Figure show charge (q) versus voltage (V) graph for series and parallel combination of two given capacitors. The capacitances are:

Equivalent capacitances in series combination (C₁) is given by

$\frac{1}{{\mathrm{C}}^1}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}\Rightarrow {\mathrm{C}}^1=\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}$

For parallel combination equivalent capacitance

${\mathrm{C}}^{11}={\mathrm{C}}_1+{\mathrm{C}}_2$

For parallel combination

$\begin{array}{l}\mathrm{q}=10\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)\\ {\mathrm{q}}_1=500\mathrm{\mu C}\\ 500=10\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)\\ {\mathrm{C}}_1+{\mathrm{C}}_2=50\mathrm{\mu F}\dots ...\left(\mathrm{i}\right)\end{array}$

For series combination

${\mathrm{q}}_2=10\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)}$

$80=10\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{50}$  from equation ……… (ii)

${\mathrm{C}}_1{\mathrm{C}}_2=400........\left(\mathrm{iii}\right)$

From equation (i) and (iii)

${\mathrm{C}}_1=10\mathrm{\mu F}{\mathrm{C}}_2=40\mathrm{\mu F}$