Find the charges stored in each of the capacitors shown in the circuit diagram.

Explanation

This given arrangement can be simplified as below:

Equivalent capacitance of two 8 `muF` capacitors connected in parallel is
`C=(8+8)muF=16muF`

The net capacitance of the network, which are connected in series, is:
`(1)/(C_(eq))=(1)/(20)+(1)/(16)`
`C_(eq)=(20xx16)/(20+16)=8.89muF`
Charge drawn from the battery,
`Q=C_(eq)V`
Here, `V=10V`
`rArr Q=8.89xx10^(-6)xx10`
`=88.9xx10^(-6)muC=88.9muC`
Charge on capacitor of `20muF=Q=88.9muC`
Charge on the two `8muF` capacitors will be distributed equally.
Charge on each `8muF` capacitor `=(Q)/(2)=44.45muC`