Rank charge density of spheres

[Delta2]

C > D = E makes sense if we remove Sphere 1. The majority of excess charge mill migrate to the far ends the 2,3 combination, that is C and the unmarked right side of 3. Density at D,E will be much less.

Then, bring Sphere 1 back. Remember, its still relatively far away and has the smallest net charge.

I suspect the right hand side of 3 has largest density, then C > A ?? E > D ?? >0 not sure where B goes.

Maybe between A & E, else between D and 0 .

I perfectly agree to what you say here.

Also I think it should be C>A>B>E>D>0, I expect E and D to be very close to zero but it might be E>D>0 due to the field effect from the sphere 1. B has to be very close to A cause the field inside the sphere 1 must be zero (B will be slightly smaller due to the fact that it is closer to the other two spheres that have net positive charge as a whole).

 

[SammyS]

A general question of clarification for @isukatphysics69 .

Referring to specific details concerning the physical situation depicted in the figures.

Initial figure for example:

Do the labels:

1) refer to individual points on the surfaces of the spheres located on the line joining the centers of the spheres?
2) refer to individual points on the surfaces of the spheres located close to the line joining the centers of the spheres (or to a small region at such a location)?​

.
I suppose I have been thinking of the labeled points in the 2nd way (small regions), but as I reconsider this, it now makes more sense to suppose that the 1st way is what was intended for this question. If this is the case, then go back to the charge density answer for the case where Spheres 1&2 touch:

which was given as: E>A>B=C=0>D .
(Charge density at) both B and C should be zero. (What the feather! That's what it did say all along !) I mis-read this too ! DUH on me !
Well this must be the case at this position with the two conductors touching along the line connecting the centers.

So...
The same must be true in the final case, where Spheres 2&3 touch, only this time D and E touch, so charge density there is zero.

My apologies to @isukatphysics69

 

[isukatphysics69]

A general question of clarification for @isukatphysics69 .

Referring to specific details concerning the physical situation depicted in the figures.

Initial figure for example:

View attachment 230401

Do the labels:

1) refer to individual points on the surfaces of the spheres located on the line joining the centers of the spheres?
2) refer to individual points on the surfaces of the spheres located close to the line joining the centers of the spheres (or to a small region at such a location)?​

.
I suppose I have been thinking of the labeled points in the 2nd way (small regions), but as I reconsider this, it now makes more sense to suppose that the 1st way is what was intended for this question. If this is the case, then go back to the charge density answer for the case where Spheres 1&2 touch:
View attachment 230402
which was given as: E>A>B=C=0>D .
(Charge density at) both B and C should be zero. (What the feather! That's what it did say all along !) I mis-read this too ! DUH on me !
Well this must be the case at this position with the two conductors touching along the line connecting the centers.

So...
The same must be true in the final case, where Spheres 2&3 touch, only this time D and E touch, so charge density there is zero.

My apologies to @isukatphysics69

I am back and I have tried a few attempts before coming back to the forum and nothing yet. @Delta² that was an incorrect answer. @SammyS my professor said to look at regions D and E as regions in this problem and not points, so I do not believe that I am supposed to have D=E=0 here but I could be wrong

 

[isukatphysics69]

I tried C>A>E>D>B>0 and was really confident about that one but incorrect

 

[SammyS]

I am back and I have tried a few attempts before coming back to the forum and nothing yet. @Delta² that was an incorrect answer. @SammyS my professor said to look at regions D and E as regions in this problem and not points, so I do not believe that I am supposed to have D=E=0 here but I could be wrong

Well, is the same true for B and C: regions rather than points.
If so, I don't see how they can be zero in the previous situation. Just saying.

I tried C>A>E>D>B>0 and was really confident about that one but incorrect

I still think E and D must be very small, and I don't think that B is very much less than A. It is stated that there is relatively large separation.

I don't know if you have tried this yet:

C > A > B > E > D > 0

My best guess.

 

[isukatphysics69]

Well, is the same true for B and C: regions rather than points.
If so, I don't see how they can be zero in the previous situation. Just saying.I still think E and D must be very small, and I don't think that B is very much less than A. It is stated that there is relatively large separation.

