An electron flew into an electric field at a speed of 10 Mm / s parallel to its lines of force and passed

| September 22, 2021 | education

An electron flew into an electric field at a speed of 10 Mm / s parallel to its lines of force and passed a potential difference of 2.3 kV. To what speed did it accelerate?

To find out the speed that the specified electron has reached, we use the equality: q * Δφ = A = ΔEk = 0.5 * me * (Vk ^ 2 – Vn ^ 2), whence we express: Vk ^ 2 – Vn ^ 2 = 2qе * Δφ / me and Vk = √ (2qe * Δφ / me + Vn ^ 2).

Const: qе – electron charge (qе = -1.6 * 10 ^ -19 C); me is the mass of an electron (me = 9.11 * 10 ^ -31 kg).

Data: Vн – the initial speed of the electron (Vн = 10 Mm / s = 10 ^ 7 m / s); Δφ is the traversed potential difference (Δφ = 2.3 kV = 2.3 * 10 ^ 3 V).

Let’s perform the calculation: Vк = √ (2qе * Δφ / me + Vn ^ 2) = √ (2 * 1.6 * 10 ^ -19 * 2.3 * 10 ^ 3 / (9.11 * 10 ^ -31) + (10 ^ 7) ^ 2) ≈ 30.13 * 10 ^ 6 m / s.

Answer: The specified electron should have accelerated to a speed of 30.13 * 10 ^ 6 m / s.



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