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In a capacitor , capacitance is given by
$$C=\frac{Q}{V}$$

I learned from various sources that V is the potential difference between charged plate and supporting plate and not just the potential of charged plate . Assuming it is true , I can't understand the principle of parallel plate capacitor ,that the supporting plate helps in decreasing V and thus increasing the capacitance because:-

Potential difference can't actually be increasing. It is more convincing to say that the potential of charged plate is decreasing while potential difference is always remaining constant according to the following video
https://www.youtube.com/watch?time_continue=1&v=HQPcOCuVG1I

Can anyone point out why the professor says more charge can be stored , even when potential difference remain constant (not decreasing )and only potential is decreasing, while potential is not actually the part of above equation.

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    $\begingroup$ Absolute values of potential don't have meaning. Potential energy is always defined only up to an additive constant, so differences are the only meaningful thing to talk about. $\endgroup$ – jacob1729 May 29 '18 at 13:33
  • $\begingroup$ @jacob1729 But, in that video potential difference always remain constant , right ? Then ,how can it help in increasing charge stored in the plate $\endgroup$ – salvin May 29 '18 at 13:37
  • $\begingroup$ I've looked at the first two thirds of this video, and wouldn't recommend using it to learn about capacitors. There's one outright mistake, and that is trying to apply the equation for the potential due to a point charge to a uniformly changed flat plate. But, quite apart from that, the whole approach via bringing up an earthed plate to a charged plate is, in my view, unnecessarily complicated. As others have said, it's the potential $difference$ between the plates that matters. Of course you may have your own reasons for approaching capacitors in this way – in which case, good luck! $\endgroup$ – Philip Wood May 29 '18 at 14:14
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"Can anyone point out why the professor says more charge can be stored, even when potential difference remain constant (not decreasing) and only potential is decreasing, while potential is not actually the part of above equation." If he says that, I believe that he is not talking about a capacitor as generally understood, that is two plates, separated by a small gap or an insulator, with equal and opposite charges on the facing surfaces of the plates.

The approach in this video dates from when capacitors (known then as 'condensers') were constructed out of plates held on insulating stands and experimented with in a Physics laboratory, but rarely seen outside a lab. These days capacitors are fully-formed devices typically consisting of two metal foils separated by an insulator, often rolled up into a small cylindrical package and sold for a few pence (or cents). Electronic devices often contain dozens (or even millions) of them.

Capacitors are, today, understood in terms of the pd between their plates, which produces an electric field between the plates, and this electric field can be related (by Gauss's theorem) to the charge on each plate. From the analysis emerges $Q=CV$, together with an expression for $C$ in terms of the geometry of the gap between the plates and the nature of the 'dielectric' between the plates. We don't bring the absolute potential of either plate into the argument.

This is the way I recommend you learn about capacitors, unless you have special reasons for using the 'earthed plate' approach. [In my opinion it is confusing to regard a capacitor made of two plates close together as a modification of a single charged plate; the transition is not a straightforward one.] And as I've said in my comment, there's quite a serious mistake made in your video: the teacher wrongly applies the formula for the potential due to a point charge to a flat charged plate.

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