An electron flies out from a point, the potential of which is 450V at a speed of 190m / s. What speed will it have at a point with a potential of 475V?
m = 9.1 * 10 ^ -31 kg.
q = 1.6 * 10 ^ -19 Cl.
V1 = 190 m / s.
φ1 = 450 V.
φ2 = 475 V.
According to the law of conservation of energy, the work of the electric field A goes to increase the kinetic energy of the electron ΔEk: A = ΔEk.
We express the work of the electric field A by the formula: A = q * (φ2 – φ1), where q is the magnitude of the charge, φ2 – φ1 is the potential difference that the charge has passed.
The change in the kinetic energy of the electron ΔEk is expressed by the formula: ΔEk = m * V2 ^ 2/2 – m * V1 ^ 2/2, where m is the electron mass, V2 is the final electron velocity, V1 is the initial electron velocity.
q * (φ2 – φ1) = m * V2 ^ 2/2 – m * V1 ^ 2/2.
m * V2 ^ 2/2 = q * (φ2 – φ1) + m * V1 ^ 2/2.
V2 ^ 2 = 2 * q * (φ2 – φ1) / m + V1 ^ 2.
V2 = √ (2 * q * (φ2 – φ1) / m + V1 ^ 2).
V2 = √ (2 * 1.6 * 10 ^ -19 C * (475 V – 450 V) / 9.1 * 10 ^ -31 kg + (190 m / s) 2) = 2.9 * 10 ^ 6 m / s.
Answer: the speed of the electron will become V2 = 2.9 * 10 ^ 6 m / s.
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