Charge in an Electric Field: A pair of charged conducting plates produces a uniform field of 12,000 N/C, directed to the right, between the plates. The separation of the plates is 40 mm. An electron is projected from plate A, directly toward plate B, with an initial speed of v0 = 2.0 × 107 m/s. What is the speed of the electron as it strikes plate B? (e = 1.6 × 10-19 C, melectron = 9.11 × 10-31 kg)

Explanation:

Force on charge in electric field = q E , q is charge and E is electric field

= 1.6 x 10⁻¹⁹ x 12000

= 19200 x 10⁻¹⁹ N

acceleration = force / mass

=  19200 x 10⁻¹⁹ / 9.1 x 10⁻³¹

a = 2109.9 x 10¹² m /s²

initial velocity of electron u = 2 x 10⁷ m /s

v² = u² + 2as  ( s is distance between plate , u is initial velocity )

v² = ( 2 x 10⁷)² + 2 x 2109.9 x 10¹² x 40 x 10⁻³

= 4 x 10¹⁴ + 1.68 x 10¹⁴

= 5.68 x 10¹⁴

v = 2.38 x 10⁷ m /s

(75°f − 32) × 5/9 = 23.889°c

the answer is a. hope this .