Electrons moving with acceleration acquire a speed of 8 * 10 ^ 6 m / s at the anode of the vacuum diode. what is the voltage between the anode and the cathode in this case? consider the initial velocity of electrons to be zero. the mass of the electron is 9.1 * 10 ^ -31 kg and the modulus of its charge is e1.6 * 10 ^ -19 cells
V0 = 0 m / s.
V = 8 * 10 ^ 6 m / s.
q = 1.6 * 10 ^ -19 Cl.
m = 9.1 * 10 ^ -31 kg.
The work of the electric field A on the movement of the electric charge q is equal to the change in the kinetic energy of the charge ΔEk: A = ΔEk.
We express the work of the electrostatic field A by the formula: A = q * U, where q is the amount of charge, U is the voltage between the anode and cathode.
ΔEk = Ek – Ek0 = m * V ^ 2/2 – m * V0 ^ 2/2 = m * V ^ 2/2.
q * U = m * V ^ 2/2.
U = m * V ^ 2/2 * q.
U = 9.1 * 10 ^ -31 kg * (8 * 10 ^ 6 m / s) ^ 2/2 * 1.6 * 10 ^ -19 C = 182 V.
Answer: the voltage between the anode and cathode is U = 182 V.
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