Calculate the speed of a proton that has moved in a uniform electric field of 180.0 n/c from rest over a distance of 12.5-cm. assume it began at rest.

Given:

Electric field = 180 N/C

Let the speed of proton be "x"

x =

the velocity of the proton is 65574.38 m/s

Explanation:

given,

uniform electric field = 180 N/C

Distance = 12.5 cm = 0.125 m

charge of proton = 1.6 × 10⁻¹⁹ C

force = E × q

         =180 ×  1.6 × 10⁻¹⁹

        F= 2.88 × 10⁻¹⁷ N

mass of proton = 1.673 × 10⁻²⁷ kg

acceleration =

                     =

                     =1.72 × 10¹⁰ m/s²

velocity =

             =65574.38 m/s

hence , the velocity of the proton is 65574.38 m/s

it is called kinetic energy