Verified Solution • 2025–26 Session moderate

QPages 121-124: Q13: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q13 Solution: NaCl molecule is bound due to the electric force between the…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often read the separation as $2.75\times10^{-8}\,\text{m}$ instead of converting the given $2.75\times10^{-8}\,\text{cm}$ to $2.75\times10^{-10}\,\text{m}$. Because Coulomb's law uses SI units, a missed factor of $10^{-2}$ in distance (and $10^{-4}$ in $r^2$) changes the force by two orders of magnitude. Always convert the distance to metres before squaring.

NaCl molecule is bound due to the electric force between the sodium and the chlorine ions when one electron of sodium is transferred to chlorine. Taking the separation between the ions to be 2.75×10−8 cm2.75 \times 10^{-8} \text{ cm}, find the force of attraction between them. State the assumptions if any) that you have made.

Step-by-Step Verified Solution

**Answer:** Force = 3.04 _ 10__ N. **Step-BY-STEP Explanation:** Given: q=1.6×10−19q = 1.6 \times 10^{-19} C, r=2.75×10−8r = 2.75 \times 10^{-8} cm =2.75×10−10= 2.75 \times 10^{-10} m.
F=kq2r2=9×109×(1.6×10−19)2(2.75×10−10)2=3.04×10−9 NF = \frac{k q^2}{r^2} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{(2.75 \times 10^{-10})^2} = 3.04 \times 10^{-9} \text{ N}
Assumption: The ions are treated as point charges .

All Questions in this Chapter

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