Two protons are released from rest when they are 0.720 nm apart. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Potential of two point charges. Part A What is the maximum speed they will reach

a) Speed of the electrons at maximum speed = (1.384 × 10⁴) m/s

The maximum speed occurs at the point where all of the initial potential energy is converted into kinetic energy.

b) Maximum acceleration of the protons = (2.660 × 10¹⁷) m/s²

The maximum acceleration occurs at the minimum distance apart for the two protons.

Explanation:

The maximum speed occurs when all the potential energy of the protons has been converted to kinetic energy.

The potential energy between the two protons at the instant of release is given by

U = (kq₁q₂/r)

k = Coulomb' s constant = (8.988 × 10⁹) Nm²/C²

q₁ = q₂ = charge on a proton = q = (1.602 × 10⁻¹⁹) C

r = separation between the two protons = 0.72 nm = (7.2 × 10⁻¹⁰) m

U = (kq²/r) = [(8.988 × 10⁹) × (1.602 × 10⁻¹⁹)²] ÷ (7.2 × 10⁻¹⁰) = (3.204 × 10⁻¹⁹) N/m or Joules

At the maximum speeds, the two protons will not possess any potential Energy, only kinetic energy.

The sum of kinetic and potential energies is always constant for the system

(Initial Kinetic Energy) + (Initial Potential Energy) = (Kinetic Energy at maximum speed) + (Potential Energy at maximum speed)

Initial Kinetic Energy of the system = 0 J (Since both protons were intially at rest)

Initial Potential Energy = (3.204 × 10⁻¹⁹) J

Kinetic Energy at maximum speed = Sum of the kinetic energies of the protons at this point = (½mv²) + (½mv²) = (mv²) J (Since theu are both protons, they have the same mass and the same speed at maximum speed)

Potential Energy at maximum speed = 0 J

0 + (3.204 × 10⁻¹⁹) = mv² + 0

mv² = (3.204 × 10⁻¹⁹)

m = mass of a proton = (1.673 × 10⁻²⁷) kg

v = speed of each of the protons at maximum speed = ?

v = √[(3.204 × 10⁻¹⁹) ÷ m]

v = √[(3.204 × 10⁻¹⁹) ÷ (1.673 × 10⁻²⁷)]

v = √(1.915 × 10⁸) = 13,838.8 m/s = (1.384 × 10⁴) m/s

b) Since the two protons repel each other and force of repulsion reduces as the dI stance between the protons increases, the maximum acceleration occurs at the minimum distance apart for the two protons.

Force of repulsion acting on each proton is given through Coulomb's law as

F = (kq₁q₂/r²)

And the force acting on each proton is obtainable using Newton's law that

F = ma

So, the acceleration of each proton at any time is obtainable through a relation of these 2 formulas.

ma = (kq₁q₂/r²)

a = (kq₁q₂/r²m)

k = Coulomb' s constant = (8.988 × 10⁹) Nm²/C²

q₁ = q₂ = charge on a proton = q = (1.602 × 10⁻¹⁹) C

r = separation between the two protons = 0.72 nm = (7.2 × 10⁻¹⁰) m

m = mass of a proton = (1.673 × 10⁻²⁷) kg

a = [(8.988 × 10⁹) × (1.602 × 10⁻¹⁹)²] ÷ [(7.2 × 10⁻¹⁰)² × (1.673 × 10⁻²⁷)]

a = (2.660 × 10¹⁷) m/s²

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