In the given circuit. C _ 1 = 2 , C _ 2 = 0 . 2 , C _ 3 = 2 , C _ 4...

From the diagram of the circuit, ${C}_{3}, {C}_{4}, {C}_{5}$ are in series, so the total capacitance is

$\dfrac{1}{C '} = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{2} ⇒ C ' = \dfrac{4}{5}\mu F$

$C ' \text{ and } {C}_{2}$ are parallel, the total capacitance is

C ' ' = \left(\dfrac{4}{5} + 0 . 2\right)\mu F = 1\mu F

{C}_{1}, C ' ' , {C}_{6} are in series. So the equivalent capacitance is,

$\dfrac{1}{{C}_{eq}} = \dfrac{1}{2} + 1 + \dfrac{1}{2} ⇒ {C}_{eq} = 0 . 5\mu F$

Using $Q = {C}_{eq}V$.

The charge through the battery, $Q$ $= 10 \times \dfrac{1}{2} = 5\mu C$

From the diagram above,

$Q ' = \dfrac{5\mu C \times 0 . 8}{0 . 8 + 0 . 2} = 4\mu C$