1
$\begingroup$

'When there are 3 capacitors connected in series, charge across each capacitor is the same.'

Okay so say we consider three caps made of plates of the same conducting materials and same dielectric materials.

                 -----||-----||-----||----
                      10mF   1mF

So when a charge is accumulated on the left plate of the first capacitor, by induction, this charge leaves the capacitor through the other plate. However, the second capacitor has a smaller capacitance value and hence less ability to store the charge provided by the first cap? So all the charge wouldn't flow through the second cap. So if it didn't what would happen to the excess charge?

$\endgroup$
  • $\begingroup$ By the way, capacitors store energy, they do not store charge. $\endgroup$ – RedGrittyBrick Jul 29 '16 at 18:18
2
$\begingroup$

You are right that the "second (smaller) capacitor has a reduced ability to store charge" but implicit in this statement is that the voltage across each capacitor is held constant. For example, the capacitance $C$, charge $Q$, and voltage drop across the capacitor $V$ are related by $Q=CV$. If $V$ is constant, larger $C$ means larger $Q$.

When the capacitors are in series this is not that case. The charge in the wire between the first and second capacitors must remain in that segment (the electrons can only move through a conductor and the gaps are dielectrics). This means that $+Q$ is on one capacitor (one side of the wire segment) then $-Q$ must be on the other capacitor (the other side of the wire). Another way to put is that there must be charge conservation on an isolated conductor.

If $Q$ is the same on both conductors (the sign isn't important), then $V_1 C_1 = Q= V_2 C_2$. So a smaller capacitor will have a larger voltage drop across it.

Remember each point of a perfect conductor must have the same potential or it will drive currents to balance out any potential differences. At the same time, the net charge on an isolated conductor is constant due to charge conservation.

Side note At a certain point, if the voltage gets too large, then the dielectric could breakdown and allow a discharge. If you put overcharge a capacitor (apply a large voltage for long enough) then the gap voltage could get too large.

$\endgroup$
  • $\begingroup$ But if the two capacitors are at different potentials, why wouldn't charge flow such that they acquired an equal potential difference across their plates? $\endgroup$ – LeroyJD Jul 29 '16 at 3:56
  • 1
    $\begingroup$ Ah great question. The capacitors are not at potentials $V_1$ and $V_2$. Those potentials describe the potential drop over each capacitor. Each wire segment is at a constant potential since we are treating them as perfect conductors. The dielectric barrier in the capacitor is not a conductor so it can support a voltage drop without driving a current. $\endgroup$ – LasersMatter Jul 29 '16 at 3:58
  • $\begingroup$ I'll edit my answer to mention that the voltages refer to voltage drops across the capacitors. $\endgroup$ – LasersMatter Jul 29 '16 at 4:04
  • 1
    $\begingroup$ So basically even though the potential drops across the capacitors are different, the potentials to which they are charged for a given value of Q is the same (dependent on the capacitors dimensions) ? So for a capacitor of smaller capacitance, according to the formula Q = CV, smaller capacitance capacitors would have a higher potential drop across them? $\endgroup$ – LeroyJD Jul 29 '16 at 4:25
  • $\begingroup$ I am not sure I follow the first question. You may have to clarify the language "the potentials to which they are charged". Each wires segment can be properly described as having a potential. Each capacitor can properly be described as having a potential drop. If I say the capacitor "has" potential $V$ it is a sloppy shorthand for referring to the drop across it. For the second question, yes if Q is the same. $\endgroup$ – LasersMatter Jul 29 '16 at 4:38
2
$\begingroup$

The battery in a closed circuit does not create any charge; it "pumps" charge around the circuit. Because of this, the negative charge on a negative capacitor plate had to be pumped there from the positive capacitor plate. This means that for only one capacitor in the circuit, assuming that the capacitor started with no charge on either plate, the final charges on the capacitor plates are equal and opposite (1 plate is positive and 1 plate is negative) because charge must be conserved. The exact same scenario applies for capacitors in series, assuming that all capacitors started with no charge on them. Because the negative charges on all capacitor plates had to come from positive capacitor plates, and because all capacitors are in series, the same amount of charge has to exist on all capacitor plates regardless of the individual capacitances, because charge must be conserved (i.e., the electrons on the negative plates had to come from somewhere).

$\endgroup$

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.