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Capacitance is measured in Farads (1 F ≡ 1 C / V). Note that ε also has units of F / m.
Capacitance depends only on the geometry of the capacitor and the permittivity. To compute the capacitance, use Gauss' law to compute E and
integrate to get ΔV.
ΔVeq = ΔV1 + ΔV2
Qeq = Q1 + Q2
= C ΔV2 / 2.
Capacitance
C ≡ Q / ΔV.
Any charge that enters one side of a series configuration must leave the other side.
Two capacitors in series have the same charge, so if we replace them with an equivalent capacitor, we must have
Qeq = Q1 = Q2
or
1 / Ceq = 1 / C1 + 1 / C2
Since conductors are equipotentials, components in a parallel configuration must be at the same potential where they are connected.
Two capacitors in parallel have the same ΔV ("voltage drop"), so if we replace them with an equivalent capacitor, we must have
ΔVeq = ΔV1 = ΔV2
or
Ceq = C1 + C2
∫ ΔV dq
If we think of this energy as being stored in the electric field within the capacitor, we can compute the energy density of the electric field to be
= Q2 / (2 C)
uE = ε E2 / 2.
(This can be shown using a parallel plate capacitor:
C ΔV2 / 2 / (area * distance) = ε ΔV2 / (2 distance2)
but it is true in general.)
= ε (E * distance)2 / (2 distance2)
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©2011, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.
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