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QPages 121-124: Q8: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q8 Solution: Two charged particles are placed 1.0 cm apart. What is the…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often think the smallest force is zero by setting one charge to zero. However, the problem asks for the minimum non‑zero force between two charged particles, so the smallest allowed charge is the elementary charge. Use $F_{min}=k,e^2/r^2$ and remember that the magnitude depends on $|q_1q_2|$, not on the sign of the charges.

Two charged particles are placed 1.0 cm apart. What is the minimum possible magnitude of the electric force acting on each charge?

Step-by-Step Verified Solution

**Answer:** Minimum force = 2.3 _ 10___ N. **Step-BY-Step Explanation:** The minimum charge on any body is the elementary charge, e=1.6×10−19e = 1.6 \times 10^{-19} C. For minimum force, both charges should be ee.
Fmin=ke2r2=9×109×(1.6×10−19)2(0.01)2=2.3×10−24 NF_{min} = \frac{k e^2}{r^2} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{(0.01)^2} = 2.3 \times 10^{-24} \text{ N}

All Questions in this Chapter

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