Hi, If I have two capacitors [itex]C_1=1 \mu F[/itex] ..... [itex]C_2=2 \mu F[/itex],and I start by taking each and connecting it to 5V battery in series, so the first cap has [itex]Q_1=C_1 V=5\mu C[/itex] then seperatley I connect the other cap to another 5V source so it gains charge [itex]Q_2=C_2 V=10 \mu C[/itex]. Now I remove the batteries and connect the two caps in series with each other, so that the opposing charge plates face each other. How would I calculate the final voltage across the 2 microfarard cap? My reasoning was if you have two caps in series and nothing else, it's basically two seperate wires. The " two wires" have 10microC at one end and -5microC at the other, so current initially flows until 7.5microC have travelled, leaving 2.5microC at both ends, similarly for the "other wire". So this reasoning would make me think on the 2microF cap the final voltage would be V=q/C_2=2.5/2=1.25V This doesnt seem to be the correct answer however, can anyone please shed some light?