Resolving the Problem: Charging Spheres A and B

[Air]

Homework Statement
Initially, sphere A has a charge of –10e and sphere B has a charge of +20e. The spheres are made of conducting material and are identical in size. If the spheres then touch, what is the resulting charge on sphere A? a) +10e, b) -10e, c) -5e, d) +5e

The attempt at a solution
Sphere B -> 20e-\left(\frac{+20e-10e}{2}\right) = 20e - 5e = 15e. So there is a loss of 5e in Sphere B. This would mean that the only -5e remains on Sphere B?

The Problem
However, the answer is +5e. Why is this? There is a loss of 5e in Sphere B. This would mean that the only -5e remains on Sphere B?

 

[Doc Al]

The attempt at a solution
Sphere B -> 20e-\left(\frac{+20e-10e}{2}\right) = 20e - 5e = 15e. So there is a loss of 5e in Sphere B. This would mean that the only -5e remains on Sphere B?

No. For some reason, you took the average charge but then subtracted it from Sphere B. Why?

Instead, ask yourself: What's the total charge on both spheres together? Once they touch, how does that total charge divide itself between the two spheres?

 

[Air]

Only negative charge moves so only negative changed one should change? After they touch, Sphere B would remain as 20e?

Together, the net change is +10e, but Sphere B contributes more than Sphere A?

 

[Doc Al]

Only negative charge moves

True.

so only negative changed one should change?

No. If charge moves between spheres, they both must change.

After they touch, Sphere B would remain as 20e?

No.

Together, the net change is +10e,

Right. (That's the equivalent of 10 missing electrons.)

but Sphere B contributes more than Sphere A?

Why?

The two spheres are identical. When they touch, the charge redistributes evenly between the two spheres.

 

[Air]

True.

That's the equivalent of 10 missing electrons. The two spheres are identical. When they touch, the charge redistributes evenly between the two spheres.

So there is a distribution of +5e on both spheres. If we add this to the previous values (original charge value) then wouldn't we get:

Sphere A: -10e+5e = -5e and not +5e.

(Sorry, I'm still confused.)

 

[Air]

I think I've understood. I've read this thread: https://www.physicsforums.com/showthread.php?t=58634

So, the net change from Sphere 1 and 2 is +10e. But, for it to be stabl;e and equlibrium, they must separate equally hence Sphere 1 and 2 will both have +5e charge? Am I correct?

 

[Doc Al]

Yes, you are correct.

What actually happens is that 15 electrons will move from A to B in order to equalize the charge distribution, thus:
A goes from -10e - (-15e) = +5e;
B goes from +20e + (-15e) = +5e.

 

[Air]

Yes, you are correct.

What actually happens is that 15 electrons will move from A to B in order to equalize the charge distribution, thus:
A goes from -10e - (-15e) = +5e;
B goes from +20e + (-15e) = +5e.

Oh I see. Thanks for the help. You've been very patient and helpful. It was encouraging.

The reason that I was getting confused is because in Fundamentals of Physics, Chapter 21, Question 4, you have to use average to find the charge left over in the positive charge. Why is that different?

 

[Doc Al]

The reason that I was getting confused is because in Fundamentals of Physics, Chapter 21, Question 4, you have to use average to find the charge left over in the positive charge. Why is that different?

If you post that question, I can see what's different.

 

[Air]

The figure below shows three pairs of identical spheres that are to be touched together and then seperated. The initial charge on them are indicated. Rank the pair according to (a) the magnitude of the charge transferred during touching and (b) the charge left on the positively charged sphere, greatest first.

 

[Air]

I understand it now because your post number 7 explain why that works but why does for positive you have to add the average and for negative you have to take away the average?

 

[Doc Al]

In all cases the average charge tells you what the final charge on each sphere will be. To find the change in charge, just subtract final - initial.

For case (1) in your last post, you start with +6e and -4e. So the final charge on each will be (+6e -4e)/2 = +1e. Thus the change in charge on left sphere will be: +1e - 6e = -5e (it gained 5 electrons); on the right sphere, it will be: +1e - (-4e) = 5e (it lost 5 electrons).

Make sense? (Since the question asks for the magnitude of the charge transferred, the answer would be 5e.)

 

[Air]

Yes, that makes sense. Thanks a million.