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In Concepts of Physics by Dr.. H.C.Verma, in the chapter on "Capacitors", in page 144, under the topic "Capacitor and Capacitance" the following statement is given:

A combination of two conductors placed close to each other is called a capacitor. One of the conductors is given a positive charge and the other is given an equal negative charge. The conductor with the positive charge is called the positive plate and the other is called the negative plate. The charge on the positive plate is called the charge on the capacitor … Note that the term charge on a capacitor does not mean the total charge given to the capacitor.

(Emphasis mine)

According to the above statement, "the charge on the capacitor" is the charge on the positive plate when both positive and negative plates of a capacitor are equally charged? How to extend this definition when the two plates are unequally charged?

Suppose we supply charges $+Q_1$ and $-Q_2$ to the positive and the negative plates respectively. After electrostatic conditions are attained, I determined the charge distribution on the four surfaces - outer surfaces and inner surfaces of the two plates. The charge on the outer surfaces of the two plates are equal to $\frac{Q_1-Q_2}{2}$. The magnitude of charge on the inner surface of the two plates is $\frac{Q_1+Q_2}{2}$.

Now coming to my doubt, if the "charge on the capacitor" is the charge on the positive plate, which surface of the positive plate should I want to consider - the outer one or the inner one? And what is the reason for this choice? It seems to me that, a capacitor with unequal charges on its plates is not a capacitor in the first place (inferred from the statement above).

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The "charge on a capacitor", in terms of circuit theory, is equal to the amount of charge that would flow down a wire connecting one plate to the other, if a wire were so connected, until the current stopped. The current will stop when there is no potential difference across the capacitor. This will happen when there is no electric field inside the capacitor. For a parallel plate capacitor this will happen when both plates carry the same charge (with the same sign).

For example, if intially the charges on the plates are $Q_1$ and $Q_2$, then an amount $(Q_1 - Q_2)/2$ will flow from plate $1$ to plate $2$, with the result that the charge then remaining on plate $1$ will be $$ Q_1 - \frac{(Q_1 - Q_2)}{2} = (Q_1 + Q_2)/2 $$ and the charge remaining on plate $2$ will be $$ Q_2 + \frac{(Q_1 - Q_2)}{2} = (Q_1 + Q_2)/2. $$ At this point both plates have the same charge with the same sign, so there is no electric field between them, and no potential difference between them. Hence the charge on the capacitor is equal to $$ \frac{Q_1 - Q_2}{2} . $$

The usual case is $Q_2 = -Q_1$ and then the final charge is zero, and the amount that flowed was $Q = (Q_1 - Q_2)/2$ which is then equal to $Q_1$.

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Two plates having unequal charges do make up a capacitor and they are no different from the common sense of capacitors that we bear in our mind.

In your case, we would consider the charge on the inner surface. There is a reason for this. To understand this, let's first characterize a capacitor. So what comes up in your mind when you hear the word "capacitor"? Well, I think of some sort of charge separation which results in an electric field which has some energy associated to it. And this energy should be our characterising criteria, because that's what capacitors are used for, storing energy.

So now, in this case, which charges are responsible for the electric field and thus the electric field energy? Clearly, the field between the plates only depends on the charge on the inner surface of the plates. And thus while discussing the charge on the capacitor, it is the inner charge which matters. If you change the outer charges such that the inner charges don't change, then the field will also remain unchanged.

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    $\begingroup$ @MGuruVishnu Change the charges such that their difference remains a constant. $\endgroup$ – user243267 Dec 16 '19 at 14:30

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