Question

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.

Answer 1

Answer:

   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

Answer 2

The distance of closest approach to the fixed particle is $2.84*10^{-21}m$

Energy :

Energy is neither be created nor be destroyed.

• The energy is always conserved.

• When proton is stopped then,

              $k\frac{q_{1}q_{2}}{r} =\frac{1}{2} mv^{2}$

Given that,

            $q_{1}=1.6*10^{-19}C,q_{2}=1.82*10^{-18} C\\m=1.67*10^{-27}kg,k=9*10^{9} Nm^{2}/C,v=3.5*10^{5} m/s$

Substitute all values in above relation.

             $r=\frac{2*1.6*10^{-19}*1.82*10^{-18} }{1.67*10^{-27}*(3.5*10^{5} )^{2} } \\\\r=2.84*10^{-21}m$

Learn more about the energy here:

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