Figure shows charge (q) versus voltage (V) graph for series and parallel

(1) Equivalent capacitance in series combination (C') is given by

$\frac{1}{\mathrm{C}},=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}} \Rightarrow \mathrm{C}^{\prime}=\frac{\mathrm{C}_{1} \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$

For parallel combination equivalent capacitance $\mathrm{C}^{\prime \prime}=\mathrm{C}_{1}+\mathrm{C}_{2}$

For parallel combination

$\mathrm{q}=10\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)$

$\mathrm{q}_{1}=500 \mu \mathrm{C}$

$500=10\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)$

$\mathrm{C}_{1}+\mathrm{C}_{2}=50 \mu \mathrm{F}$            ....(1)

For Series Combination-

$\mathrm{q}_{2}=10 \frac{\mathrm{C}_{1} \mathrm{C}_{2}}{\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)}$

$80=10 \frac{C_{1} C_{2}}{50}$ From equation      ....(2)

$\mathrm{C}_{1} \mathrm{C}_{2}=400$             ....(3)

From equation (i) and (ii)

$\mathrm{C}_{1}=10 \mu \mathrm{F} \quad \mathrm{C}_{2}=40 \mu \mathrm{F}$