Question

Two free protons are separated by a distance of 1 Å. If they are released, the kinetic energy of each proton when at infinite separation is

• (A) 5.6 × 10^-19J
• (B) 11.5 ×10^-19J
• (C) 23.0 × 10^-19 J
• (D) 46.0 × 10^-19 J

Answer 1

The Correct Option is B

Explanation:
Given:Distance between the free protons, r = 1Å = 1 x 10^-10 mInitially kinectic energy of the proton is zero and electric potential energy is maximum.At infinite separation, potential energy is zero and all the energy is converted into kinetic energy (using law of conservation of energy )i.e, 2K = Uwhere, K is kinetic energy of each proton.$\Rightarrow K=\frac{1}{2}U=\frac{1}{2}\frac{e^2}{4\pi {ϵ}_0\mathrm{r}}$where,$e=1.6\times {10}^{-19}\mathrm{C}:$ charge on the proton${ϵ}_0=8.85\times {10}^{-12}{\mathrm{C}}^2{\mathrm{N}}^{-1}{\mathrm{m}}^{-2}$ : permittivity of free space$\Rightarrow K=\frac{1}{2}\times \frac{{(1.6\times {10}^{-19}\mathrm{C})}^2}{\left({10}^{-10}\mathrm{m}\right)}\times \frac{1}{4\pi {ϵ}_0}$Also, $\frac{1}{4\pi {ϵ}_0}=9\times {10}^{9}N{m}^2{C}^{-2}$$\Rightarrow K=\frac{1}{2}\times \frac{{(1.6\times {10}^{-19}\mathrm{C})}^2}{\left({10}^{-10}\mathrm{m}\right)}\times 9\times {10}^9\mathrm{N}{\mathrm{m}}^2{\mathrm{C}}^{-2}$$\Rightarrow K=11.5\times {10}^{-19}J$Hence, the correct option is (B).

Answer 2

Potential Energy of a System of Two Charges

We know, when a charge q is brought from infinity to a point where electric potential V due to any source charge is present, then Work done is given by

W = qV

Let two charges q₁ and q₂ initially lie at infinity.

Initially, when charge q₁ is brought from infinity to a particular point, due to absence of electric potential at that point no work is done i.e. 

W₁ = q₁ V = q₁ x 0 = 0

Now when charge q₂ is brought from infinity to a point at distance r from the charge q₁, the electric potential is present at that point due to charge q₁, then work done in bringing q₂ is given by

W₂ = q₂ x V = q₂ x 1/4πϵ₀ q₁/r

⇒ W₂ = 1/4πϵ₀ q₁q₂/r

Total work done

W = W₁ + W₂

⇒ W = 0 + 1/4πϵ₀ q₁q₂/r

⇒ W = 1/4πϵ₀ q₁q₂/r

This work done is equal to the potential energy (U) of the system of the two charges. Hence

U = 1/4πϵ₀ q₁q₂/r