Question

An electron moves 4.5 m in the direction of an electric field of strength of 325N/C. Determine the change in electrical potential energy associated with the electron

Answer 1

The change in electrical potential energy of the electron is given by:
$\Delta U = e \Delta V$
where
e is the electron charge
$\Delta V$ is the potential difference between the initial and final point of the electron.

The electron moves by a distance d=4.5 m in the electric field of intensity $E=325 N/C$, so the potential difference between the initial and final location of the electron is
$\Delta V= -E d= -(325 N/C)(4.5 m)=-1462.5 V$
where the negative sign is due to the fact that the electron is moving in the direction of the electric field, so it is moving from a point at higher potential to a point of lower potential, so the $\Delta V$ must be negative.

Therefore, the change in electrical potential energy of the electron is
$\Delta U= e \Delta V=(-1.6 \cdot 10^{-19} C)(-1462.5 V)=2.34 \cdot 10^{-16} J$