**Capacitors
in parallel and series**

Here we discuss the
effective capacitance of connecting two capacitors together. In
the figure on the left the capacitors are connected in * parallel*,
while in the right-hand figure the capacitors are connected
in

First, let us consider the left-hand figure where the two capacitors are in parallel. The charge on the combined left-hand element is:

,

where *V* is
the voltage across both elements. Remember that the voltage along
a conductor does not change, and that a wire is a conductor, so
both capacitors have the same voltage drop across them. From
inspection of the equation above, one sees that the effective
capacitance, *C* is:

This makes sense
because for parallel plate capacitors,
the capacitance is *e*_{0}*A/d*, and
putting two identical capacitors side by side should look like
one capacitor with double the area and therefore double the
capacitance.

Next, consider the
situation on the right, where the capacitors are in series.
Remember that when we refer to the charge on a capacitor, it means the
charge on one plate; the opposite charge appears on the
other plate. Thus, the charge
*q*_{d} = *q*_{c}.For
two capacitors in series, the charges are identical, but the
voltages are different. The sum of the voltage drops across the
two capacitors must equal the net voltage drop *V*.

Using the fact that
the effective charge *Q* on one plate of the combined system
equals *q*_{c}
which equals *q*_{d}, we see that

Thus we have found that
**
**

- For capacitors in parallel, the capacitances add
- For capacitors in series, their inverses add