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I have a question that is bothering me, Lets assume that I have a parallel plates capacitor with distance $d$ between them. Now connect it to DC voltage source $V$. according to equation $$V=\frac{Q}{C}$$ $$C\approx\frac{1}{d}$$ by getting the plates smaller I will decrese $Q$ because $V$ is constant in this case (I am not plug out the battery) my question is where the charge $Q$ going? the capacitor is connected only to voltage source so intuitively I do not understand why would I will lose charge?

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When the separation distance $d$ increases, all other things being equal, the capacitance is less, as you already know. But the charge doesn't "go anywhere". You don't "lose charge".

With no voltage applied to the capacitor, the charge on the metal plates consists of an equal number of mobile free electrons and protons. The total charge on the capacitor is the sum of the two and is constant. But there is no net positive or negative charge on each plate until a voltage is applied.

When a voltage source (e.g., battery) is connected it does work to separate the existing charge into plus on one plate and minus on the other. It doesn't add any charge to the capacitor. A lower capacitance simply means that less charge $Q$ can be separated per volt $V$ applied across the plates. No charge is "lost". Or, to put it another way, it requires more work per unit charge (more voltage) to separate the charge the greater the distance between the plates.

I am still confused than what does is it mean when the value of 𝑄 is decrease /increase while changing the distance between plates. and when you say the net charge is constant is it necessary zero?

It means that the amount of the charge on the capacitor that is separated into plus charge on one plate and minus charge on the other plate is less. Perhaps the following rough analogy will help.

Let's say I have 20 marbles of which 10 are black (the electrons) and 10 are red (the protons) on each side of a table (analogous to each plate of the capacitor). There is no net amount of red or black on either side of the table. There is a total of 40 marbles on the table. This is our capacitor with no voltage applied between the two plates of the capacitor.

A force (analogous to electric field) does work per marble (analogous to voltage) comes along and pushes (or pulls) 5 black marbles from one side of the table and moves them to the other side of the table . Now there is an excess of 5 red marbles on one side of the table (the positively charged plate of the capacitor) and an excess of 5 black marbles on the other side of the table (the negatively charged plate of the capacitor). Note that there are still a total of 40 marbles on the table.

Now the distance between the two sides of the table is increased (greater $d$ in the capacitor). It now requires more work per marble (voltage) to move the marbles from once side of the table to the other. Maybe we can only do enough work per marble (if there is no increase in voltage) to move only 3 black marbles (equivalent to a lower $Q$) from one side to the other. Now there is only an excess of 3 red marbles on one side of the table and 3 black marbles on the other side of the table. The increase in distance across the table is analogous to the increase in plate separation. It requires more work per unit charge to move charge from one plate to the other.

In all of this, regardless of how many marbles we are able to move from one side of the table to the other, the total number of marbles is the same. For the capacitor, the total amount of charge is the same.

Hope this helps.

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  • $\begingroup$ I am still confused than what does is it mean when the value of $Q$ is decrease /increase while changing the distance between plates. and when you say the net charge is constant is it necessary zero? $\endgroup$
    – Sagigever
    Sep 2 '20 at 22:04
  • $\begingroup$ @Sagigever I have updated my answer to respond to your follow up question. Hope it helps. $\endgroup$
    – Bob D
    Sep 2 '20 at 23:39
  • $\begingroup$ This is great explantion, thank you very much! $\endgroup$
    – Sagigever
    Sep 3 '20 at 8:01
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The electric field on one plate is "felt" by the other.

A simple way to think about why the distance between the plates matters, is that the closer the plates are, the more strongly will the field of one plate help pull charges towards the other plate. Thus, more charge can be accumulated on the plates when they are closer.

When the charges are not concentrated/accumulated at the plate, then they are distributed somewhere else along the wire/circuit leading up to the plate. Or possibly, the voltage source, such as the battery, has not yet released them to the battery terminal yet (the electrochemical reaction that releases the charge has not yet happened).

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You connect the battery which puts a net charge $+Q$ on the top plate and $-Q$ on the bottom plate. The plates are attracted to each other and when you pull them apart you are doing work on the system.

My guess is that this work ends up forcing some of the charge on the plates to flow back to the battery.

Maybe try this calculation:

Separate the plates by some small amount $ds$. You have to do work $Fds=(VQ/s) ds$ (where $s$ is the current plate separation, and $V/s$ is the electric field between the plates; all of this is happening at constant $V$ set by the battery).

The hypothesis is that this work is used to transport charge off the top plate, through the battery "the wrong way" and onto the bottom plate. We can write this as, $$ \frac{VQ}{s}ds = - VdQ, $$ where the minus sign is because losing charge from the top plate ($dQ<0$) corresponds to doing work on the system (i.e. forcing the charge "uphill" through the battery).

Therefore, $dQ/Q = -ds/s$.

This is consistent with the two equations you started with: $Q=CV$ and $C \propto 1/s$. So maybe this scenario is plausible.

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  • $\begingroup$ According to what you wrote, I understand that I lose the net charge on the capacitor even though it doesn't connect to a circuit wich is really confusing because the other 2 answers says that the net charge on capacitor is constant as long it doesn't connet to a circuit $\endgroup$
    – Sagigever
    Sep 3 '20 at 8:01
  • $\begingroup$ Well the net charge on the capacitor is zero at all times: $+Q$ on the top plate and $-Q$ on the bottom plate, no matter what value $Q$ actually takes. And there is a complete circuit here. Positive charge on the top plate can flow to the (+) terminal of the battery and out the (-) terminnal to the bottom plate of the capacitor. $\endgroup$
    – Alex
    Sep 3 '20 at 15:13

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