Question

Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 554 N/C. If the particles are free to move, what are their speeds (in m/s) after 51.6 ns

Answer 1

the speed of electron is 5.021 x 10⁶ m/s

the speed of proton is 2733.91 m/s

Step-by-step explanation:

Given;

magnitude of electric field, E = 554 N/C

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

time of motion, t = 51.6 ns = 51.6 x 10⁻⁹ s

The force experienced by each particle is calculated as;

F = EQ

F = (554)(1.6 x 10⁻¹⁹)

F = 8.864 x 10⁻¹⁷ N

The speed of the particles are calculated as;

`F = ma\\\\F = (mv)/(t) \\\\v = (Ft)/(m) \\\\v_e = (Ft)/(m_e)\\\\v_e = ((8.864 * 10^(-17))(51.6* 10^(-9)))/(9.11 * \ 10^(-31))\\\\v_e = 5.021 * 10^(6) \ m/s`

`v_p = (Ft)/(m_p)\\\\v_p = ((8.864 * 10^(-17))(51.6* 10^(-9)))/(1.673 * \ 10^(-27))\\\\v_p = 2733.91 \ m/s`