Question
A 100 pF capacitor is charged to a potential difference of $50 \mathrm{V},$ and the charging battery is disconnected. Thecapacitor is then connected in parallel with a second (initiallyuncharged) capacitor. If the potential difference across the first capacitor drops to $35 \mathrm{V},$ what is the capacitance of this secondcapacitor?
Step 1
Step 1: The initial charge on the charged capacitor is given by $Q = C_1 \times V_{\text{initial}}$, where $C_1 = 100 \, \text{pF}$ is the capacitance and $V_{\text{initial}} = 50 \, \text{V}$ is the initial potential difference. Show more…
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