Answer to Question #142767 in Electricity and Magnetism for Kgokane Sekgoka

Question

A 3.6 μC charge is moved from a position where its electric potential energy is 7.2 mJ to a position where its potential energy is 1.8 mJ. What is the potential at the first position?

Answer 1

The potential energy of an electrostatic field is

$E_{p,1} = q\cdot \varphi_1,$ where $\varphi$ is the potential at the position in question. Therefore,

$\varphi_1 = \dfrac{E_{p,1}}{q} = \dfrac{7.2\cdot10^{-3}\,\mathrm{J}}{3.6\cdot10^{-6}\,\mathrm{C}} = 2\cdot10^3\,\mathrm{V}.$

Worth noting that the potential is defined up to a constant, but it is often assumed that the potential is 0 when the distance tends to infinity.