A proton moving to the right at 6.2 × 105 ms-1 enters a region where there is an

Question

A proton 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)

Explanation / Answer

The acceleration of the particle F = qE ma = qE ==> a = qE / m We know that v = u + at = 0 + at ==> v = qEt / m velocity acquired by charged particle after time t is v = q E t / m ==> t = v m / q E = 6.2 x 10^5 * 1.67 x 10^-27 / 1.6 x 10^-19 * 62 x 10^3 = 0.1043 x 10^-6 s

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