Find the charges on the three capacitors shown in figure

Explanation

Take the potential at the juction of the batteries to be zero. Let the left battery supply a charge `
`Q_1`
` and the right battery a charge `
`Q_2`
`. The charge on the `
` 5muF`
` capacitor will be `
`Q_1+ Q_2`
`. Let the potential at the junction of the capacitors be `
`V_1`
`. The charges at different plates and potentials at different points are shows in . Note that the charge on the three plates which are in contact add to zero. It should be so, because , these plates taken together form an isolated system which cannot receive charges from the batteries. Applying the equation Q =CV to the three capacitors, we get, `
`Q_1 = 2 muF(6V - V_1)`
` `
`Q_2 = 4 muF(6V-V_1)`
` `
` and `
`Q_1+Q_2 = 5muF(V_1 - 0)`
`. From (ii) and (iii), `
`5Q_2 +4(Q_1 +Q_2 ) = 20 muF X 6V`
` or, `
`4Q_1 + 9Q_2 = 120 muC`
` or, `
`4Q_1 + 18Q_1 = 120muC`
` or, `
` Q_1 = 5.45 muC`
` and `
` Q_2 = 10.9 muC`
`. Thus, the charges on the `
`2muF, 4muF`
` and `
`5muF capacitors are `
`5.45 mu`
`, 10.9 muC`
` and `
`16.35 muC`
` respectively