Calculate the work done to move an electron from the positive to the negative terminal of a 12 V battery.

The potential difference of the battery is $\Delta V_{battery} = 12 \, \text{V}$.
The charge of an electron is $q = -1.602 \times 10^{-19} \, \text{C}$.

The electron is carried from the positive terminal to the negative terminal.
The positive terminal is at a higher potential than the negative terminal.
Let $V_+$ be the potential of the positive terminal and $V_-$ be the potential of the negative terminal.
We know that $V_+ - V_- = 12 \, \text{V}$.
The initial potential is $V_{initial} = V_+$.
The final potential is $V_{final} = V_-$.
The change in potential experienced by the electron is $\Delta V = V_{final} - V_{initial} = V_- - V_+$.
Therefore, $\Delta V = -12 \, \text{V}$.

The work $W$ required to move a charge $q$ through a potential difference $\Delta V$ is given by the formula:

Substitute the values:

Rounding to two significant figures, consistent with the given voltage:

The work required is positive because an electron (negative charge) is being moved from a higher potential (positive terminal) to a lower potential (negative terminal), which is against the direction it would naturally move under the influence of the electric field.

The work required is $1.9 \times 10^{-18} \, \text{J}$.