A proton is fired from very far away directly at a fixed particle with charge q = 1.82 ✕ 10−18 C. If the initial speed of the proton is 3.5 ✕ 105 m/s, what is its distance of closest approach to the fixed particle? The mass of a proton is 1.67 ✕ 10−27 kg.

   r = 2.56 10⁻¹² m

Explanation:

For this exercise let's use energy conservation

Starting point, proton too far

             Em₀ = K = ½ m v²

Final point. When proton is stop

             Emf = U = k q₁ q₂ / r

How energy is conserved

               Em₀ = Emf

              ½ m v² = k q₁ q₂ / r

              r = 2k q₁ q₂  / m v²

Let's calculate

             r = 2 8.99 10⁸ 1.6 10⁻¹⁹ 1.82 10⁻¹⁸ / (1.67 10⁻²⁷ (3.5 10⁵)² )

             r = 2.56 10⁻¹² m

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