Question

A proton initially at rest is accelerated by a uniform electric field. The proton moves 5.62 cm in 1.15 x 10^-6 s. Find the voltage drop through which the proton moves. (Answer should be positive)

Answer 1

Answer:

49.85 V

Explanation:

u = 0, s = 5.62 cm, t = 1.15 x 10^-6 s

Let the electric field is E and voltage is V.

Use second equation of motion

s = ut + 1/2 a t^2

5.62 x 10^-2 = 0 + 0.5 a x (1.15 x 10^-6)^2

a = 8.5 x 10^10 m/s^2

m x a = q x E

E = m x a / q

E = (1.67 x 10^-27 x 8.5 x 10^10) / (1.6 x 10^-19)

E = 887.19 V/m

V = E x s

V = 887.19 x 5.62 x 10^-2 = 49.85 V