Charge distribution among connected metal spheres -- Simple question

[lightlightsup]

Homework Statement

See image.

Relevant Equations

See image.

My answer was +Q/3.
I was assuming that the charges would distributed themselves completely.
But, apparently, I'm wrong?
For example, if there were 12$e^-$s on Sphere C, then, in the first step in the system: the $e^-$s would balance out until each sphere has 4 $e^-$s each?
What am I getting wrong here?

Similar problem here?:

 

[mitochan]

If there are only A and C, the both have charge Q/2 by connection and subsequent separation. Then pull A left and right as to form original A and B and separate them finally. Each side shares charge in half.

 

[kuruman]

Homework Statement:: See image.
Relevant Equations:: See image.

My answer was +Q/3.
I was assuming that the charges would distributed themselves completely.
But, apparently, I'm wrong?
For example, if there were 12$e^-$s on Sphere C, then, in the first step in the system: the $e^-$s would balance out until each sphere has 4 $e^-$s each?
What am I getting wrong here?

I agree with you that the answer should be Q/3. Look at the second step in 18.4.4 where all three conductors are connected. They all are at the same potential. If the spheres are identical (no reason to assume otherwise), they must all carry the same fraction of the total charge, Q/3. Disconnecting the wire between A and C just traps the charge on C and the AB combination that is still connected. No charge will flow from A to B or B to A because they already have equal amounts of charge. The answer in 21.3.7 is not (b) because of similar considerations. (On edit) The answer given for the second situation is (b).

Your assumption that the charges are distributed equally is correct. By symmetry if two identical conductors are connected, there is no reason for one conductor to have more charge than the other (we neglect any charge on the connecting wire). You may wish to ask your teacher about this.

 

[lightlightsup]

I agree with you that the answer should be Q/3. Look at the second step in 18.4.4 where all three conductors are connected. They all are at the same potential. If the spheres are identical (no reason to assume otherwise), they must all carry the same fraction of the total charge, Q/3. Disconnecting the wire between A and C just traps the charge on C and the AB combination that is still connected. No charge will flow from A to B or B to A because they already have equal amounts of charge. The answer in 21.3.7 is not (b) because of similar considerations.

Your assumption that the charges are distributed equally is correct. By symmetry if two identical conductors are connected, there is no reason for one conductor to have more charge than the other (we neglect any charge on the connecting wire). You may wish to ask your teacher about this.

LOL. Here I was, hoping someone would tell me I was wrong about my understanding of something.
Those are official answers from Wiley.

 

[kuruman]

Wiley has been known to be in error, mostly typos from poorly proofed copy by poorly paid editors. Your teacher is the ultimate authority.

 

[rude man]

As kuruma n pointed out you have to assume the radii of all spheres is the same or you can't do the problem.
Under this assumption , looks like the answer should be Q/3 all right.

 

[haruspex]

As kuruma n pointed out you have to assume the radii of all spheres is the same or you can't do the problem.
Under this assumption , looks like the answer should be Q/3 all right.

Also, you need to assume that that radius is much smaller than the distances between the spheres.
@lightlightsup , what are you getting for the second problem? I get the book answer. Note that A starts with a charge of 2Q here.

 

[kuruman]

The answer to the second problem is the book's answer. I cannot reconstruct from memory what I was thinking when I posted the original version of #3 about it.

 

[alan123hk]

I'm probably wrong, but I still tend to believe that the answer to 18.44 is Q/4.

 

[rude man]

Also, you need to assume that that radius is much smaller than the distances between the spheres.
@lightlightsup , what are you getting for the second problem? I get the book answer. Note that A starts with a charge of 2Q here.

I was referring to problem 18.4.4 for which the answer I got was Q/3. Prob. 21.3.7 was labeled "similar problem" by the OP. Anyway I did not look at 21.3.7.

 

[rude man]

The answer to the second problem is the book's answer. I cannot reconstruct from memory what I was thinking when I posted the original version of #3 about it.

Like others among us I believe you confused prob. 18.4.4 with 21.7.3.

 

[kuruman]

Like others among us I believe you confused prob. 18.4.4 with 21.7.3.

I was right in the first place to say that the book's answer is incorrect for 21.3.7. I restored post #3 and crossed out my comments in #8.

@lightlightsup , what are you getting for the second problem? I get the book answer. Note that A starts with a charge of 2Q here.

And C starts with charge +Q, not 0. The label for C could be mistaken for 0 although +0 doesn't make much sense.

Question 21.3.7:
Initial distribution
QA = +2Q; QB = 0; QC = +Q
Connect A and B
QA = +Q; QB = +Q; QC = +Q
$\dots$

 

[haruspex]

And C starts with charge +Q, not 0.

Ah yes, I should have expanded the image.