Answer
The increase in the potential difference across capacitor 1 is $~~78.9~V$
Work Step by Step
We can find the equivalent capacitance of $C_{12}$:
$C_{12} = C_1+C_2 = 10.0\mu F+5.00~\mu F = 15.0~\mu F$
We can find the equivalent capacitance of all three capacitors:
$\frac{1}{C_{eq}} = \frac{1}{15.0~\mu F}+\frac{1}{4.00~\mu F}$
$\frac{1}{C_{eq}} = \frac{4}{60.0~\mu F}+\frac{15}{60.0~\mu F}$
$C_{eq} = \frac{60.0~\mu F}{19}$
We can find the charge $q$:
$q = C_{eq}~V = (\frac{60.0~\mu F}{19})(100.0~V) = \frac{6000}{19}~\mu C$
Note that the charge $q_2$ on capacitor 2 is $q-q_1$
We can find the charge on capacitor 1:
$\frac{q_1}{C_1} = \frac{q_2}{C_2}$
$\frac{q_1}{C_1} = \frac{q-q_1}{C_2}$
$C_2~q_1 = C_1~(q-q_1)$
$q_1~(C_1+C_2) = C_1~q$
$q_1 = \frac{C_1~q}{C_1+C_2}$
$q_1 = \frac{(10.0~\mu F)~(\frac{6000}{19}~\mu C)}{15.0~\mu F}$
$q_1 = 211~\mu C$
We can find the potential difference across capacitor 1:
$V_1 = \frac{q_1}{C_1} = \frac{211~\mu C}{10.0~\mu F} = 21.1~V$
After capacitor 3 is like a conducting wire, the potential difference across capacitor 1 is $100.0~V$
We can find the increase in the potential difference across capacitor 1:
$(100.0~V) - (21.1~V) = 78.9~V$
The increase in the potential difference across capacitor 1 is $~~78.9~V$
