Respuesta :
Answer:
Magnitude of electric field is 1.06 x [tex]10^5[/tex] V/m along negative X-direction
Explanation:
Given: initial velocity of proton = u = 3.5 x [tex]10^6[/tex] m/s
final velocity of proton = v = 0 m/s
initial point [tex]l_i[/tex] = 0.2 m and final point is [tex]l_f[/tex] = 0.8 m
According to conservation of energy:
change in in kinetic energy = change in potential energy of proton
⇒[tex]\frac{m}{2}(v^2-u^2 ) = qE(l_i - l_f)[/tex]
where q and m is the charge and mass of proton E is the electric field , [tex]l_i[/tex] and [tex]l_f[/tex] is the initial and final position of proton
on substituting the respected values we get,
1.023 x [tex]10^-^1^4[/tex] = 9.6 x [tex]10^-^2^0[/tex] x E
⇒ E = 1.06 x [tex]10^5[/tex] 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 [tex]10^5[/tex] V/m along negative X-direction