I don't know if you have tried this yet:

C > A > B > E > D > 0

My best guess.

That was actually the most recent one that I tried and it was incorrect, and there was a reason I believe that the previous one was 0, I think it had something to do with both spheres having equal net charge, now both spheres in contact do not have equal net charge

 

[SammyS]

That was actually the most recent one that I tried and it was incorrect, and there was a reason I believe that the previous one was 0, I think it had something to do with both spheres having equal net charge, now both spheres in contact do not have equal net charge

The spheres did have equal net charge in the previous case, but Sphere 1 is larger, so it has greater surface area, so I would expect the surface density would less at C than at B (because of signs), but less in magnitude at B than at C, although, I would expect both to be very small in magnitude.

We can speculate some more. Right now I'm out of guesses.

Got any ideas you want to run up the flag pole?

EDIT ed above as shown in red

 

[isukatphysics69]

The spheres did have equal net charge in the previous case, but Sphere 1 is larger, so it has greater surface area, so I would expect the surface density would less at C than at B (because of signs), but less in magnitude at B than at C, although, I would expect both to be very small in magnitude.

We can speculate some more. Right now I'm out of guesses.

Got any ideas you want to run up the flag pole?

EDIT ed above as shown in red

I am absolutely stumped, so sphere 3 has the most net charge, as sphere 2 is moving toward sphere 3, id imagine that at around the three quarters mark nearest sphere 1, the power of sphere 3's polarization will take over and start really pushing the positives to point C and negatives to point D, so now when they come in contact the negatives are at D and positives are at point C and then they will neutralize. after neutralization the net positive will push more positives to A and B will still be positive but not as positive as A

 

[isukatphysics69]

I am just absolutely stumped I've never spent 10 hours on a single problem before this is crazy

 

[Delta2]

I am intrigued to see what's the correct answer here, don't forget to write the answer here please, after your teacher tells it to you.

 

[isukatphysics69]

I am intrigued to see what's the correct answer here, don't forget to write the answer here please, after your teacher tells it to you.

I will, I don't think he will tell me the right answer until the submission option is closed tho which won't be until September 14th. I will continue trying until then

 

[isukatphysics69]

Also guys before I go to sleep I will leave the messages that my professor has sent to me to try to help without giving away too much, maybe there is a clue I am missing here...

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.
Second:

The questions all occur in sequence, unless it specifically says otherwise. Sphere 2 was moved to contact sphere 1 for questions 2 and 3. That made sphere 1 have net + charge and sphere 2 net -charge. Now, for questions 4 and 5, sphere 2 is moved from contact with sphere 1 and made to touch sphere 3. Here D and E don't represent the same point. D is the right side of sphere 2 and E is the left side of sphere 3, not just the one point of contact. Also, note that the + charge in spheres 2 and 3 will not be concentrated in the middle. The + charges there repel each other and want be as far away for each other as possible, which would put them mostly on the outside parts of the spheres with less in the middle.
Third:

Consider what happens bit by bit and then put it together. When 1 and 2 were touching, 1 had net + and 2 had net -. After separating them, this is still true. When 2 touches 3, charge will transfer. 3 had more + than 2 has -, so the combination is net +. If 1 were absent, this charge would be shared equally between 2 and 3, with more charge on the outside (left side of 2 and right side of 3) and less in the middle. With 1 present, the + in 1 will push some + from 2 to 3, but not a lot (+ in 1 would be much less than originally in 3 and it's pretty far away compared to the size of 3). The same thing happens in 1. 2 and 3 are both +, so that will push some + in 1 from B to A, but not a lot (not so much that B is made -). I hope that helps. Let me know if you have more questions.

 

[isukatphysics69]

@SammyS @Delta² Messages from professor above

 

[Delta2]

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.

 

[SammyS]

I am absolutely stumped, so sphere 3 has the most net charge, as sphere 2 is moving toward sphere 3, id imagine that at around the three quarters mark nearest sphere 1, the power of sphere 3's polarization will take over and start really pushing the positives to point C and negatives to point D, so now when they come in contact the negatives are at D and positives are at point C and then they will neutralize. after neutralization the net positive will push more positives to A and B will still be positive but not as positive as A

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..

 

[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.