140 Week 4: Capacitance
+Q −QC 1 C 2 C 3Ctot
Vtot1+Q −Q +Q −QV V 2 V 3
+Q −QFigure 42: Find the total capacitance of a much of capacitorsin series.Let’s use this symbol (and our knowledge thatC=Q/V) and compute thetotalcapacitance of
seriesandparallelarrangements of capacitors. We’ll start with series.
In figure 42 we see two arrangements. The top arrangement consists of three capacitors, labelled
C 1 , C 2 , C 3 , in aline, so that the tail of each is connected to the head of the next one by aconducting
wire(which appears as a simple straight line in the figure). This arrangement is calledseriesas each
capacitor “follows” the next. Underneath this is a single capacitor labelledCtot.
We need to find whatCtothas to be for these two arrangements to behaveidenticallyin an
electrical circuit. That is, when our devil-dude moves a chargeQfrom oneendto the otherend,
we want the potential differencebetween the endsto be exactly the same. Here’s how you can
understand what goes on.
Suppose you have a charge +Qon the leftmost plate as shown (which came from the rightmost
plate in either arrangment, leaving behind a charge of−Q). This pair of charges creates afieldin
between. However, there canbe no fieldin the conducting plates and wires in the middle of the
top row – they are in equilibrium! To cancel the field produced by the first plate, a charge−Qis
attracted to the plate facing it. But it cannot come from any part of the conducting plates or wires
in between, it has to come from the surface of the next plate (leftmost of capacitorC 2 ) charging
itup to +Q. This in turn attracts−Qto the right plate ofC 2 , leaving a charge +Qon the left
plate ofC 3. At this point (and you should check this) the capacitors should all be happy. Each one
has a charge±Qon it, with a field confined to live only between its plates. The field is zeroinside
the plates themselves and in the connecting wires. Note that all we really used in this reasoning
ischarge conservation– we couldn’t create charges anywhere, only move charges around– and the
idea that conductors in equilbrium can have no field inside.
Now consider thepotential differencesacross each capacitor on top. Clearly the potential dif-
ference acrossC 1 isV 1 =Q/C 1 , the potential difference acrossC 2 isV 2 =Q/C 2 , acrossC 3 is
V 3 =Q/C 3. Similarly the potential difference across our desired total capacitance isVtot=Q/Ctot,
since it has to have thesamecharge on its left plate as the arrangement on top.
Each wire between the capacitors isequipotential, because conductors in electrostatic equilibrium
have no field inside and are thus equipotential. If we want to find thetotalpotential difference across
the top row of capacitors, we just have toadd up the potential difference across each capacitor. You
can think of this as doing a piecewise continuous integral across thewire at one end (get zero), the
gap (pick up potential differenceV 1 ), across the next wire (get zero), across the next capacitor’s
gap, (getV 2 ) etc. We end up with thetwoequations for the upper and lower arrangements:
Vtot = V 1 +V 2 +V 3 +...=Q
C 2 +
Q
C 2 +
Q
C 3 +... (234)
Vtot = Q
Ctot(235)
where the dots indicate that there was nothing special aboutthreecapacitors in a row – there could