Answer to Question #103408 in Electricity and Magnetism for bjbnl

Question

How much work is required to move a positive charge of 3.6 µC between two points that have a potential difference of 69 V?

Answer 1

The work to be done is equal to the product of the potential difference on the charge of the moving particle $W=U\cdot Q=69 V\cdot 3.6\cdot10^{-6}C==248J$

This follows from the determination of the potential energy of a charged particle in an electric field. The potential energy for a positive particle increases towards positive values of the potential of electric field. Indeed, when a positive particle moves freely, it repels from positive charges and is attracted to negative ones, i.e. it acquires kinetic energy at the expense of potential (decreasing) energy.

Answer:  The work is required to move a positive charge of 3.6 µC between two points that have a potential difference of 69 V is 248 J.