Answer
The magnitude of the maximum potential difference that can exist between points A and B is $~~190~V$
Work Step by Step
We can find the charge on each capacitor if the potential difference across each capacitor is $100~V$
$q_1 = C_1~V = (10.0~\mu F)(100~V) = 1000~\mu C$
$q_2 = C_2~V = (20.0~\mu F)(100~V) = 2000~\mu C$
$q_3 = C_3~V = (25.0~\mu F)(100~V) = 2500~\mu C$
Since the capacitors are in series, all three capacitors have the same charge. Therefore, the charge on each capacitor is limited to a maximum of $1000~\mu C$, otherwise the potential difference across capacitor 1 would be more than 100 V
We can find the potential difference across each capacitor with a charge of $1000~\mu C$
$V_1 = \frac{q}{C_1} = \frac{1000~\mu C}{10.0~\mu F} = 100~V$
$V_2 = \frac{q}{C_2} = \frac{1000~\mu C}{20.0~\mu F} = 50.0~V$
$V_3 = \frac{q}{C_3} = \frac{1000~\mu C}{25.0~\mu F} = 40.0~V$
The magnitude of the maximum potential difference that can exist between points A and B is $~~190~V$
