Electric Potential And Potential Energy
Two protons are brought from very far apart to a separation of 1 femtometre inside a fusion reactor model, so what is their potential energy?
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About This Question
- Subject
- physics
- Chapter
- electrostatics
- Topic
- electric potential and potential energy
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
2.3×10−13J
The electrostatic potential energy of two point charges is U=4πε01rq1q2, taking the reference of zero energy at infinite separation, as presented in NCERT Class 12, Chapter 2 (Electrostatic Potential and Capacitance). Each proton has charge e=1.6×10−19 C and the separation is r=1×10−15 m. Substituting, U=9×109×10−15(1.6×10−19)2. The charge product is 2.56×10−38; dividing by 10−15 gives 2.56×10−23, and multiplying by 9×109 yields 2.3×10−13 J. The value 1.15×10−13 J is wrong because it halves the result as though only one proton contributes. The value 2.3×10−16 J is wrong because it uses r=10−12 m instead of 10−15 m. The value 4.6×10−13 J is wrong because it doubles the correct energy without cause. Expressed in electron-volts this energy is roughly 1.4 MeV, which is precisely the scale of the repulsive barrier that thermal energies in a star or reactor must overcome for the protons to reach fusion range. A magnitude check confirms fusion-scale confinement of protons requires energies of order 10−13 J, and the result is correctly positive since both interacting charges are positive and thus repel.
This hard difficulty physics question is from the chapter electrostatics, covering the topic of electric potential and potential energy. It appeared in the 2025 exam.
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