The electron flew into a uniform field with a strength of 20 kV / m in the direction of its lines of force.

The electron flew into a uniform field with a strength of 20 kV / m in the direction of its lines of force. The initial velocity of the electron is 1.2 Mm / s. Find the acceleration acquired by the electron in the field, and the velocity in a time of 0.1 ns.

q = 1.6 * 10 ^ -19 Cl.

m = 9.1 * 10 ^ -31 kg.

E = 20 kV / m = 20 * 10 ^ 3 V / m.

V0 = 1.2 Mm / s = 1.2 * 10 ^ 6 m / s.

t = 0.1 ns = 0.1 * 10 ^ -9 s.

a -?

V -?

Since the electron has a negative electric charge q = 1.6 -10 ^ -19 C and it flies in the direction of the lines of force that come out of the positive charge, it will be decelerated. 2 Newton’s law will have the form: m * a = q * E.

a = q * E / m.

a = 1.6 * 10 ^ -19 C * 20 * 10 ^ 3 V / m / 9.1 * 10 ^ -31 kg = 3.5 * 10 ^ 5 m / s2.

Let us write down the formula for acceleration during deceleration of an electron: a = (V0 – V) / t.

V0 – V = a * t.

V = V0 – a * t.

V = 1.2 * 10 ^ 6 m / s – 3.5 * 10 ^ 5 m / s2 * 0.1 * 10 ^ -9 s = 1.2 * 10 ^ 6 m / s.

Answer: the acceleration of the electron was a = 3.5 * 10 ^ 5 m / s2, the speed of the electron in such a short time will decrease by Δ V = 0.35 * 10-5 m / s and practically will not change.