two capacitors of capacitance 3 and 6 are charged to a potential of 12V each they are now connected to each other with the positive plate of each joined to the negative plate of the other the potential difference across each will be...........please answer urgent

Here,we have to apply formula Q=CV,where Q is charge,C is capacitance
and V is the potential difference.
Q1=3*12=36uC
Q2=6*12=72uC

Now, they are connected in series,so equivalent capacitance=3+6=9uF
Potential difference across 3uF=36/9=4V
Potential difference across 6uF=72/9=8V

Hope you understood ?.