A potential difference of V=80.0V is applied across two capacitors in series,with capacitances 17.0 nF and 8.00 nF.What is the magnitude of the electric potential across the 17.0 nF capacitor?

Question

Quizplus · Accepted Answer

The total capacitance of capacitors in series is given by the formula $1/C_{\text{total}} = 1/C_1 + 1/C_2$. For capacitors in series, the charge $Q$ is the same on both, and the voltage across each capacitor is given by $V = Q/C$. The voltage across the 17.0 nF capacitor can be found using the proportion of the total voltage based on the inverse of the capacitances, leading to $V_{17nF} = V \times (C_{\text{total}}/C_{17nF})$, where $C_{\text{total}}$ is the equivalent series capacitance. Calculating $C_{\text{total}}$ and then $V_{17nF}$ yields 25.6 V.

A Potential Difference of V=80