question_answer

A capacitor of capacitance \[\text{C}=\left( \text{2}.0\text{ }\pm \text{ }0.\text{1} \right)\mu \text{ F}\] is charged t6 a voltage \[\text{V}=\left( \text{2}0\text{ }\pm \text{ }0.\text{5} \right)\text{volt}\]. Calculate the charge Q with error limits.

Answer:

                \[\text{Q}=\text{CV}=\text{2}.0\times \text{2}0=\text{4}0\]micro coulomb = \[\text{4}0\times \text{1}{{0}^{-\text{6}}}\text{ C}\]                 \[\frac{\Delta Q}{Q}=\frac{\Delta C}{C}+\frac{\Delta V}{V}=\frac{0.1}{2.0}+\frac{0.5}{20}=\frac{3}{40}\]                 \[\Delta Q=\frac{3}{40}\times Q=3.0\mu C\]    Hence, \[Q=\left( 40\pm 3.0 \right)\times {{10}^{-6}}C\]