A `4muF` capacitor is charged to a potential 10V using a battery. The battery is then removed and then this capacitor is connected parallel to an unchanged capacitor of capacity `6muF`. What is the common potential?

Explanation

Data supplied, `C_(1)=4muF, V_(0)=10V, C_(2)=6muF`
Common potential =V
By the law of conservation of charges, `C_(1)V_(0)=C_(1)V+C_(2)V=(C_(1)+C_(2))V`
`V=(C_(1)V_(0))/(C_(1)+C_(2))=(4 xx 10^(-6) xx 10)/((4+6) 10^(-6))=4V`