# Connecting a charged capacitor to an uncharged capacitor

I was attending a lecture about capacitors and something confused me.

If I charge a capacitor using a DC supply, the capacitor will gain charge $Q_0$.

Now, if I discharged it along an uncharged capacitor in this arrangement, according to the lecture notes, the capacitors share the total charge $Q_0$.

Now, I had a question. Aren't there electrons on the uncharged capacitor, such that they flow between the two capacitors to cause equal p.d. on both capacitors hence the total charge in this circuit greater than $Q_0$?

What you suggest would be true if the upper component were a battery, which maintains a certain definite voltage and can act as a supplier of charge. But a capacitor has only a finite charge, here $Q$, and as it flows off to the other capacitor it is not replenished.

The usual confusion with this particular circuit is to ask what happens to the energy...

When we charge a capacitor, it gains charge q on one of the plates and loses charge q from the other plate, i.e., its total charge remains zero.

Capacitors differ, in that sense, from other objects, like our bodies or spheres and rods used in various electrostatic devices and experiments, which actually gain a net charge, when they are charged.

So the charged capacitor on the top of your diagram, initially has charge $+Q_0$ on the left plate and charge $-Q_0$ on the right plate and its total charge is zero.

When the switch closes, $Q_0/2$ worth of electrons are flowing from the right plate of the top capacitor to the right plate of the bottom capacitor and $Q_0/2$ worth of electrons are flowing from the left plate of the bottom capacitor to the left plate of the top capacitor. The resulting total charge remains zero.

When we say that a capacitor is uncharged it means that the net charge on each plate of the capacitor is zero ie equal numbers of positively charged ions and negatively charged electrons.
The charged capacitor also has a net zero charge it just so happens that there is a net surplus of electrons on one plate and an equal net deficit of electrons on the other plate. The magnitude of the surplus/deficit you have called $Q_0$.

Overall the net charge on the system of the two capacitors before connection is zero and stays zero after connection.
If you start with a surplus/deficit of $Q_0$ then it will stay as such because there is no way that the charges can neutralise one another as they reside on two different parts of the circuit separated by insulators between the plates.

Finally found the answer box! Now to the point. My dear friend I would suggest you to get yourself enlightened on conservation laws.

https://en.wikipedia.org/wiki/Conservation_law

• conservation of charge
• conservation of mass-energy (Its two approximate conservation laws are conservation of mass and conservation of energy )
• conservation of linear momentum
• conservation of angular momentum

Yes, the electric potentials do become the same due to discharging and that does cause a transient current. But current is itself the flow of charges and it is charge which is flowing from higher to lower potential (In case of positive charge, conventional current; the reality is the opposite, electrons are the only mobile charges and thus move from lower to higher potential).

Hence the net charge in any circuit (closed system of charges) will always remain the same. Instead, it is only that the charges are in motion. The current usually drops when the emf provided decreases due to loss of free energy in the cell as the chemical reaction driving the battery further moves towards an chemically equilibrium state leaving no energy remaining to provide the potential for the charge flow in the circuit.