Answer
The potential difference between the points $a$ and $b$ is $~~50~V$
Work Step by Step
We can find the charge on capacitor 1:
$q_1 = C_1~V_1 = (1.0~\mu F)(100~V) = 100~\mu C$
We can find the charge on capacitor 2:
$q_2 = C_2~V_2 = (3.0~\mu F)(100~V) = 300~\mu C$
When the switches are closed, the net charge on the bottom half of the circuit is $300~\mu C-100~\mu C$ which is $200~\mu C$
Similarly, the net charge on the top half of the circuit is $100~\mu C-300~\mu C$ which is $-200~\mu C$
Note that the equivalent capacitance is $4.0~\mu F$ since the capacitors are in parallel.
We can find the potential difference:
$V = \frac{q}{C_{eq}} = \frac{200~\mu C}{4.0~\mu F} = 50~V$
The potential difference between the points $a$ and $b$ is $~~50~V$.
