Answer to Question #181906 in Mechanics | Relativity for Jay

Question

An electric charge of 2.10^-3C at a point X in an electric field had electric potential energy of 4.10^-2 J. It was moved to another point Y where it’s potential energy was 6.10^-2J. 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

Solution:

ΔEPE = 0.0610J - 0.0410J = 0.0200 J, so

a) ΔV_xy = $\dfrac{0.02J}{2.1*10^{-3}C}=9.52V$

b) E = $\dfrac{ΔV_{xy}}{d}=\dfrac{9.52V}{0.25m}=38.1(\frac{V}{m})$

assuming the motion was parallel to the field.