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Suppose if we supply two plates of parallel plate capacitor with charges $Q_1$ and $Q_2$.What would be the charge of the capacitor? As the field inside plate should be zero and hence using gauss law I can understand that charges on inner surface should be equal in magnitude and of opposite nature.Further charges on outer surface would be $\frac {Q_1+Q_2}2$ but what do we call the charge of capacitor in these cases?

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    $\begingroup$ I'm not totally clear on what you're asking. Usually we talk about a capacitor with charge $+Q$ on one plate and charge $-Q$ on the other plate. Are you asking about the case where the two charges are not exactly equal and opposite, and the capacitor has a net charge? $\endgroup$ – rob May 12 '14 at 13:10
  • $\begingroup$ @rob I want to ask what if the charges on the two plates of parallel plates capacitor is not equal.what will be the charge of the capacitor in this case? $\endgroup$ – user117638 May 12 '14 at 13:35
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First, to be clear, when we speak of a charged capacitor, we almost never mean electrically charged but, rather, 'charged' with energy in the same way we speak of a charged battery.

If the plates of a capacitor with capacitance $C$ have equal and opposite electric charge $Q$, the capacitor is electrically neutral but stores an energy

$$U = \frac{Q^2}{2C} = \frac{1}{2}QV$$

where $V$ is the potential difference between the plates.

If the plates of a capacitor have unequal charge, there is now energy stored in more than one capacitance. There is the capacitance that exists between the two plates (the mutual capacitance) as well as the capacitance of each plate to the environment.

Let $V_1$ be the potential of plate 1 with charge $Q_1 = Q$.

Let $V_2$ be the potential of plate 2 with charge $Q_2 = -Q + \delta Q$.

The energy of system is then

$$U = \frac{1}{2}Q_1V_1 + \frac{1}{2}Q_2V_2 = \frac{1}{2}Q(V_1 - V_2) + \frac{1}{2}\delta QV_2 = \frac{Q^2}{2C} + \frac{\delta Q^2}{2C_{20}}$$

where $C_{20}$ is the capacitance of plate 2 to the environment.

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For an example of a real system where this occurs, consider the capacitor in a high-pass filter. Here's a diagram from Wikipedia:

high-pass filter

If $V_\text{in}$ is a constant, the capacitor will charge up, but $V_\text{out}$ will discharge through the resistor. In that case one side of the capacitor is at ground.

However if $V_\text{in}$ oscillates, it takes some time for the capacitor to charge up, and so an oscillating signal appears at $V_\text{out}$. There is a phase shift between $V_\text{in}$ and $V_\text{out}$, so for much of the oscillation neither end of the capacitor is at the ground potential. This is exactly the situation that you're talking about, except perhaps that you weren't imagining a transient net charge.

Alfred Centauri says, correctly, that there will be an "extra" capacitance between the net charge on $C$ and the ground plane of the circuit. However for all but the most sensitive of measurements this "stray" capacitance can be neglected. Typically stray capacitances have values of a few picofarads, while intentional capacitors may carry nanofarads or microfarads.

In analyzing a real circuit, then, we always describe the "charge on the capacitor" $Q$ as if the capacitor had zero total charge, so that the voltage drop across the capacitor is $V = Q/C$.

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enter image description here

From your conclusion, which is correct, $$\large q=\frac {Q_1+Q_2}2$$ and thus as charge is conserved the charge on inner surface of plate $\mathrm{I}$ is $$Q_1-q=\frac {Q_1-Q_2}2$$ and similiarly on inside of other plate $\mathrm{II}$ is $$Q_2-q=\frac {Q_2-Q_1}2$$ Note that these two are just equal but negative in charge, which poses a similiarity to a "normal capacitor", when the outside charge is zero and the inside charges are equal and opposite in sign; The charge of the capacitor can be said as $$\left|\frac {Q_2-Q_1}2\right|$$

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