Why Does a Switch Between Two 5V Capacitors in Parallel Show 5V Across It?

Hi guys, I know we were in a debate about voltage and I'm totally understandable of what does voltage mean .. but yeah still some gaps ! I need your help to overcome with ..and hope so.
let's assume I have one capacitor with 5v voltage ! and another capacitor with 5v voltage, between those two capacitors (they are connected in parallel ) there's a switch, so once the switch is off (opened) on two capacitors there's 5v voltage, but once I close the switch ten in the bridge of the switch itself there would be also 5v voltage , why? I MEAN ONCE I closed the switch why on the switch itself (lets assume switch looks like a wire) there's voltage of 5v ...?5v on first capacitors + switch is on + 5v on the second capacitors, so we get 5v transferred to 5v which mean nothing ... so the switch itself must be 5v?!
to be clear , switch is like a wire like this ....    .....
thanks alot

Hi Ryan -- with regard to the post you just made on your Current Divider question (https://www.eeweb.com/forum/current-divider) -- in which you say that you are not attending any formal training course, but are instead learning by yourself in order to work as an analog designer.Some people learn well by reading books and bouncing around the Internet. Others -- like myself -- learn better in a more structured environment where there is a teacher who can map out a path through the materials, starting with the fundamentals and then adding on the more complex concepts; also -- and very importantly -- who is available to answer questions.Having looked at all of your questions from the past, it seems like you are bouncing around from one topic to another -- every time you land on a new topic you fire off a bunch of questions -- and then you bounce off to the next topic.A big problem is that some of your questions are meaningless because you don't have a good grasp of the underlying concepts.I'm more of a digital guy -- less good at analog -- I can;t imagine trying to learn it by myself. I really would advise you to start looking for a course you could attend -- or an online training course -- but one that starts at the very beginning with the base-level concepts.

if you are more of "digital guys" here is a question, and I didn't think it's meaning less.
assume I have output from specific circuit which it's "1" and on the output connected serial an ideal switch, on the other side othe switch connected another capacitor which the voltage there is constant as "1" , so my question is: now I have like this 1 -----    ------1once I close the switch, how is 1 transferred to 1? how really the wire/bridge between one and one wil be also 1 ? I mean the voltage on the wire/switch once closed will be the same voltage as its sides, how is that happen? is the 1 traverse to the other 1?

That's a fantastic question!I'm glad you asked!
As I previously noted, the best answer I can give is to advise you to start looking for a course you can attend with a professional teacher
-- or an online training course -- but one that starts at the very
beginning with the base-level concepts.

but once again ! why I need to take course for that question?! if you answer this question and I didn't understand what you say, then I should take course for understanding what you're saying.
So please lets give attempt answer and afterwards ask me a question that implies if I understand what you explain or not ! I personally asked this questions because it's a lil confusion !
thanks alot

>> but once again ! why I need to take course for that question?!
Well, one reason might be that you don't appear to have understood any of the numerous answers posted by any of the EEWeb Experts to any of your myriad questions

>> So please lets give attempt answer and afterwards ask me a question that implies if I understand what you explain or not!I'm sorry, but I've already spend a humongous amount of time posting responses to your cornucopia of questions, but based on your follow-up questions, and your follow-up to your follow-up questions, it appears that nothing I say makes any sense to you.

>> ...if you answer this question and I didn't understand what you say, then I
should take course for understanding what you're saying...YES! That's what I've been trying to say -- I've answered many of your countless questions -- and you didn't understand what I said -- so that's why you should "take a course for understanding what I'm saying"

And one more thing -- a lot of people post questions on these forums -- our EEWeb Experts spend a lot of time answering these questions -- may of the people posting the questions are kind enough to post a follow-up saying something like ... oh, I don't know ... "Thank You"Looking at the back-and-forth between you and everyone who has posted answers to your questions, I don't recall a single time that you mentioned your appreciation for all the effort so many people have put in on your behalf.

Max to be fair he did say thanks ONCE to the best of my knowledge:https://www.eeweb.com/forum/voltage-concept#comment-223241
But I will agree it is pretty rare.

Ryan I think in many cases you are confusing yourself.  Just learn the rules of what happens, and don't try to understand exactly how things work - you've been asking questions which get into quantum physics where you want to learn analogue electronics basics.For example - just accept that copper wires conduct electricity.  You don't need to know how they do this, just accept that if you connect a resistance to a battery with copper wires, current will flow in the circuit.  Do you know Ohms law yet?  There is a good tutorial here. You NEED to know this well - it is far more important for electronics than what you have been asking.  Then  go on to Thevenin and Norton circuits, etc. With Memory, if you want to learn to program, just accept that when you write a value into a memory location or variable, that's what goes there.  Don't worry about what happens to the previous value.  If you have a piece of paper and your write "3" on it, then you rub the 3 out and write "4", the paper has 4 on it now.  Don't worry about where the "3" has gone. It's gone, you don't need to know any more than that.The www.electronics-tutorials.ws website will help you a lot, if there is something there you don't understand, give us the web page you have looked at and tell us what you don't understand.  Show your calculations as well, then we will be able to see where you went wrong.I will agree with Max that you should take a proper course on electronics.  I learned a lot by myself but I learned a lot more in the classroom training I did, because if I had a question I could ask a knowledgeable teacher right away and he would give me an answer.  A course will teach you the right things in the right order and it will make a lot more sense to you.You will not learn everything straight away, nor do you need to.  You're not asking stupid questions, but a lot of your questions have nothing to do with what you need to know to learn electronics.  Show us that you are trying to learn the basics, show us where you are learning from and try some practical work yourself with a breadboard, some components and a meter. You can't learn electronics just in your head.  Help yourself a bit and we will be happy to help you further.If you ask any more questions that are not relevant to what you are trying to learn I will be happy to point this out. 

Ryan, your question can be answered with Ohms law, if you know it.If you have two capacitors on either side of a switch, both charged to 5V, and you close the switch, what happens?There is 5V on both sides of the switch, when you close it there will be a very low resistance, almost 0 ohms.If you have the same voltage on both sides of the switch, there is 0v across the switch.By Ohms law, 0V across 0 Ohms = 0 Amps, so no current will flow.NOW, what happens if one capacitor is at 5V and one is at 0V (ie discharged).  What will happen if you close the switch?  The capacitors are both the same value).If you cannot work this out you need to know about charge on capacitors.  There is a good article on this in Electronics tutorials Here.  It will tell you that Q=CxV, charge = capacitance times voltage..  When you close the switch, the charge stays the same  but the capacitance doubles (the two capacitors are now in parallel).  So what will happen to the voltage?These are the sorts of things you so should be worrying about, not how current flows along a wire or through a switch..