Answer
Positive charge will decrease on the left-hand capacitor in circuit (3)
Work Step by Step
The capacitors in each circuit are wired in parallel. After the switch is closed, the potential difference across each capacitor in the circuit will be equal.
We can find the potential difference across each capacitor before the switch is closed:
(1)
left capacitor: $V = \frac{6q}{2C} = \frac{3q}{C}$
right capacitor: $V = \frac{3q}{C}$
(2)
left capacitor: $V = \frac{6q}{3C} = \frac{2q}{C}$
right capacitor: $V = \frac{3q}{C}$
(3)
left capacitor: $V = \frac{6q}{2C} = \frac{3q}{C}$
right capacitor: $V = \frac{3q}{2C} = \frac{1.5q}{C}$
After the switch has been closed, in order to balance the potential difference across each capacitor in the circuit, positive charge will decrease on the left-hand capacitor in circuit (3)
