doubt on capacitors the two plates

Hi All,

I have this doubt about capacitors. suppose one plate of capacitor is charged with say 5V and the other plate of the capacitor with say 3V, then the two plates are at different potentials? or both of them will adjust to a common voltage? what is the voltage across the capacitor? another doubt is if i connect a capacitor across a voltage source and if the voltage raises suddenly to some voltage what is the voltage across the capacitor? According to the theory the capacitor voltage raises slowly, but using multimeter or any instrument if I measure a voltage across a capacitor it will always give me the reading of the supply voltage then how anyone can prove that the capacitor voltage raises slowly? (It can be shown with a resistor, but without resistor how can it be proved?).

Thanks in advance,
regards

If they have different values, they charge or store different voltages, at all times make sure Cap specification voltage is always bigger than that of the supply!

Do not think of a single plate of a capacitor being charged to a potential. It does not work like that.

A capacitor is energized by a potential difference across its plates, to a charge given by Q=CV. Each plate is energized to the same magnitude of charge, Q, but with opposite polarity.

So, if you connect a capacitor between 5V and 3V, all that matters is the potential difference of 2V. The capacitor's plates will energize to a charge given by Q=CV. Assuming you give the system enough time to settle. Conversely, if you now remove the power supply, the capacitor is left with a charge, Q, on its plates. This gives rise to a potential difference across the plates again given by Q=CV. If nothing has changed (assuming a perfect/ideal capacitor with no leakage, etc) then the potential difference across it will be the same, 2V, as you energized it with in the first place.

The time taken to fully energize depends on how quickly enough energy can be transferred; that depends on the resistance in the circuit between the plates (and even of the plates themselves) and the power supply.

There is nothing to say it must be 'slow'. 'Slowly is an ambiguous and relative term; it has no meaning here without giving it context.

The charge on the capacitor with time is given by:

Q = Q_max ( 1-e^-t/RC)

Since Q is proportional to V, you can also say:

V = V_max ( 1-e^-t/RC)

So, you can see that as R gets smaller, the -e^-t/RC term becomes less significant, so the capacitor reaches full energy quicker. In theory, it never quite gets to full energy in a real system (with non-zero resistance), but in practice it gets close enough that we don't care.

Thank you very much for the reply, one final question before I start studying about capacitors once again. As you said that equal charge is stored on the two plates, is it because of diffusion of electrons where it wants to make the conductors two plates neutral ? Basic question is how equal charges are stored on the two plates that is the process involved and why?

Thanks in advance,
regards

One more thing form your first post the Capacitor voltage does vary in the case of AC supply till it reaches steady state that as Foxy said depends on the Time constant

Mostly the time constant is too small for human eye to note unless the capacitor has very small value

electronic_satya, when the capacitor is connected to the voltage source, electrons (charge) flows around the circuit. When the voltage across the capacitor is the same as the voltage source, you have an excess of electrons on one plate and an equal deficit on the other. A surplus of electrons is a - charge and the deficit is a + charge, these two charges attract each other, so the charges remain stationary. If you put a resistor across the capacitor, the electrons flow and cancel out the deficit until the voltage across the capacitor is zero when no more electrons flow (zero current)
Frank