS]

I have a new thought, A could potentially be the greatest, yes it has the least net charge in that sphere but the charges in the 2/3 system come to equilibrium spreading out the charge equally

Not likely.
See item 1 (First) from your prof., particularly the highlighted part.

First:

I stepped out for a couple minutes around that time, I'm back now. For 5, there is a transfer of charge between spheres 2 and 3 when they are in contact. The net charge on both spheres would be 0 if both spheres had the same amount of charge, with one + and the other - (and if 1 were not present). However, the charges are not equal. 2 got its charge from 1 through polarization while they were in contact. 1 and 2 were pretty far away (compared with the radius the spheres) at that time, so there was only a small amount of charge separation, much smaller than the net + charge on 3.​

Basically, the magnitude of the net charge on Spheres 1 & 2 is much less than the magnitude of the net charge on Sphere 3. (When 1 & 2 have separated.) so when Sphere 2 contacts Sphere 3, the small amount of negative charge on Sphere 2 only neutralizes a small fraction of the charge on Sphere 3. Besides look at Answer 4., Magnitudes after Spheres 2 & 3 touch.

 

[isukatphysics69]

Not likely.
See item 1 (First) from your prof.), particularly the highlighted part.
First:

I stepped out for a couple minutes around that time, I'm back now. For 5, there is a transfer of charge between spheres 2 and 3 when they are in contact. The net charge on both spheres would be 0 if both spheres had the same amount of charge, with one + and the other - (and if 1 were not present). However, the charges are not equal. 2 got its charge from 1 through polarization while they were in contact. 1 and 2 were pretty far away (compared with the radius the spheres) at that time, so there was only a small amount of charge separation, much smaller than the net + charge on 3.​

Basically, the magnitude of the net charge on Spheres 1 & 2 is much less than the magnitude of the net charge on Sphere 3. (When 1 & 2 have separated.) so when Sphere 2 contacts Sphere 3, the small amount of negative charge on Sphere 2 only neutralizes a small fraction of the charge on Sphere 3. Besides look at Answer 4., Magnitudes after Spheres 2 & 3 touch.

Good point, thank you. Back to drawing board

 

[SammyS]

Only other quick idea now is do the same order with E = D .

Personally I don't like it.

What I meant here was:

Modify C > E > D > A > B > 0, as suggested by Δ², but he didn't like (although I do), and it didn't work for you.

So try this but with E and D being equal: C > E = D > A > B > 0 . (I don't really like it because you can make a good case for E > D .) Also, as I said before, D & E should be really small, unless each is big enough to account for something like an entire hemisphere of their respective spheres. Charge on surfaces of conductors tends to accumulate more at the ends and very little in nooks and crannies. I don't know how much electrostatics you have covered.

 

[isukatphysics69]

FINALLY GOT THE RIGHT ANSWER!

 

[isukatphysics69]

C>E>D>A>B>0

@SammyS @Delta²

 

[SammyS]

C>E>D>A>B>0

@SammyS @Delta²

Golly! I thought you had tried that previously !

 

[Delta2]

C>E>D>A>B>0

@SammyS @Delta²

I had suggested that back at post #49 and I thought you had tried it.

Still I don't like it that much. The charge on sphere 1 must be much much much smaller than the charge in sphere 3 in order for this to be true.