Charge Conservation: A+2 & B+7 → A+5 & B+5

[Ed Quanta]

Now suppose we have two conductors A and B. A has a charge of +2 and B has a charge of +8. If we let the two conductors touch and then separate them, each will have a charge of +5. This is because charge is conserved and will flow between the two conductors until each have equal charges.

Ok, so how would this change if instead of B having a charge of +8, it had an initial charge of +7. Now if let the conductors touch and separate them, each of them cannot have a charge of +4.5 because charge is quantized and there is no such thing as half of an electron. So what would happen? What would be the charge on each conductor?

 

[neu]

When you say a charge of do you mean in units of e?

This is confusing. If you are talking in units of e then you describing basically two nuclei of Z=2 and 8 "touching". This is nothing like touching 2 macroscopic objects of some charge.

can you explain a bit more what you mean?

 

[neu]

Wait, sorry do you mean a macroscopic conductor with net charge of +2e and +8e

 

[Ed Quanta]

Yeah

 

[neu]

Well, the net charge for A and B in isolation would be +8e and +7e. This is the sum over ALL the charges of the system where all the positive nuclei are canceled by all but 8 and 7 electrons.

So you are not dealing with an 8e and 7e system, you have to take into account all the electrons.

When A and B approach in the first case as you say, the electrons will equilise the charge on each body. In the second case where there is an odd number of electrons and and even number of protons for the whole system (A+B) then the charge distribution of A and B will shift to accommodate the charge imbalance.

 

[Dale]

There would be a charge of 4 electrons on one and a charge of 5 on the other. Assuming that your spheres were absolutely perfect identical spheres then there would be a 50/50 chance of each having the "extra" charge. It would be determined simply by thermal noise.