$C=C_{1}+C_{2}$
$=10+6=16 \mu F$
The charge stored in first capacitor
$q=C_{1} V$
$=10 \times 1000=10^{4} \mu C$
Final potential difference across each capacitor
$V^{\prime}=q / C=10^{4} / 16$
$=625$ volt


Reason : The electric field between the plates of a charged isolated capacitance increases when dielectric fills whole space between plates.

$V = 6x - 8x{y^2} - 8y + 6yz - 4{z^2}$
Then electric force acting on $2\,C$ point charge placed on origin will be......$N$