Physics - Capacitor Question with Solution | TestHub

Charge on capacitor,
$q=CV$

When it is connected with another uncharged capacitor.

$V_c=\frac{q_1+q_2}{C_1+C_2}=\frac{q+0}{C+C}$
$V_c=\frac{V}{2}$
Initial energy,
$U_i=\frac{1}{2}C{V}^2$
Final energy,
$U_f=\frac{1}{2}C{\left(\frac{V}{2}\right)}^2+\frac{1}{2}C{\left(\frac{V}{2}\right)}^2$

$=\frac{C{V}^2}{4}$

Loss of energy $=U_i-U_f$
$=\frac{C{V}^2}{4}$,
i.e., decreases by a factor of $2$.