Question

In the figure below, if C₁ = C₂ = 2C₃ = 21.0 µF, how much charge is stored on each capacitor when V = 43.6 V?

C1? C2? C3?

Answer 1

$C_{1}=21.0 \mu F$

$C_{2}=21.0 \mu F$

$C_{3}=10.5 \mu F$

Charge in C1:

$Q_{C1}=C_{1}V=21.0*10^{-6}F*43.6V=9.156*10^{-4}C$

Charge in C2 and C3 is the same because the capacitor are in series:

$C_{eq}=\frac{1}{\frac{1}{C_{2}}+\frac{1}{C_{3}}}$

$C_{eq}=7*10^{-6}F$

So the charge in C2 and C3 is:

$Q=C_{eq}*V=7*10^{-6}F*43.6V=3.052*10^{-4}C$