An electron flies into a uniform electric field at a speed of 0.5 * 10 ^ 7 m / s and moves

An electron flies into a uniform electric field at a speed of 0.5 * 10 ^ 7 m / s and moves in the direction of the field strength lines. what is the distance the electron will fly until the complete loss of speed, if the modulus of the field strength is 300v / m?

E = 300 V / m.

V0 = 0.5 * 10 ^ 7 m / s.

V = 0 m / s.

q = 1.6 * 10 ^ -19 Cl.

m = 9.1 * 10 ^ -31 kg.

S -?

An electron in an electrostatic field is acted upon by a force F, the value of which is determined by the formula: F = q * E, where q is the electron charge, E is the strength of the electric field.

The work of the electric field A goes to change (decrease) its kinetic energy ΔEk: A = ΔEk.

A = F * S * cos180 ° = – q * E * S.

ΔEk = m * V2 / 2 – m * V0 ^ 2/2 = m * (V2 – V0 ^ 2) / 2.

Since the electron stopped V = 0 m / s, then ΔEk = – m * V0 ^ 2/2.

– q * E * S = – m * V0 ^ 2/2.

The path S to the stop of the electron will be determined by the formula: S = m * V0 ^ 2/2 * q * E.

S = 9.1 * 10 ^ -31 kg * (0.5 * 107 m / s) ^ 2/2 * 1.6 * 10 ^ -19 C * 300 V / m = 0.237 m.

Answer: to a complete stop, the electron will travel the path S = 0.237 m.