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

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

• The speed of proton is 2,225.82 m/s

Step-by-step 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 = (m(v-u))/(t) \\\\F = (m(v-0))/(t)\\\\F = (mv)/(t) \\\\mv = Ft\\\\v = (Ft)/(m) \\\\For \ electron;\\\\v_e = (Ft)/(m_e) \\\\v_e = (7.68 * 10^(-17) \ * \ 48.4 * 10^(-9))/(9.11 * 10^(-31)) \\\\v_e = 4.08 * 10^6 \ m/s`

`For \ proton;\\\\v_p = (Ft)/(m_p) \\\\v_p = (7.68 * 10^(-17) \ * \ 48.4 * 10^(-9))/(1.67 * 10^(-27)) \\\\v_p = 2,225 .82 \ m/s`