A 4 μF capacitor is charged by a 200V supply. It is then disconnected from the supply, and is connected to another uncharged 2 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Given, a 4 μF capacitor is charged by a 200 V supply.
Then,
Charge on the capacitor-Q= CV
                                    = 4 x 10^–6 x 200

                                    = 8 x 10^–4 C

Also given, it is then disconnected and, connected to another uncharged 2μF ($2 \times {10}^{-6} \mathrm{F}$) capacitor. 
Thus, total capacitance-C'= (4 + 2)μF =6 μF

Until both the capacitor acquire a common potential, charge on the first capacitor is shared between them.
After the combination, the common potential-V' =Q/C
           $$\begin{gathered} \mathrm{V}\text{'} = \frac{8 \times {10}^{-4}\mathrm{C}}{6 \times {10}^{-6}\mathrm{F}} = 1.33 \times {10}^2\mathrm{V} \\ \mathrm{V}\text{'} = 133 \mathrm{V} \end{gathered}$$ 

Electrostatic energy of the first capacitor, before the combination,
            $$\begin{gathered} {\mathrm{U}}_1 = \frac{1}{2}{\mathrm{CV}}^2 = \frac{1}{2}\left(4\times {10}^{-6}\right) {\left(200\right)}^2 \\ {\mathrm{U}}_1 = 8 \times {10}^{-2}\mathrm{J} \end{gathered}$$

Electrostatic energy of the system, after the combination,
$$\begin{gathered} {\mathrm{U}}_2 = \frac{1}{2}\mathrm{C}\text{'}\mathrm{V}{\text{'}}^2 = \frac{1}{2}\left(6\times {10}^{-6}\right) (133{)}^2 \\ {\mathrm{U}}_2 = 5.30 \times {10}^{-2}\mathrm{J} \end{gathered}$$

Now, electrostatic energy lost by the first capacitor in the form of heat and electromagnetic radiation is

                   $$\begin{gathered} \mathrm{U} = {\mathrm{U}}_1-{\mathrm{U}}_2 \\ = (8\times {10}^{-2}-5.3 \times {10}^{-2}) \\ = 2.7 \times {10}^{-2}\mathrm{J} \end{gathered}$$