Question

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 1.7 cm, and the electric field within the capacitor has a magnitude of 2.9 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?

Answer 1

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!!!