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
