Question 16.5: Capacitors C1 = 6.0 nF and C2= 3.0 nF are connected in serie...
Capacitors C_{1}=6.0 nF \text { and } C_{2}=3.0 nF are connected in series. (a) What’s the equivalent capacitance of this combination? (b) Find the charge on each capacitor and the potential difference across each when the series pair is connected across a 9.0-V battery.
ORGANIZE AND PLAN We start by drawing a circuit diagram (Figure 16.19).
For the two capacitors in series, the equivalent capacitance follows from the reciprocal rule, Equation 16.12a. Each capacitor carries the same charge, which follows from Q=C_{ s } V Knowing Q, we can get the individual potential differences from V = Q/C.
\text { Known: } C_{1}=6.0 nF ; C_{2}=3.0 nF .
\frac{1}{C_{ S }}=\frac{1}{C_{1}}+\frac{1}{C_{2}} (16.12 a).
(a) The equivalent capacitance follows from Equation 16.12a:
\frac{1}{C_{ S }}=\frac{1}{C_{1}}+\frac{1}{C_{2}} (16.12 a).
\frac{1}{C_{ s }}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{6.0 nF }+\frac{1}{3.0 nF }=\frac{1}{2.0 nF }.
\text { Therefore, } C_{ s }=2.0 nF .
(b) The charges on the two capacitors are the same:
Q=C_{ s } V=(2.0 nF )(9.0 V )=18 nC.
Then the potential differences are
V_{1}=\frac{Q}{C_{1}}=\frac{18 nC }{6.0 nF }=3.0 C / F =3.0 V.
V_{2}=\frac{Q}{C_{2}}=\frac{18 nC }{3.0 nF }=6.0 C / F =6.0 V.
REFLECT Here’s a check to see if the last result makes sense: The potential differences V_{1} \text { and } V_{2} should add up to the potential difference across the battery terminals, and they do -it’s 9.0 V.
Note that now the equivalent capacitance is smaller than the capacitance of any individual capacitor -which follows because their reciprocals add. Notice throughout this example how we repeatedly applied the capacitance definition C = Q/V first solving for one quantity, then for another.