Given :

Capacitance C= 5.0 μF

Voltage V= 50V

Formula used :

1) If switch S is closed, it will be a short circuit. Current flow always chooses a low resistance path. No current will flow through capacitor at switch S., So we don’t need to consider it. Finally, we will left with two capacitor which are in parallel. We add the capacitance when the capacitors are in parallel.

C_eqv=C₁+C₂

Where

C₁, C₂ are the capacitance of the capacitors

2) Charge supplied by the battery

Where

C is the capacitance of the capacitor

V is the voltage across the capacitance

Q is the charge on the capacitor

Putting the values in the formula 1, we get

C_eqv= 5+5=10μF

Putting the values in the formula 1, we get

Charge supplied by the battery Q=500μC.

But when the switch has not connected the charge Q=C_eq×V

The capacitors are connected in series connection, we get

After switch S is closed the initial charge stored in the capacitor will discharge.

Note: Q¹ will be negative because the capacitor is discharging.)

So the net charge flows from A to B is

Total Charge will flow through A and B when switch S is closed.