What makes the charge on a capacitor change?

Question

I have just realized that I actually do not have a through understanding of capacitors and hence, I am asking this basic question.

I was trying to understand how an astable multivibrator works using this source and I have come across the following part (diagram is in the link, and below):

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The next significant thing which happens is when the voltage on the right hand plate of C1, and therefore TR2’s base, approaches +0.6V.

TR2 begins to conduct.

The collector voltage of TR2, which by now will have increased in value as C2 started to charge through R4, will fall and go down to +100mV.

The voltage across C2 can’t change instantaneously so when the voltage at its right hand plate goes down the voltage at its left hand plate goes down by the same magnitude. If TR2’s collector voltage had risen to, say, +3V, then when it reduces to +100mV (a negative-going change of 2.9V) the left hand plate of C2, and the base of TR1, will also go 2.9V more negative, but starting from +0.6V.

They will end up at –2.3V. This will cut TR1 off.

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So, as far as I understand, the following happens:

\$ C = Q / V \$ equation must hold for capacitors at all times. When we consider a particular capacitor, \$ C \$ is constant. This means that only \$ Q \$ and \$ V \$ may change in order to make the equation hold at all times.

However, there is a quirk with the capacitors. In a capacitor, \$ Q \$ cannot change instantaneously. That is, it takes time to change \$ Q \$.

Hence, when the voltage at one plate of a capacitor undergoes a sudden change (while the voltage on the other plate remains untouched), this event creates a situation where the equation \$ C = Q/V \$ does not hold. The reason is, \$ V \$ has changed, while \$ C \$ and \$ Q \$ are the same. (\$ Q \$ is the same because charge on a capacitor cannot change instantaneously)

However, we know that the equation \$ C = Q / V \$ must hold at all times. Hence, the capacitor responds this change by altering the voltage at its other plate, putting the equation \$ C = Q / V \$ in balance once again.

Here comes my question: Since the capacitor has changed V at its other plate in response to a sudden voltage change on the opposite plate, the equation \$ C = Q / V \$ is once again satisfied. However, we observe that the capacitor starts charging or discharging after this event.

Since the equation \$ C = Q / V \$ has been put in balance by the voltage change at the opposite plate, the reason for (dis)charging cannot be to balance the equation.

So if a capacitor does not (dis)charge in order to satisfy the equation \$ C = Q / V \$, then why does a capacitor (dis)charge?

Answer 1

The basic rule is you can't instantly change the voltage ACROSS a capacitor.

If you suddenly change the voltage on ONE PLATE then to maintain the voltage across the capacitor the other plate must instantly rise by the same amount.

Answer 2

This statement: -

In a capacitor, Q cannot change instantaneously. That is, it takes time to change Q

Is contradicted by this: -

the voltage at one plate of a capacitor undergoes a sudden change

Because: -

You cannot change the voltage instantaneously without infinite current being sunk into the capacitor. If infinite current is sunk (or sourced) by the capacitor then Q will change instantaneously.

The remainder of your question is based on this incorrect (or contradictory) premise.

Q = CV and the rate of change of Q (dQ/dt) = C dV/dt.

This of course equals current because rate of change of charge IS current.

Therefore if voltage changes instantly, dV/dt is infinite therefore current flow is infinite.

Answer 3

V is voltage across the capacitors plates, and that voltage can not be changed instantly. BUT potential of the plates can be changed instantly. When one plate of the capacitor, V1 ,is charged with some Q and because there is an electric field between plates, potential of other plate of the capacitor,V2, is instantly changed with same difference as first plate ( that makes V=V1-V2=Const.) When that process is done, the charge of the other plate V2 is going to flow through the resistor making a current with exponential form (because potential of that plate is falling down--that is discharge).

I hope this will help you to understand difference between voltage across capacitor and potentials of the plates, and what is discharge.

Charge of the capacitor is the same, if first plate of capacitor is connected to some voltage source via resistor (then you would have some current flow, and current is bringing charge to the plate of capacitor). At same time other plate is changing it potential making voltage across capacitor constant.

Answer 4

As Q=CV, the charge storage will depend upon the capacitance value which is factory built and specifies and also the voltage applied across its terminals.