91Ó°ÊÓ

a) Capacitance of each capacitor is C.

b) There are three capacitors.

Using the formulae for the equivalent capacitance, for capacitors is in series and parallel, we can find the equivalent capacitance for series and parallel arrangement respectively. Using Eq.25-1, we can find out which arrangement has more charge on the equivalent capacitance.

Formulae:

If three capacitors are in series, the equivalent capacitanceis given by,

$\frac{1}{C_{eq}}=\underset{}{∑}\frac{1}{C}$ …(¾±)

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

$C_{eq}=\underset{}{∑}C$ …(¾±¾±)

The charge between the plates of the capacitor,$q=CV$ …(¾±¾±¾±)

We are given that capacitance on each capacitor is given by,$C_1=C_2=C_3=C$

Thus, the equivalent capacitance$C_{eq}$ of these capacitors in series is given by:

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

Therefore, the equivalent capacitance ${\mathrm{C}}_{\mathrm{eq}}\mathrm{is}\frac{\mathrm{C}}{3}$.

We are given that capacitance on each capacitor is given by,$C_1=C_2=C_3=C$

Thus,the equivalent capacitance$C_{eq}$of these capacitors in series is given by:

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

Therefore, the equivalent capacitance$C_{eq}$ is 3C.

If capacitors are in series, then we get the charge using equivalent capacitance in series using equation (iii) as follows:

$q=\frac{1}{3}CV$

If capacitors are in parallel, then we get the charge using equivalent capacitance in series using equation (iii) as follows:

$q=3CV$

Therefore, we can say that the parallel will end up having more charge on the equivalent capacitance.