# A $0.4 \mu F$ capacitor is charged to $1200 v.$ After removing battery a $0.2 \mu F$ capacitor is connected in parallel to it. The voltage on capacitor will become :

$(A)\;200 \\ (B)\;600 \\ (C)\; 1200 \\ (D)\;800$

Total voltage $=\large\frac{Total \;charge}{Total\; capacitance}$
$\qquad= \large\frac{C_1V_1}{C_1+C_2}$
$\qquad= \large\frac{0.4 \times 1200}{0.4 +0.2}$
=> $V=800 V$
Hence D is the correct answer.