Answer
\[\underline{9.4\times {{10}^{13}}}\] excess electrons and \[\underline{8.5\times {{10}^{-17}}\text{ kg}}\]
Work Step by Step
The electronic charge is \[-1.60218\times {{10}^{-19}}\text{ C}\]. Calculate the number of electrons as follows:
\[\begin{align}
& \text{Number of electrons}=\left( -15\,\text{ }\!\!\mu\!\!\text{ C} \right)\times \frac{{{10}^{-6}}}{1\,\text{ }\!\!\mu\!\!\text{ }}\times \left( \frac{1\text{ electron}}{-1.60218\times {{10}^{-19}}\,\text{C}} \right) \\
& =9.4\times {{10}^{13}}\,\text{electrons}
\end{align}\]
The electron has a mass of \[0.00091\times {{10}^{-27}}\text{ kg}\]. Calculate the total mass as follows:
\[\begin{align}
& \text{Mass}=\frac{0.00091\times {{10}^{-27}}\text{ kg}}{1\,\text{electron}}\times 9.4\times {{10}^{13}}\,\text{electrons} \\
& =\text{8}\text{.5}\times \text{1}{{\text{0}}^{-17}}\,\text{kg}
\end{align}\]
The number of excess electrons acquired is \[\underline{9.4\times {{10}^{13}}}\] and the collective mass is \[\underline{8.5\times {{10}^{-17}}\text{ kg}}\].
