A 600   pF capacitor is charged by 200   V supply. It is then disconnected from the supply and is…

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

$U_i=\frac{1}{2}C{V}^2 ...\left(1\right)$

The formula to calculate the final energy stored when two capacitors are connected is given by 

$\begin{array}{rcl}{U}_f& =& \frac{1}{2}C{\left(\frac{V}{2}\right)}^2\times 2\\ & =& \frac{1}{4}C{V}^2 ...\left(2\right)\end{array}$

Subtract equation (2) from equation (1) to obtain the energy loss $\left(∆U\right)$.

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

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

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