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(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*10^(19)C)\cdot(0.3) \\ \Delta E=-4.8*10^(-20)J \end{gathered}`

Answer:

-4.8x10⁻²⁰J