In the figure, a 12 V battery is joined to three capacitors, C₁ = 6.0 µF, C₂ = 3.0 µF and C₃ = 6.0…

(a) Equivalent capacitance. $C_2$ and $C_3$ are in parallel: $3.0 + 6.0 = 9.0\ \mu\text{F}$. This is in series with $C_1$: $C_{eq} = \dfrac{6.0 \times 9.0}{6.0 + 9.0}\ \mu\text{F} = 3.6\ \mu\text{F}$.

(b) Charge on $C_1$. The total charge from the battery is $Q = C_{eq}V = 3.6\ \mu\text{F} \times 12\ \text{V} = 43.2\ \mu\text{C}$. It all flows through $C_1$, which is in series with the pair.

(c) Across $C_3$. $C_1$ has $\dfrac{43.2\ \mu\text{C}}{6.0\ \mu\text{F}} = 7.2\ \text{V}$ across it. The rest is across the parallel pair: $12 - 7.2 = 4.8\ \text{V}$.

(d) Energy in $C_3$. $U = \tfrac12 C_3 V^2 = \tfrac12 \times 6.0\ \mu\text{F} \times (4.8\ \text{V})^2 = 69.1\ \mu\text{J}$.

(e) With the slab. $C_2$ becomes $4 \times 3.0 = 12\ \mu\text{F}$. The pair is now $12 + 6.0 = 18\ \mu\text{F}$, and $C_{eq} = \dfrac{6.0 \times 18}{6.0 + 18}\ \mu\text{F} = 4.5\ \mu\text{F}$. The battery now supplies $Q' = 4.5 \times 12 = 54.0\ \mu\text{C}$. The extra charge is $54.0 - 43.2 = 10.8\ \mu\text{C}$.