A 600pF capacitor is charged by 200V supply. It is then...

The formula to calculate the initial energy stored in the first capacitor is given by

${U}_{i} = \dfrac{1}{2}C{V}^{2} . . . \left(1\right)</p><p>The formula$ to calculate the final energy stored when two capacitors are connected is given by

$\begin{cases}{U}_{f} & = & \dfrac{1}{2}C{\left(\dfrac{V}{2}\right)}^{2} \times 2 \\ & = & \dfrac{1}{4}C{V}^{2} . . . \left(2\right)\end{cases}$$</p><p>Subtract equation (2) from equation (1) to obtain the energy loss \left(∆ U\right)$.

$\begin{cases}∆ U & = & \dfrac{1}{2}C{V}^{2} - \dfrac{1}{4}C{V}^{2} \\ & = & \dfrac{C{V}^{2}}{4} . . . \left(3\right)\end{cases}$

Substitute the values of the known parameters into equation (3) to calculate the required loss.

$\begin{cases}∆ U & = & \dfrac{600 \times {10}^{- 12} \times ( 200{)}^{2}}{4}J \\ & = & 6\mu J\end{cases}$