Determine the potential difference between two points in the electric field, if the speed

Determine the potential difference between two points in the electric field, if the speed of the electron when moving between these points increases from 1000 km / s to 3000 km / s.

m = 9.1 * 10 ^ -31 kg.

q = 1.6 * 10 ^ -19 Cl.

V1 = 1000 km / s = 10 ^ 6 m / s.

V2 = 3000 km / s = 3 * 10 ^ 6 m / s.

φ2 – φ1 -?

Since the electron has an elementary charge q = 1.6 * 10 ^ -19 C, the electric field does work A to accelerate the electron, the values ​​of which are expressed by the formula: A = q * (φ2 – φ1).

According to the law of conservation of energy, the work of the electrostatic field A goes to increase the kinetic energy of the electron: A = ΔEk.

A moving electron has kinetic energy: Ek = m * V ^ 2/2, where m is its mass, V is the speed of its movement.

The increase in the kinetic energy of the electron ΔEk is expressed by the formula: ΔEk = m * V2 ^ 2/2 – m * V1 ^ 2/2 = m * (V ^ 22 – V1 ^ 2) / 2.

q * (φ2 – φ1) = m * (V2 ^ 2 – V1 ^ 2) / 2.

φ2 – φ1 = m * (V2 ^ 2 – V1 ^ 2) / 2 * q.

φ2 – φ1 = 9.1 * 10 ^ -31 kg * ((3 * 10 ^ 6 m / s) ^ 2 – (1 * 10 ^ 6 m / s) ^ 2) / 2 * 1.6 * 10 ^ -19 C = 22.75 V.

Answer: an electron in an electric field has passed the accelerating potential difference at φ2 – φ1 = 22.75 V.