When a particle with a mass of 5 g passes with a charge q = 20 nC between two points of the electrostatic field

When a particle with a mass of 5 g passes with a charge q = 20 nC between two points of the electrostatic field, the modulus of its velocity increases from V1 = 40 cm / s to V2 = 90 cm / s. Determine the voltage between these points.  

m = 5 g = 5 * 10 ^ -3 kg.

q = 20 nC = 20 * 10 ^ -9 C.

V1 = 40 cm / s = 0.4 m / s.

V2 = 90 cm / s = 0.9 m / s.

U -?

The work of the electric field A goes to increase the kinetic energy ΔEk of a charged particle: A = ΔEk.

We express the work of the electric field A by the formula: A = q * U.

The change in kinetic energy ΔEk is expressed by the formula: ΔEk = Ek2 – Ek1 = m * V2 ^ 2/2 – m * V1: 2/2 = m * (V2 ^ 2 – V1 ^ 2) / 2.

q * U = m * (V2 ^ 2 – V1 ^ 2) / 2.

U = m * (V2 ^ 2 – V1 ^ 2) / 2 * q.

U = 5 * 10 ^ -3 kg * ((0.9 m / s) 2 – (0.4 m / s) 2) / 2 * 20 * 10 ^ -9 C = 81.25 * 10 ^ 3 V = 81.25 kV.

Answer: the voltage between the points is U = 81.25 kV.