An electron is released from rest at a distance of 9.00 cm from a proton. If the proton is 11) held in place, how fast will the electron be moving when it is 3.00 cm from the proton?

Vf = 1.43 m/s

Explanation:

From Coulomb's Law, the electrostatic force between electron and proton is given as:

F = kq₁q₂/r²

F = Electrostatic force = ?

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

q₂ = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C

r = distance between electron and proton = 9 cm = 0.09 m

Therefore,

F = (9 x 10⁹ N.m²/C²)(1.6 x 10⁻¹⁹ C)(1.6 x 10⁻¹⁹ C)/(0.09 m)²

F = 2.84 x 10⁻²⁶ N

but, from Newton's second law:

F = 2.84 x 10⁻²⁶ N = ma

where,

m = mass of electron = 9.1 x 10⁻³¹ kg

a = acceleration of electron = ?

Therefore,

2.84 x 10⁻²⁶ N = (1.67 x 10⁻²⁷ kg)(a)

a = 2.84 x 10⁻²⁶ N/1.67 x 10⁻²⁷ kg

a = 17.03 m/s²

Now, we apply 3rd equation of motion to the motion of electron from a distance of 9 cm to 3 cm near to the proton:

2as = Vf² - Vi²

where,

s = distance traveled = 9 cm - 3 cm = 6 cm = 0.06 m

Vf = speed of electron when it is 3 cm from proton = ?

Vi = Initial speed of electron = 0 m/s

Therefore,

2(17.03 m/s²)(0.06 m) = Vf² - (0 m/s)²

Vf = √2.04 m²/s²

Vf = 1.43 m/s

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