How much work is needed to carry an electron from the positive terminal to the negative terminal of a $9.0 - \mathrm { V }$ battery. $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} \right)$

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The work needed to move a charge through a potential difference (voltage) is given by $W = qV$, where $q$ is the charge and $V$ is the potential difference. For an electron, $q = 1.60 \times 10^{-19} \, \mathrm{C}$ and $V = 9.0 \, \mathrm{V}$, so $W = (1.60 \times 10^{-19} \, \mathrm{C})(9.0 \, \mathrm{V}) = 14.4 \times 10^{-19} \, \mathrm{J}$.

How Much Work Is Needed to Carry an Electron from the Positive