Answer
$C=43pF$
Work Step by Step
The initial charge on the first capacitor is given as:
$q_{i1}=C_1V_i$
$q_{i1}=100\times10^{-12}\times 50=5000\times 10^{-12}=5000pC$
The final charge on the first capacitor is given as:
$q_{f1}=C_1V_f$
$q_{f1}=100\times10^{-12}\times 35=3500\times 10^{-12}=3500pC$
When the two capacitors are connected then charge on the second capacitor is
$q_2=5000-3500=1500pC$
We know that:
$C_2=\frac{q_2}{V_2}$
$C_2=\frac{1500}{35}=43pF$
