Aproton moves at 4.50 × 105 m/s in the horizontal direction. it enters a uniform vertical electric field with a magnitude of 9.60 × 103 n/c. ignoring any gravitational effects, find
(a) the time interval required for the proton to travel 5.00 cm horizontally,
(b) its vertical displacement during the time interval in which it travels 5.00 cm horizontally,
(c) the horizontal and vertical components of its velocity after it has traveled 5.00 cm horizontally.

(a) t = 1.1 x 10^-7 s

(b) 5.56 mm

(c) vx = 4.5 x 10^5 m/s, vy = 10^5 m/s

Explanation:

v = 4.5 x 10^5 m/s       along horizontal direction

E = 9.60 x 10^3 N/C     along vertical direction

(a) Let t be the time taken.

Horizontal distance, x = 5 cm = 0.05 m

Horizontal distance = horizontal velocity x time

0.05 = 4.5 x 10^5 x t

t = 1.1 x 10^-7 s

(b) Let the vertical displacement is y.

y = 1/2 a t^2

Here a be the acceleration.

a = force/ mass = q E / m = (1.6 x 10^-19 x 9.6 x 10^3) / (1.67 x 10^-27)

a = 9.19 x 10^11 m/s^2

So, y = 0.5 x 9.19 x 10^11 x (1.1 x 10^-7)^2 = 5.56 x 10^-3 m = 5.56 mm

(c) Let vx be horizontal component of velocity and vy be the vertical component of velocity after it travels for 5 cm horizontally.

vx is same as v because the acceleration in horizontal direction is zero.

vy = 0 + a t = 9.19 x 10^11 x 1.1 x 10^-7 = 10.109 x 10^4 m/s

vy = 1 x 10^5 m/s