The connnections shown in figure are established with the swithc S open. How much charge will flow through the switch if it is closed?

Explanation

When the switch is open, capacitors (2) and (3) are in series. Their equivalent capacitance is (2)/(3) muF. The charge appearing on each of these capacitors is, therefore , 24 X(2)/(3)muF = 16muC. The equivalent capacitance of (1) and (4), which are also connected in series, is also (2)/(3)muF and the charge on each of these capacitors is also 16muC. The total charge on the two plates of (1) and (4) connected to the switch is, therefore, zero. The situation when the switch S is closed is shown in . Let the charges be distributed as shown in the figure. `
`Q_1`
` and `
`Q_2`
` are arbitrarily chosen for the positive plates of (1) and (2). The same magnitude of charges will appear at the nagative plates of (3) and (4). Take the potential at the nagative terminal to be zero and at the switch to be `
`V_0`
`. Writing equations for the capacitors (1), (2), (3)and (4), `
` Q_1 = (24 V-V_0) X1 muF`
` `
` Q_2 = (24 V-V_0) X2 muF`
` `
` Q_1 = V_0 X1 muF`
` `
` Q_2 = V_0 X2 muF`
` From (i) and (iii), V_0 = 12 V, `
` Q_1 = 12muC and Q_2 = 24 muC`
`. The charge on the two plates of (1) and (4) which are connected to the switchis, therefore, Q_2 - Q_1 = 12muC`
`. When the switch was open, this charge was zero. Thus, `
`12muC`
` of charge has passed through the switch after it was closed.