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हीलियम परमानु के नाभिक में दो प्रोटोन होते हैं, जो 3.0 x 10-15 m दूरी द्वारा पृथक होते हैं। उस स्थिर वैद्युत बल का परिमाण क्या होगा जो प्रत्येक प्रोटोन अन्य पर आरोपित करता है?

March 22, 2026Physics

ScanSolve AI Answer

Answer

25.6 N

5 steps

The electrostatic force between two protons is given by Coulomb's law.
$F = k \frac{q_1 q_2}{r^2}$
where $k = 9.0 \times 10^9 \, \mathrm{N \cdot m^2 / C^2}$ , $q_1 = q_2 = e = 1.6 \times 10^{-19} \, \mathrm{C}$ , and $r = 3.0 \times 10^{-15} \, \mathrm{m}$ .
Compute $q_1 q_2 = e^2$ .
$e^2 = (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \mathrm{C^2}$
Compute $r^2$ .
$r^2 = (3.0 \times 10^{-15})^2 = 9.0 \times 10^{-30} \, \mathrm{m^2}$
Compute $\frac{q_1 q_2}{r^2}$ .
$\frac{q_1 q_2}{r^2} = \frac{2.56 \times 10^{-38}}{9.0 \times 10^{-30}} = \frac{2.56}{9.0} \times 10^{-38+30} = 0.2844 \times 10^{-8} = 2.844 \times 10^{-9} \, \mathrm{C^2 / m^2}$
Compute $F$ .
$F = (9.0 \times 10^9) \times (2.844 \times 10^{-9}) = (9.0 \times 2.844) \times 10^{9-9} = 25.6 \times 10^0 = 25.6 \, \mathrm{N}$
25.6 N

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