Question

Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 480 N/C. If the particles are free to move, what are their speeds (in m/s) after 48.4 ns

Answer 1

Answer:

• The speed of electron is 4.08 x 10⁶ m/s

• The speed of proton is 2,225.82 m/s

Explanation:

Given;

electric field, E = 480 N/C

mass of electron, Me = 9.11 x 10⁻³¹ kg

mass of proton, Mp = 1.67 x 10⁻²⁷ kg

time of motion, t = 48.4 ns = 48.4 x 10⁻⁹ s

initial velocity of the particles, u = 0 (initially at rest)

let the speed of each particle after 48.4 ns = v

the magnitude of charge of the particles, q = 1.6 x 10⁻¹⁹ C

The force experienced by each particle is calculated as;

F = Eq

F = (480 N/C) x (1.6 x 10⁻¹⁹ C)

F = 7.68 x 10⁻¹⁷ N

The speed of each particle after 48.4 ns is calculated as;

$F = ma\\\\F = \frac{m(v-u)}{t} \\\\F = \frac{m(v-0)}{t}\\\\F = \frac{mv}{t} \\\\mv = Ft\\\\v = \frac{Ft}{m} \\\\For \ electron;\\\\v_e = \frac{Ft}{m_e} \\\\v_e = \frac{7.68 \times 10^{-17} \ \times \ 48.4 \times 10^{-9}}{9.11 \times 10^{-31}} \\\\v_e = 4.08 \times 10^6 \ m/s$

$For \ proton;\\\\v_p = \frac{Ft}{m_p} \\\\v_p = \frac{7.68 \times 10^{-17} \ \times \ 48.4 \times 10^{-9}}{1.67 \times 10^{-27}} \\\\v_p = 2,225 .82 \ m/s$