Question

A particle with a charge of q = 15.0 µC travels from the origin to the point (x, y) = (20.0 cm, 50.0 cm) in the presence of a uniform electric field E = 270î V/m. Determine the following.

(a) the change in the electric potential energy (in J) of the particle–field system = -8.1*10e-4 (my answer)

Incorrect: Your answer is incorrect. How is the change in electric potential energy of the charge related to the work done on the charge by the conservative force? Write the force acting on the charge and the displacement of the charge in unit vector notation. Since the electric field and, therefore, the electric force on the charge are both constant, how can you determine the work done on the charge? J

Answer 1

answer) we know that the change in electrical potential energy is given by

P=-qExx

we only need to consider the electric field and displacement in x direction

so we have all the values and just need to plug in

P=-15*10^-6m*270V/m*0.2m=-8.10*10^-4J

answer is -8.10*10^-4J

( your answer is correct , but i think they need 3 significant figures because in the question every number is in three significant figures, try this one and write like i have written)