Question

Part b Assume the electric field has a magnitude of E-61.3 vi, and the particle has a net positive charge of Q-+5.40 C. Calculate the magnitude of the electric potential difference, AVI-|VB - VA Enter answer here 69.8 Volts 69.8Volts of 6 attempts used Part c CHECK ANSWER We increase the magnitude of the electric field to Enew183 , and repeat the experiment with the same Q +5.40 C particle. Calculate the net change in the electric potential energy, APE PE- PEA. [Be sure the sign of your answer accurately reflects whether the particle gained or lost potential energy in moving from point (A) to point(B) in this experiment Enter answer here Joules

Answer 1

Given is:-

Given is: - Vertical-bar E_new Vertical-bar = 183V/m, Q = + 5.40 C Now, the potential energy in going from point A to B is given by delta PE = (PE)_B - (PE)_A or delta PE = q (V_B - V_A) delta PE = q (-2E) by plugging all the values we get delta PE = (5.40) (-2 (183)) which gives us delta PE = -1976.4 J

Now,

the potential energy in going from point A to B is given by

or

by plugging all the values we get

which gives us