Five identical capacitors each of magnityde `5 muF` are arranged with a battery of emf `20V` as shown in . Initially, the switch is open. The charge that flows from point ` A` to point `B`, when the switch is closed.
.

Answer

`-(400)/7 muC`

Explanation

When switch is open
`20q_(0)/C_(0)-q_(0)/C_(0)-q_(0)/C_(0)=0`
or `20(3epsilon_(0))/C_(0)q_(0)=(100)/3xx10^(-6)C`
.
In loop (1)
`(-q_(0)+q-q_(1)-q_(2))/C_(0)-((q_(0)+q-q_(2))/C_(0))=0`
In loop `(2)`
`(q_(0)+q-q_(1)-q_(2))/C_(0)+((q_(0)+q-q_(1)))/C_(0)+((q_(0)+q-q_(1)))/C_(0)=0`
or `3q_(0)+3q-3q_(1)-q_(2)=0` (ii)
In loop `(3)`
`V_(0)-((q_(0)+q)/C_(0))-((q_(0)+q-q_(1))/C_(0))=0`
or `2q_(0)+2q_(1)=20xx5xx10^(-6)=100xx10^(-6)` (iii)
After ssolving Eqs. `(i),(ii)`, and `(iii)`, we get
`q_(1)=(400)/7muC` from `B` to `A`.
Thus, charge flowing from `A` to `B` is `-400//7 muC`.