Verified Solution • 2025–26 Session moderate

QPages 121-124: Q4: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q4 Solution: Two equal charges are placed at a separation of 1.0 m. What…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently stop at the intermediate result $q^2 = 5.44\times10^{-8}\,\text{C}^2$ and report it as the charge, forgetting to take the positive square root. Always compute $q=\sqrt{q^2}$ to obtain the actual magnitude of the charge.

Two equal charges are placed at a separation of 1.0 m. What should be the magnitude of the charges, so that the force between them equals the weight of a 50 kg person?

Step-by-Step Verified Solution

**Answer:** q=2.3×10−4q = 2.3 \times 10^{-4} C. **Step-BY-Step Explanation:** Given: r=1.0r = 1.0 m, F=50×9.8=490F = 50 \times 9.8 = 490 N.
F=kq2r2⇒q2=Fr2k=490×19×109=5.44×10−8F = \frac{k q^2}{r^2} \Rightarrow q^2 = \frac{F r^2}{k} = \frac{490 \times 1}{9 \times 10^9} = 5.44 \times 10^{-8}
q=2.33×10−4 Cq = 2.33 \times 10^{-4} \text{ C}

All Questions in this Chapter

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