Good evening. Is there someone for a fast help?

@Qmechanic @rob Good evening kind moderators. Can I have a fast help?

@Sebastiano Perhaps. What’s up?

@rob Thank you for your availability. It is a small curiosity on an exercise.

Two objects possess an electrical charge of -8.0 μC and -2.0 μC, respectively. How many electrons must be transferred from the first object to the second object in order for both to have the same electrical charge?

But to have the same electrical charge, don't they have to be put in contact?

@Sebastiano I interpret the question as “later both objects have charge -5µC, how many electrons moved?”

You could put them in contact to do so

(But that would bring them to the same potential, not necessarily to the same charge)

Or you could hit the more-negative one with UV light so that electrons are ejected, and collect the electrons on the less-negative one.

@rob In fact. I have done the calculus and I have done $(q_1+q_2)/2$.

That is -5µC: I have converted all in MKS

@rob and I have found 3.125 number of electrons.

Is that the European decimal, so “three thousand and change”? Or the American decimal, so “three and one-eighth?”

Neither is correct.

But the result is $1,9\times 10^{13}$ instead of $3,125 \times 10^{13}$.

Oh, that 10^13 is the part I was cuffed about.

@rob I use or the fullstop or the comma :-( (a bit European decimal and a bit American decimal with the fullstop.)

je te comprends le europe un peu, but my accent is terrible in all the languages

So, you moved 3µC from one object to the other

@rob ahahahahahaahahaha

You know that 1 e = 1.6e-19 C

The correct answer is: $1{,}9\times 10^{13}$. Yes yes....you use the math-notation

It’s a unit conversion. 3 µC / (1.6e-19 C / e) * (10^6 µC / 1 C) = 1.875e13 e.

So how have you gotten 3.125 as your initial significant figures?

@rob Are 6

Okay, skip the electron stuff. How much charge did you move to equilibrate the two charges?

@rob They are six

@rob But why I must done -8-(-2) if I must done the algebriac sum?

There’s your factor of two. Do it with beans in a cup. Only 3µC moves to equalize.

@rob But the formula what is? Why you moves to equalize only 3µC moves.

3µC=[-8-(-2)/2]µC. Why I must do the difference?

If they are in contact the charge equalizes: hence in total I have -10/2 µC

Ten passengers escape from a sinking ship. Alice, Bob, Charlie, Dawn, Edgar, Florence, George, and Harriet (total: 8) get in the first lifeboat. Isaac and Jennifer (total: 2) get in the second lifeboat.

The first lifeboat is too heavy and starts to sink.

The team decides to adjust so that there are five people in each lifeboat.

How many people move? Three.

@rob oK!

I have understood

@rob I thank you immensely.

Thank you very much for your precious time for me.

You’re welcome!

@rob I reciprocate your kindness. On my sheet of paper it said 5/1.6=3.125 $\ddot\smile$. Thank you again.