A 190 pF capacitor is charged to a potential difference of 110 V, and the charging battery is disconnected. The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the measured potential difference drops to 100 V, what is the capacitance of this second capacitor? (pF)

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A 190 pF capacitor is charged to a potential difference of 110 V, and the charging battery is disconnected. The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the measured potential difference drops to 100 V, what is the capacitance of this second capacitor? (pF) 

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Step 1

Given value---

The capacitor is C1=190 pF

The potential difference of V1=110 V

The measured potential difference drops to V2=100 V

 

It is required to determine the capacitance of this second capacitor.

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