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

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To find the capacitance of the second capacitor in parallel, first calculate the initial charge on the 100 pF capacitor using the formula Q = C × V, resulting in 5 nC. When connected in parallel with the second capacitor, the total charge remains the same, but the voltage drops to 35 V. The charge on the first capacitor at 35 V is 3.5 nC, indicating that the second capacitor must account for the difference in charge, which is 1.5 nC. Using the voltage across both capacitors, the capacitance of the second capacitor can be calculated as C = Q/V, yielding a capacitance of approximately 42.86 pF. The discussion emphasizes the principles of charge conservation and parallel capacitance calculations.
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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) Q\ =\ C\ \times\ V

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

3) Charge is always conserved.
 

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