How Do You Calculate the Capacitance of a Second Capacitor in Parallel?

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A 100 pF capacitor is charged to a potential difference of 50 V, and the charging battery is disconnected. The
capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the potential
difference across the first capacitor drops to 35 V, what is the capacitance of this second capacitor?

I am totally stuck! Please help! I know the voltage of both will be the same..
 
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Why do you think the voltage across the first capacitor dropped from 50V to 35V?
 
I shall give you three hints.

1) [tex]Q\ =\ C\ \times\ V[/tex]

2) For capacitors connected in parallel: [tex]C_{T}\ =\ C_{1}\ +\ C_{2}\ +\ C_{3}\ + ...[/tex]

3) Charge is always conserved.