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

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SUMMARY

The discussion focuses on calculating the capacitance of a second capacitor connected in parallel with a 100 pF capacitor initially charged to 50 V, which then drops to 35 V after connection. Using the formula for charge (Q = C × V) and the principle of charge conservation, the total charge before and after connection can be equated. The capacitance of the second capacitor can be determined to be 57.14 pF, confirming the conservation of charge in parallel capacitors.

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  • Understanding of capacitor charge equations (Q = C × V)
  • Knowledge of parallel capacitor configurations (C_{T} = C_{1} + C_{2} + ...)
  • Familiarity with the concept of charge conservation
  • Basic principles of electrical circuits
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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.
 

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