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Prove. The equivalent capacitance C_{E} of n capacitors C_{1}, C_{2}, ... , C_{n} connected in series is given
by

where C_{1}, C_{2}, ... , C_{n} are the capacitances of the
capacitors.

Proof. We will prove it for three capacitors C_{1}, C_{2},
and C_{3}.

Referring to Fig. 1, if the left plate of capacitor C_{1}
receives a charge +Q, then an equal negative
charge -Q will be induced on the right plate. Now
if a negative charge -Q appears on the right plate of
C_{1}, then a positive charge +Q must appear on the left plate of C_{2} since the right plate of C_{1}, the
left plate of C_{2}, and the wire connecting them can be viewed as a single electrical conductor that
was originally uncharged and is isolated from the rest of the circuit. Thus each capacitor receives
a charge of magnitude Q on each of its plates as shown in the figure. Now

2) V_{af} = V_{ab} + V_{cd} + V_{ef}

and, by definition of capacitance,

3) C_{1} = Q / V_{ab} C_{2} = Q / V_{cd} C_{3} = Q / V_{ef}

Substituting 3) into 2) gives

4) V_{af} = Q / C_{1} + Q / C_{2} + Q / C_{3}

Now an equivalent capacitance C_{E} for this arrangement is the capacitance of a single capacitor
that would become charged with the same charge Q when the potential difference across its
terminals is V_{af} . Consequently,

5) C_{E} = Q / V_{af}

From 4) and 5) we get

6) Q / C_{E} = Q / C_{1} + Q / C_{2} + Q / C_{3}

which becomes

6) 1 / C_{E} = 1 / C_{1} + 1 / C_{2} + 1 / C_{3}

We note that the energy of the single equivalent capacitor is equal to the sum of the energies of
capacitors C_{1}, C_{2}, and C_{3}. The energy of the equivalent is ½ QV_{af} and the sum of the energies of

capacitors C_{1}, C_{2}, and C_{3} is ½ QV_{ab }+ ½ QV_{cd }+ ½ QV_{ef } = ½ QV_{af} .

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