In the circuit shown in figure initially the switch is opened. The switch is closed now.

The charge that will flow in direction 2 is

Options

A. `-(C_(2)^(2)epsilon)/(C_1 + C_2)`
B. `((C_1C_2)/(C_1+C_2)) epsilon`
C. `(C_(1)^(2)epsilon)/(C_1+C_2)`
D. `C_2epsilon`

Answer

D

Explanation

Charge on capacitors `C_(1)` and `C_(2)` before closing the switch S is
`q_(0)=((C_(1)C_(2))/(C_(1)+C_(2)))epsilon`
After closing S, charge in `C_(2)` (final charge) is
`q_(2)=C_(2)epsilon`
In loop `ABBDEFGA,epsilon=0` or `q_(1)=0` ltbr. final charge on `C_(1),q_(1)=0` to make final charge in `C_(1)` the charge hence charge flown towards direction 2 is
`triangle q_(2)=-((C_(1)C_(2))/(C_(1)+C_(2)))epsilon`
To make charge on capacitor `C_(2)` to final value `q_(2)` the charge flow into capacitor is
`triangleq=C_(2)epsilon-(C_(1)C_(2))/(C_(1)+C_(2))epsilong=C_(2)epsilon[1-(C_(1))/(C_(1)+C_(2))]`
`triangleq_(B)=(C_(2)^(2)epsilon)/(C_(1)+C_(2))` ltbr. at junction F,
`triangleq_(1)=triangleq-triangleq_(2)=(C_(2)epsilon)/(C_(1)+C_(2))[C_(2)+C_(1)]=C_(2)epsilon`
Alternative method: fig
in loop `ABDEG,-((q_(0)+triangleq_(1)))/(C_(1))-epsilon+epsilon=0`
or `triangleq_(1)=-q_(0)=-(C_(1)C_(2)epsilon)/(C_(1)+C_(2))`
in loop BFEDB, `-((q_(0)+triangleq_(1)))/(C_(2))+epsilon=0`
or `triangleq=C_(2)epsilon-q_(0)`
At junction F
`triangleq_(2)=triangleq_(1)-triangle_(q)=-q_(0)-(C_(2)epsilon-q_(0))=-C_(2)epsilon`
Hence charge flow in the direction of
`(1) is -(C_(1)C_(2)epsilon)/(C_(1)+C_(2))`
(2). is `C_(2)epsilon`