Answer:
Magnitude of electric field is 1.06 x V/m along negative X-direction
Explanation:
Given: initial velocity of proton = u = 3.5 x m/s
final velocity of proton = v = 0 m/s
initial point = 0.2 m and final point is = 0.8 m
According to conservation of energy:
change in in kinetic energy = change in potential energy of proton
⇒
where q and m is the charge and mass of proton E is the electric field , and is the initial and final position of proton
on substituting the respected values we get,
1.023 x = 9.6 x x E
⇒ E = 1.06 x V/m
external force is opposite to the motion as velocity of proton decreases with distance.
Therefore, magnitude of electric field is 1.06 x V/m along negative X-direction