Question

Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 478 N/C. If the particles are free to move, what are their speeds (in m/s) after 52.4 ns

Answer 1

Answer:

the speed of electron is 4.42 x 10⁶ m/s

the speed of proton is 2406.7 m/s

Explanation:

Given;

electric field strength, E = 478 N/C

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

mass of proton, Mp = 1.673 x 10⁻²⁷ kg

mass of electron Me = 9.11 x 10⁻³¹ kg

time of motion, t = 54.2 ns = 54.2 x 10⁻⁹ s

The magnitude of charge experienced by the particles is calculated as;

F = EQ

F = 478 x 1.6 x 10⁻¹⁹

F = 7.648 x 10⁻¹⁷ N

The speed of the particles is calculated as;

$F = \frac{mv}{t} \\\\v = \frac{Ft}{m} \\\\v_e = \frac{(7.684 \times 10^{-17})(52.4\times 10^{-9})}{9.11\times 10^{-31}} \\\\v_e = 4.42 \ \times \ 10^6 \ m/s$

$v_p = \frac{Ft}{m_p} \\\\v_p = \frac{(7.684 \times 10^{-17})(52.4\times 10^{-9})}{1.673\times 10^{-27}} \\\\v_p = 2406.7 \ m/s$