Figure 25-22 shows an open switch, a battery of potential difference, a current-measuring meter A, and three identical uncharged capacitors of capacitance C. When the switch is closed and the circuit reaches equilibrium, what is (a) the potential difference across each capacitor and (b) the charge on the left plate of each capacitor? (c) During charging, what net charge passes through the meter?

1. The potential difference of the battery is V.
2. The capacitance of each capacitor is C.

Using Eq.25-9, we can find the capacitance of each capacitor from the given values of area and plate separation. Then using Eq.25-1, we can compare the capacitance from the slopes of the plots. Then, comparing the predictions about the capacitance yielded from both the equations, we can find out which plot goes with which capacitor.

Formulae:

The charge within the plates of the capacitor, q = CV …(i)

If capacitors are in series, the equivalent capacitance $C_{eq}$is given by,

$\frac{1}{C}=\sum _{}\frac{1}{C}$ …(ii)

Analyzing given circuit, we can note that all the three capacitors are connected in series. From the properties of series capacitor circuit and they have same capacitance, the potential difference across each capacitor must be same and total capacitance across them should be equal to the electromotive force of the battery.

Hence, the potential difference across each capacitor is$\frac{V}{3}$

Since the capacitors are connected in series, from Eq.25-1, we can note that each capacitor must have the same charge as the equivalent capacitor of the circuit.

If the capacitors are in series, the equivalent capacitance$C_{eq}$is given using equation (ii) by: (For,$C_1=C_2=C_3=C$)

$$\begin{gathered} \frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C} \\ \frac{1}{C_{eq}}=\frac{3}{C} \\ {C}_{eq}=\frac{C}{3} \end{gathered}$$

The charge on each capacitor is given using equation (i) by:

$q=\frac{CV}{3}$

Since positive end of the battery is connected to the left plate of capacitor. From the left plate of each capacitor, electrons are removed. So, it must be positively charged.

Hence, the charge on the left plate of each capacitor is$+\frac{CV}{3}$.

Since the charge on the left plate of each capacitor is$+\frac{CV}{3}$

Hence, the net charge flowing through ammeter is$-\frac{CV}{3}$