Aproton is traveling horizontally to the right at 4.60×106 m/s . part a find (a)the magnitude and (b) direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.50 cm . part c how much time does it take the proton to stop after entering the field? part d what minimum field ((a)magnitude and (b)direction) would be needed to stop an electron under the conditions of part (a)?

Explanation:

u = 4.6 x 10^6 m/s

Let E be the electric field

s = 3.5 cm = 0.035 m

v = 0

a = qE / m

So use third equation of motion

v^2 = u^2 - 2 a s

0 = (4.6 x 10^6)^2  - 2 x q E / m x 0.035

21.16 x 10^12 = 2 x 1.6 x 10^-19 x E / (1.67 x 10^-27 x 0.035)

E = 3865 N/C

(a) The magnitude of electric field is 3865 N/C

(b) the direction of electric field is opposite to the direction of motion of proton, i.e., towards left.

(c) Let t be the time taken

v = u + a t

0 = 4.6 x 10^6 - (1.6 x 10^-19 x 3865) t / (1.67 x 10^-27)

t  = 1.24 x 10^-5 sec

(d) For electron, the direction of electric field is same the direction of electron, i.e., rightwards.

Use third equation of motion

v^2 = u^2 - 2 a s

0 = (4.6 x 10^6)^2  - 2 x q E / m x 0.035

21.16 x 10^12 = 2 x 1.6 x 10^-19 x E / (9.1 x 10^-31 x 0.035)

E = 2.1 N/C

multiply 2kg by latent heat of vap

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