Answer :
(a) In series: The charge on each capacitor is 2.5[tex]μC[/tex].
(b) In parallel: The charge on each capacitor is 6[tex]μC[/tex].
Let's correct the calculation and format it properly with LaTeX:
(a) When capacitors are connected in series across a battery, the total capacitance [tex]\( C_{\text{series}} \)[/tex] is given by the reciprocal of the sum of the reciprocals of individual capacitances:
[tex]\[ C_{\text{series}} = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2}} \][/tex]
Substituting the given values, we get:
[tex]\[ C_{\text{series}} = \frac{1}{\frac{1}{2.5 \times 10^{-6}} + \frac{1}{6.25 \times 10^{-6}}} \][/tex]
[tex]\[ C_{\text{series}} = \frac{1}{\frac{1}{2.5 \times 10^{-6}} + \frac{1}{6.25 \times 10^{-6}}} = \frac{1}{0.4 + 0.16} = \frac{1}{0.56} \][/tex]
Therefore, [tex]\( C_{\text{series}} = 1.785 \times 10^{-6} \) F.[/tex]
Using [tex]\( Q = CV \),[/tex] where [tex]\( Q \)[/tex] is the charge, [tex]\( C \)[/tex] is the capacitance, and [tex]\( V \)[/tex] is the voltage, the charge on each capacitor in series is:
[tex]\[ Q = C_{\text{series}} \times V = 1.785 \times 10^{-6} \, \text{F} \times 6 \, \text{V} = 10.71 \times 10^{-6} \, \text{C} = 10.71 \, \mu\text{C} \][/tex]
(b) When capacitors are connected in parallel across a battery, the total charge on each capacitor is equal to the charge on the battery. Thus, each capacitor receives the full charge of the battery.
Therefore, the charge on each capacitor in parallel is simply equal to the charge of the battery, which is[tex]\( 6 \, \text{V} \times (2.5 \times 10^{-6} \, \text{F} + 6.25 \times 10^{-6} \, \text{F}) = 6 \, \text{V} \times 8.75 \times 10^{-6} \, \text{F} = 52.5 \, \mu\text{C} \).[/tex]
