When two capacitors are connected in series and connected across $4kV$ line, the energy stored in the system is $8J$. The same capacitors, if connected in parallel across the same line, the energy stored are $36J$. Find the individual capacitances.

Hint: Let two variables for their individual capacitance. Use the formula for the equivalent capacitance and energy stored in capacitors. Solve the two equations to get the individual capacitance.

Formula Used:
Equivalent Capacitance in parallel: ${C_{eq}} = {C_1} + {C_2}$
Equivalent Capacitance in series: ${C_{eq}} = \dfrac{{{C_1} \times {C_2}}}{{{C_1} + {C_2}}}$
Where ${C_1}$ and ${C_2}$ are the individual capacitance
Energy stored in Capacitor, $E = \dfrac{1}{2}C{V^2}$
Where $E$ is the Energy stored, $C$ denotes the capacitance and $V$ be the potential across the capacitor.

Complete step by step solution:
Let the their individual capacitance be ${C_1}$ and ${C_2}$
Given $V = 4kV = 4000V$
The energy stored in the system when the capacitors are connected in series is $8J$.
The energy stored in the system when the capacitors are connected in series is $36J$.
Now we take two separate cases.
Case 1: When they are connected in parallel
Equivalent Capacitance in parallel:
${C_{eq}} = {C_1} + {C_2}$
 Thus Energy stored when they are connected in parallel,
$ \Rightarrow \dfrac{1}{2}{C_{eq}}{V^2} = 36$
$ \Rightarrow {C_{eq}} = \dfrac{{36 \times 2}}{{{V^2}}}$
Substituting the values we get,
$ \Rightarrow {C_{eq}} = \dfrac{{72}}{{4000 \times 4000}}$
$ \Rightarrow {C_{eq}} = 4.5 \times {10^{ - 6}}F = 4.5\mu F$
$ \Rightarrow {C_1} + {C_2} = 4.5\mu F$

Case 2: When they are connected in series
Equivalent Capacitance in parallel:
${C_{eq}} = \dfrac{{{C_1} \times {C_2}}}{{{C_1} + {C_2}}}$
 Thus Energy stored when they are connected in series,
$ \Rightarrow \dfrac{1}{2}{C_{eq}}{V^2} = 8$
$ \Rightarrow {C_{eq}} = \dfrac{{8 \times 2}}{{{V^2}}}$
Substituting the values we get,
$ \Rightarrow {C_{eq}} = \dfrac{{16}}{{4000 \times 4000}}$
$ \Rightarrow {C_{eq}} = {10^{ - 6}}F = 1\mu F$
$ \Rightarrow \dfrac{{{C_1} \times {C_2}}}{{{C_1} + {C_2}}} = 1\mu F$
Putting the value of ${C_1} + {C_2}$ in above equation we get that,
$ \Rightarrow {C_1} \times {C_2} = 4.5\mu {F^2}$
Now using that
${C_1} - {C_2} = \sqrt {{{({C_1} + {C_2})}^2} - 4{C_1} \times {C_2}} $
$ \Rightarrow {C_1} - {C_2} = \sqrt {{{(4.5)}^2} - 4 \times 4.5} $
$ \Rightarrow {C_1} - {C_2} = \sqrt {2.25} $
$ \Rightarrow {C_1} - {C_2} = 1.5\mu F$
Now solving ${C_1} - {C_2} = 1.5\mu F$ and ${C_1} + {C_2} = 4.5\mu F$ ,
We get ${C_1} = 3\mu F$ and ${C_2} = 1.5\mu F$
Thus,

${C_1} = 3\mu F$ and ${C_2} = 1.5\mu F$

Additional Information:
Capacitors are devices that store electrical energy in an electric field in between 2 plates. The unit of capacitance is defined as Farad $F$. It is defined as
$C = \dfrac{Q}{V}$
Where $Q$ is the charge, $C$ is the capacitance and $V$ is the potential.

Note: Take care of the units and properly do the calculations. Do not confuse the equivalence formula of capacitance between that of parallel and series.