Verified Solution • 2025–26 Session moderate

QPages 121-124: Q12: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q12 Solution: Two insulating small spheres are rubbed against each other and…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently forget to take the square root of the charge squared ($q^2$) derived from Coulomb's law, leading to a massive error in the final electron count. Always isolate $q$ by taking the square root before dividing by the elementary charge $e$.

Two insulating small spheres are rubbed against each other and placed 1 cm apart. If they attract each other with a force of 0.1 N, how many electrons were transferred from one sphere to the other during rubbing?

Step-by-Step Verified Solution

**Answer:** Number of electrons transferred = 2.08 _ 10__. **Step-by-Step Explanation:** Given: F=0.1F = 0.1 N, r=10−2r = 10^{-2} m.
F=kq2r2⇒q2=Fr2k=0.1×10−49×109=1.11×10−15F = \frac{k q^2}{r^2} \Rightarrow q^2 = \frac{F r^2}{k} = \frac{0.1 \times 10^{-4}}{9 \times 10^9} = 1.11 \times 10^{-15}
q=3.33×10−8 Cq = 3.33 \times 10^{-8} \text{ C}
Number of electrons = q/e=3.33×10−8/1.6×10−19=2.08×1011q/e = 3.33 \times 10^{-8} / 1.6 \times 10^{-19} = 2.08 \times 10^{11} .

All Questions in this Chapter

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