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Notely Physics 12
Class-12 Physics

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1 - N u m e r i c a l   1

Problem

What is the magnitude of the force of attraction between an iron nucleus bearing charge q=26eq = 26e and its innermost electron, if the distance between them is 1×10−12 m1 \times 10^{-12} \, \mathrm{m}?

Data

  • Charge on electron, q1=1.6×10−19 Cq_1 = 1.6 \times 10^{-19} \, \mathrm{C}
  • Charge of nucleus, q2=26e=26×1.6×10−19 C=41.6×10−19 Cq_2 = 26e = 26 \times 1.6 \times 10^{-19} \, \mathrm{C} = 41.6 \times 10^{-19} \, \mathrm{C}
  • Separation distance, r=1×10−12 mr = 1 \times 10^{-12} \, \mathrm{m}
  • Coulomb’s constant, k=9×109 Nm2C−2k = 9 \times 10^9 \, \mathrm{Nm^2C^{-2}}

Prerequisite Concepts

  • Coulomb’s Law:
    The force between two charges is given by:
F=kq1q2r2 F = \frac{k q_1 q_2}{r^2}

Answer

Substituting the values:

F=(9×109)(1.6×10−19)(41.6×10−19)(1×10−12)2F = \frac{(9 \times 10^9) (1.6 \times 10^{-19})(41.6 \times 10^{-19})}{(1 \times 10^{-12})^2}

Simplifying:

F=5.99×10−3 NF = 5.99 \times 10^{-3} \, \mathrm{N}

Final answer:

F=6×10−3 N.F = 6 \times 10^{-3} \, \mathrm{N}.