Answer:
(7.90 × 10⁻¹⁵) J
Explanation:
The electric force exerted on the elecrron by rhe electric field is given by
F = qE
where |q| = charge on the particle = (1.602 × 10⁻¹⁹) C
E = magnitude of the electric field = (2.9 × 10⁶) V/m or N/C
F = 1.602 × 10⁻¹⁹ × 2.9 × 10⁶ = (4.646 × 10⁻¹³) N
From Newton's first law of motion relation, we can obtain the acceleration this force confers on the electron
F = ma
m = mass of the electron = (9.11 × 10⁻³¹) kg
a = acceleration of the electron caused by the electric force = ?
(4.646 × 10⁻¹³) = (9.11 × 10⁻³¹) × a
a = (4.646 × 10⁻¹³)/(9.11 × 10⁻³¹)
a = (5.10 × 10¹⁷) m/s²
Now, using the equations of motion, we can obtain the velocity with which the electron reaches the positive plate
u = initial velocity of the electron = 0 m/s (since the electron was initially at rest)
v = final velocity of the electron = ?
a = acceleration of the electron = (5.10 × 10¹⁷) m/s²
y = distance covered by the electron = 1.7 cm = 0.017 m
v² = u² + 2ay
v² = 0² + 2(5.10 × 10¹⁷)(0.017)
v² = (1.734 × 10¹⁶)
v = 131,677,182.5 m/s = (1.32 × 10⁸) m/s
Kinetic energy with which the electron hits the positive plate = (1/2)(m)(v²) = (1/2)(9.11 × 10⁻³¹)(1.32 × 10⁸)² = (7.90 × 10⁻¹⁵) J
Hope this Helps!!!
