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

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

A) $14.4 \times 10 ^ { - 19 } \mathrm {~J}$
B) $9.0 \mathrm {~J}$
C) $1.6 \times 10 ^ { - 19 } \mathrm {~J}$
D) $17 \times 10 ^ { - 19 } \mathrm {~J}$
E) $14.4 \times 10 ^ { - 19 } \mathrm {~J} / \mathrm { C }$

Correct Answer:

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