Question

An electron moves 0.10 m along the direction of an electric field of a magnitude of 3.0 N/C. What is the change in electric potential energy between the electron’s initial and final points?

Answer 1

$V=E\cdot d$

Where:

$\begin{gathered} d=0.10m \\ E=3.0\frac{N}{C} \end{gathered}$

Therefore:

$V=3\cdot0.1=0.3V$

The change in the electric potential is:

$\begin{gathered} \Delta E=qV \\ \Delta E=(1.6\times10^{19}C)\cdot(0.3) \\ \Delta E=-4.8\times10^{-20}J \end{gathered}$

Answer:

-4.8x10⁻²⁰J