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
The Correct Option is B
Approach Solution - 1
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.where, charge on the proton : permittivity of free spaceAlso, Hence, the correct option is (B).
Approach Solution -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
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