Answer to Question #181900 in Mechanics | Relativity for Jay

Question

An electric charge of 2.10^-3 C at a point X in an electric field had electric potential energy of 4.10^-2J. It was moved to another point Y where it’s potential energy was 6.10^-2 J. What is the potential difference between X and Y and the electric field strength if the points X and Y and seperated by distance 0.25m?

Answer 1

Answer

Using energy conservation

$\Delta U=q\Delta V$

So potential difference

$\Delta V=\frac{\Delta U}{q}\\=\frac{ (6\times10^{-2 } -4\times10^{-2 }) }{2\times10^{-3 }}\\=10Volts.$