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Archy [21]
3 years ago
5

An electron passing through Point Ain the electric fieldhad thespeed of 107 m/s and atPoint Bthe speed of2×107m/s.

Physics
1 answer:
bearhunter [10]3 years ago
4 0

Answer:

a) Electric potential = 853 V

b) Electron speed at point B, if at Point A, the speed were zero = 1.732 × 10⁷ m/s

Explanation:

For an electron moving in an electric field with potential V,

Work done = qV where q is the charge on the electron

And the Work done is equal to the change in kinetic energy of the electron

qV = m(v₂² - v₁²)/2

V = m(v₂² - v₁²)/2q

q = 1.602 × 10⁻¹⁹C

m = 9.11 × 10⁻³¹ kg

v₁= 10⁷ m/s

v₂ = 2 × 10⁷ m/s

Putting these values in for the variables and solving

V = 853 V

b) If the electron started from rest,

qV = mv²/2

v = √(2qV/m) =√((2 × (1.602 × 10⁻¹⁹) × 853)/(9.11 × 10⁻³¹)) = 1.732 × 10⁷ m/s

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