Question
A capacitor of capacitance $\text{C}=(2.0\pm0.1)\mu\text{F}$ is charged to a voltage$\text{V}=(20\pm0.5)\text{V}$ Calculate the charge Q with error limits.
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Answer
$\text{Q}=\text{CV}=2.0\times20=40\mu\text{C}$ $=40\times10^{-6}\text{C}$ $\frac{\Delta\text{Q}}{\text{Q}}=\frac{\Delta\text{C}}{\text{C}}+\frac{\Delta\text{V}}{\text{V}}$ $=\frac{0.1}{2.0}+\frac{0.5}{20}=\frac{3}{40}$ $\Delta\text{Q}=\frac{3}{40}\times\text{Q}=\frac{3}{40}\times10^{-6}=3\mu\text{C}$ Hence, $\text{Q}=(40\pm3.0)\times10^{-6}\text{C}$
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