Aproton moving to the right at 6.2 × 105 ms-1 enters a region where there is an electric field of 62 knc-1 directed to the left. describe qualitatively the motion of the proton in this filed. what is the time taken by the proton to come back to the point where it entered the field? (use the standard values of the mass and charge of a proton)

2.06 x 10^-7 s

Explanation:

For going opposite to the applied electric field

u = 6.2 x 10^5 m/s, v = 0 ,  q = 1.6 x 19^-19 C, E = 62000 N/C,

m = 1.67 x 106-27 Kg, t1 = ?

Let a be the acceleration.

a = q E / m = (1.6 x 10^-19 x 62000) / (1.67 x 10^-27) = 6 x 10^12 m/s^2

Use first equation of motion

v = u + a t1

0 = 6.2 x 10^5 - 6 x 10^12 x t1

t1 = 1.03 x 10^-7 s

Let the distance travelled is s.

Use third equation of motion

V^2 = u^2 - 2 a s

0 = (6.2 x 10^5)^2 - 2 x 6 x 10^12 x s

s = 0.032 m

Now when the proton moves in the same direction of electric field.

Let time taken be t2

use second equation of motion

S = u t + 1/2 a t^2

0.032 = 0 + 1/2 x 6 x 10^12 x t2^2

t2 = 1.03 x 10^-7 s

total time taken = t = t1 + t2

t =  1.03 x 10^-7 +  1.03 x 10^-7 =  2.06 x 10^-7 s

thermal energy is the answer

this is an example of "committed" time.

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