Two parallel plates capacitors A and B having capacitance of `1 muF and 5 muF` are charged separately to the same potential of 100 V. Now, the positive plate of A is connected to the negative plate of B and the negative plate of A to the positive plate of B. Find the final charges on each capacitors.

Explanation

As for a capacitor q =CV so initially the charge on each capacitor .
`q_(1)=C_(1)V_(1)=(1xx10^(-6))xx100=100muC`
and `q_(2)=C_(2)V_(2)=(5xx10^(-6))xx100=500muC`
Now when two capacitors are joined to each other such that positive plate of one is connected with the negative of the other by conservation of charge `q=q_(1)+q_(2)=|q_(1)|-|q_(2)|=(500-100)muC = 400 muC.` So common potential
`V=((q_(1)+q_(2)))/((C_(1)+C_(2)))=(400xx10^(-6))/((1-:5)xx10^(-6))=(200)/(3)V`
And hence after sharing charge on each capacitor
`q_(1)=C_(1)V=(1xx10^(-6))xx(200)/(3) =(200)/(3)muC`
`q_(2)=C_(2)V=(5xx10^(-6))xx(200)/(3)=(1000)/(3) muC`