Question

In the figure below, C_1 = 4.00 mu F, C_2 = 8.00 mu F, and V_ab = +56.0 V. Calculate the charge on each capacitor. Q_1 = C Q_2 = C Calculate the potential difference across each capacitor. V_c1 = V V_c2 = V

Answer 1

Given data

$C_{1}=4\mu F$

$C_{2}=8\mu F$

$V_{ab}=56Volt$

(a)

We know   

$V=\frac{Q}{C}$

$Q=CV$

$Q_{1}=VC_{1}=(56\times 4\times 10^{-6})C=224\times 10^{-6}C$

$Q_{2}=VC_{2}=(56\times 8\times 10^{-6})C=448\times 10^{-6}C$

(b)

Capacitors are connected in parallel so potential difference at each capacitor will be same

Hence $V_{C1}=V_{C2}=56Volt$