Answer:
The magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 1.01 x [tex]10^{-4}[/tex] N/C
Explanation:
given information,
kinetic energy, KE = 3.25 x [tex]10^{-15}[/tex] J
proton's mass, m = 1.673 x [tex]10^{-27}[/tex] kg
charge, q = 1.602 x [tex]10^{-19}[/tex] C
distance, d = 2 m
to find the electric field that will stop the proton, we can use the following equation:
E = F/q
= (KE/d) / q , KE = Fd --> F = KE/d
= KE/qd
= (3.25 x [tex]10^{-15}[/tex] J) / (1.602 x [tex]10^{-19}[/tex] C)(2 m)
= 1.01 x [tex]10^{-4}[/tex] N/C
