Question

A proton moves at 4.50 3 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.60 3 103 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, and (c) the horizontal and vertical components of its velocity after it has traveled 5.00 cm horizontally. 26. Protons are projected with an initial

Answer 1

Answer:

Explanation:

Horizontal velocity V_x = 4.5 x 10⁵ m /s

vertical electric field E_y = 9.6 x 10³ N/C

acceleration in vertical direction a_y = force on proton / mass

= 9.6 x 10³ x 1.6 x 10⁻¹⁹ / 1.67 x 10⁻²⁷

= 9.2 x 10¹¹ m /s²

a ) In horizontal direction it will move with uniform velocity

time required = distance / velocity

= 5 x 10⁻² / 4.5 x 10⁵

= 1.11 x 10⁻⁷ s

b )

vertical displacement in time 1.11 x 10⁻⁷ s

h = 1/2 x at² , initial vertical velocity is zero.

= .5 x 9.2 x 10¹¹ x ( 1.11 x 10⁻⁷ )²

= 5.66 x 10⁻³ m

5.66 mm

c )

Horizontal velocity will be unchanged ie 4.5 x 10⁵ m /s

vertical velocity will change due to acceleration

= u + at

0 + 9.2 x 10¹¹ x1.11 x 10⁻⁷

= 10.21 x 10⁴ m / s

= 1.021 x 10⁵ m /s