0
$\begingroup$

There are already a few questions which ask why charge is equal in two capacitors connected in series.

But what I can't get my head around is:

Isn't the capacitance how much charge a capacitor can store?

So by definition, wouldn't capacitors with different capacitors have a different charge?


Or is it that each capacitor won't be completely full?

$\endgroup$
2
$\begingroup$

Isn't the capacitance how much charge a capacitor can store?

No and in fact, capacitors don't store charge, capacitors store energy. The amount of charge $Q$ on one plate is balanced by the charge $-Q$ on the other plate. The total charge stored is $Q + (-Q) = 0$.

The energy stored in a capacitor is related to the voltage across the capacitor by

$$W = \frac{1}{2}CV^2$$

where $C$ is the capacitance and $V = \frac{Q}{C}$ is the voltage (potential difference) across the plates of the capacitor.

And so we see that the greater the capacitance of a capacitor, the greater the stored energy for a given voltage across.


Or is it that each capacitor won't be completely full?

For an ideal capacitor, there is no limit to the value of $Q$ and thus the voltage across and energy stored. For physical capacitors, there is a maximum voltage across before dielectric breakdown occurs.

$\endgroup$
  • $\begingroup$ "No and in fact, capacitors don't store charge, capacitors store energy", well yes, but they do it by separating charges, so you could well say they store (balanced) charges (at a distance from each other). $\endgroup$ – Gyro Gearloose Nov 20 '16 at 21:55
1
$\begingroup$

Capacitors with different capacity hold different amount of charges if they are held across the same potential. If Q(1) = C(1)V and Q(2)= C(2)V then off course Q1 is greater than Q2 if C1 is greater than Q2. In series connections charge cannot get lost. The capacitors of different capacity hold same charges because the potential difference across them is not same. Q=C(1)*V(1) (for 1 capacitor) and Q=C(2)*V(2) for the other capacitor. The V's vary in such way so as to give the same charge since Q charge leaving the battery has to return back to it.

$\endgroup$
1
$\begingroup$

My answer is an attempt to amplify and perhaps slightly change the emphasis of @AlfredCentauri 's answer.

There are lots of examples in Physics where a simplification is made for the sake of brevity.
A good example is the statement that a mass a certain height above the Earth's surface has a certain gravitational potential energy when in fact one should be stating that it the Earth and the mass that have the gravitational potential energy.

Capacitors are not immune to such short cuts.

Capacitors are all to do with separating charges and keeping them separated - storing separated charges.
If you start with two uncharged conductors and start moving charge $+Q$ from one conductor to the other (directly or indirectly) there will be set up an imbalance of charge on the two conductors with one conductor having a net charge of $-Q$ and the other conductor having a net charge $+Q$.
The total charge on the two conductors is $+Q+(-Q)=0$ but there is a net charge of magnitude $Q$ on each of the conductors.

So the term "charge stored" can be thought of as short hand for "magnitude of charge stored on one conductor" or "magnitude of charge moved from one conductor to the other".

The two conductors which store separated charge are called a capacitor and a measure of the ability of the two conductors to store separated charge is called capacitance.

It is found that the magnitude of the charge transferred between the two conductors or the magnitude of the charge stored on one of the conductors, $Q$, is proportional to the potential difference between the two conductors, $V$, and the constant of proportionality is defined as the capacitance $C$

So $\rm capacitance = \dfrac{\text{charge stored (on one of the conductors)}}{\text{potential difference (between the conductors)}}$

Now to move the charge $dQ$ from one conductor to the other conductor when the potential difference between them is $V$ work $(=VdQ)$ has to be done and thus the capacitor stores electric potential energy $\frac 1 2 QV = \frac 12 CV^2 = \frac 12 \frac{Q^2}{C}$.

If capacitance is the ability to store charge, doesn't this mean two capacitors of different capacitance will store different amounts of charge?

The comparison has to be made correctly in that if the potential difference across capacitors with different capacitances is the same then the charge stored (.......... etc) on the two capacitors will be different.
An example would be two capacitors with unequal capacitances connected in parallel.

In the case of two capacitors with unequal capacitance connected in series then the charge stored (.......... etc) on the two capacitors will be the same but the potential difference across each of the capacitors will be different.

As long as charge can be moved from one conductor to the other so the magnitude of the charge on one of the conductors will increase and so will the potential difference between the conductors.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.