Moving 2.0 coulombs of charge at constant speed a distance of 6.0 meters from point a to pointb within a uniform electric field requires a 5.0-newton force. what is the electric potentialdifference between points a and b?

The force produced on a particle of charge q by an electric field of intensity E is

in our problem, the force is F=5.0 N while the charge is q=2.0 C, so we can find the intensity of the electric field:


The relationship between electric field intensity and potential difference  between two points A and B is

where d is the distance between the two points. By using d=6.0 m, we find

where the negative sign means that the initial point, VA, is at higher potential than the final point VB.

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward